Intermediate Algebra MATH 083 Textbook
First Edition (2016)
Attribution This textbook, Intermediate Algebra, is licensed under a Creative Commons Attribution 3.0 Unported License. Faculty members at The Community College of Baltimore County in Baltimore, Maryland remixed and edited Beginning and Intermediate Algebra by Tyler Wallace, also licensed under a Creative Commons Attribution 3.0 Unported License.
Table of Contents Chapter 1: Factoring Section 1.1: Greatest Common Factor Section 1.2: Factor by Grouping Section 1.3: Factoring Trinomials Whose Leading Coefficient Is 1 Section 1.4: Factoring Trinomials Whose Leading Coefficient Is Not 1 Section 1.5: Factoring Special Products Section 1.6: Factoring Strategy Section 1.7: Solve Equations by Factoring Chapter 2: Rational Expressions and Equations Section 2.1: Reduce Rational Expressions Section 2.2: Multiply and Divide Rational Expressions Section 2.3: Least Common Denominator Section 2.4: Add and Subtract Rational Expressions Section 2.5: Proportions Section 2.6: Solving Rational Equations Section 2.7: Motion and Work Applications Section 2.8: Variation Chapter 3: Radical Expressions and Equations Section 3.1: Square Roots Section 3.2: Higher Roots Section 3.3: Add and Subtract Radical Expressions Section 3.4: Multiply and Divide Radical Expressions Section 3.5: Rationalize Denominators Section 3.6: Rational Exponents Section 3.7: Solving Radical Equations Section 3.8: Complex Numbers Chapter 4: Quadratic Equations and Graphs Section 4.1: Solving Equations with Exponents Section 4.2: Completing the Square Section 4.3: Quadratic Formula Section 4.4: Parabolas Section 4.5: Quadratic Applications Chapter 5: Functions and Exponential and Logarithmic Equations Section 5.1: Functions Section 5.2: Operations on Functions Section 5.3: Exponential Functions and Equations Section 5.4: Logarithmic Functions and Equations Section 5.5: Compound Interest i
2 7 12 18 22 28 32 40 47 52 58 64 72 79 86 94 99 104 108 113 117 123 129 138 143 150 157 165 172 184 190 197 205
CCBC ~ Math 083
Chapter 1
Chapter 1: Factoring Section 1.1 - Greatest Common Factor ............................................................2 Section 1.2 - Factor by Grouping .....................................................................7 Section 1.3 - Factoring Trinomials Whose Leading Coe cient Is 1.................12 Section 1.4 - Factoring Trinomials Whose Leading Coe cient Is Not 1..........18 Section 1.5 - Factoring Special Products ........................................................22 Section 1.6 - Factoring Strategy .....................................................................28 Section 1.7 - Solve Equations by Factoring ....................................................32
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CCBC ~ Math 083
Section 1.1
Section 1.1 - Greatest Common Factor Objective: Find the greatest common factor of a polynomial and factor it out of the expression. The inverse of multiplying polynomials together is factoring polynomials. There are many bene ts of a polynomial being factored. We use factored polynomials to help us solve equations, study behaviors of graphs, work with fractions and more. Because so many concepts in algebra depend on us being able to factor polynomials, it is very important to have very strong factoring skills. In this lesson, we will focus on factoring using the Greatest Common Factor or GCF of a polynomial. When multiplying monomials by polynomials, such as 4x2(2x2 ¡ 3x + 8); we distribute to get a product of 8x4 ¡ 12x3 + 32x2. In this lesson, we will work backwards, starting with 8x4 ¡ 12x3 + 32x2 and factoring to write as the product 4x2(2x2 ¡ 3x + 8). We will rst introduce this idea by looking at nding the GCF of several numbers. To nd a GCF of several numbers, we look for the largest number that can divide each number without leaving a remainder. This can often be done with quick mental math. See the example below. Example 1. Determine the GCF. Find the GCF of 15; 24; and 27 24 27 15 = 5; = 6; =9 3 3 3 GCF =3
Each of the numbers can be divided by 3 Our Answer
When there are variables in our problem, we can rst nd the GCF of the numbers as above. Then we take any variables that are in common with each term, using the lowest exponent. This is shown in the next example. Example 2. Determine the GCF. GCF of 24x4 y 2z; 18x2 y 4; and 12x3 yz 5 24 18 12 = 4; = 3; =2 6 6 6 x2 y
Each number can be divided by 6 x and y are in all 3 terms; using lowest exponents Our Answer
GCF = 6x2 y
To factor out a GCF from a polynomial, we rst need to identify the GCF of all the terms. This is the part that goes in front of the parentheses. Then we divide each term by the GCF; the answer is what is left inside the parentheses. This is shown in the following examples.
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CCBC ~ Math 083
Section 1.1
Example 3. Factor using the GCF. 4x2 ¡ 20x + 16
=4(x2 ¡ 5x + 4)
GCF of 4x2; ¡20x; and 16 is 4; divide each term by 4 ¡20x 16 4x2 = x2; = ¡5x; =4 4 4 4 The result is what is left inside the parentheses Our Answer
With factoring, we can always check our answers by multiplying (distributing); the resulting product should be the original expression. Example 4. Factor using the GCF. 25x4 ¡ 15x3 + 20x2
GCF of 25x4; ¡15x3; and 20x2 is 5x2; divide each term by 5x2 3 25x4 20x2 2 ¡15x = 5x ; = ¡3x; =4 5x2 5x2 5x2 The result is what is left inside the parentheses
=5x2(5x2 ¡ 3x + 4)
Our Answer
Example 5. Factor using the GCF. 3x3 y 2z + 5x4 y 3z 5 ¡ 4xy 4
GCF of 3x3 y 2z; 5x4 y 3z 5; and ¡4xy 4 is xy 2; divide each term by xy 2 3x3 y 2z 5x4 y 3z 5 ¡4xy 4 = 3x2z; = 5x3 yz 5; = ¡4y 2 2 2 xy xy xy 2
The result is what is left in parentheses =xy 2(3x2z + 5x3 yz 5 ¡ 4y 2)
Our Answer
World View Note: The rst recorded algorithm for nding the greatest common factor comes from Greek mathematician Euclid around the year 300 BC! Example 6. Factor using the GCF. 21x3 + 14x2 + 7x
=7x(3x2 + 2x + 1)
GCF of 21x3; 14x2; and 7x is 7x; divide each term by 7x 21x3 14x2 7x = 3x2; = 2x; =1 7x 7x 7x The result is what is left inside the parentheses Our Answer
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CCBC ~ Math 083
Section 1.1
It is important to note in the previous example, that when the GCF was 7x and 7x was one of the terms, dividing gave an answer of 1. Factoring will never make terms disappear. Anything divided by itself is 1; be sure not to forget to put the 1 into the solution. Example 7. Factor using the GCF. ¡12x5 y 2 + 6x4 y 4 ¡ 8x3 y 5
=¡2x3 y 2(6x2 ¡ 3xy 2 + 4y 3)
GCF of ¡12x5 y 2; 6x4 y 4; and ¡8x3 y 5 is (¡2x3 y 2) because the rst term is negative; divide each term by (¡2x3 y 2) 4 4 ¡8x3 y 5 ¡12x5 y 2 2 6x y 2 = 6x ; = ¡3xy ; = 4y 3 ¡2x3 y 2 ¡2x3 y 2 ¡2x3 y 2 The result is what is left inside the parentheses Our Answer
Often the second line is not shown in the work of factoring the GCF. We can simply identify the GCF and put it in front of the parentheses as shown in the following example. Example 8. Factor using the GCF. 18a4 b3 ¡ 27a3b3 + 9a2b3 =9a2b3(2a2 ¡ 3a + 1)
GCF is 9a2b3; divide each term by 9a2b3 Our Answer
Again, in the previous problem, when dividing 9a2b3 by itself, the answer is 1. Be very careful that each term is accounted for in your nal solution. It is possible to have a problem where the GCF is 1. If there is no other common factor between all of the terms, then we cannot factor the polynomial; that is, we say the polynomial is prime. This is shown in the following example. Example 9. Factor using the GCF. 8ab ¡ 17c + 49 prime
GCF is 1 because there are no other factors in common to all 3 terms Our Answer
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CCBC ~ Math 083
Section 1.1
1.1 Practice - Greatest Common Factor Factor using the GCF. If the expression cannot be factored, state that it is prime. 2) x ¡ 5
1) 9 + 8b2
4) 1 + 2n2
3) 45x2 ¡ 25
6) 50x ¡ 80y
5) 56 ¡ 35p
8) 27x2 y 5 ¡ 72x3 y 2
7) 7ab ¡ 35a2b
10) 8x3 y 2 + 4x3
9) ¡3a2b + 6a3b2
12) ¡32n9 + 32n6 + 40n5
11) ¡5x2 ¡ 5x3 ¡ 15x4
14) 21p6 + 30p2 + 27
13) 20x4 ¡ 30x + 30
16) ¡10x4 + 20x2 + 12x
15) 28m + 40m + 8 4
3
18) 27y 7 + 12y 2x + 9y 2
17) 30b + 5ab ¡ 15a 9
2
20) 30m6 + 15mn2 ¡ 25
19) ¡48a b ¡ 56a b ¡ 56a b 2 2
3
5
22) 3p + 12q ¡ 15q 2r2
21) 20x y z + 15x y z + 35x y z 8 2 2
5 2
3 3
24) 30y 4z 3x5 + 50y 4z 5 ¡ 10y 4z 3x
23) 50x2 y + 10y 2 + 70xz 2
26) 28b + 14b2 + 35b3 + 7b5
25) 30qpr ¡ 5qp + 5q
28) 30a8 + 6a5 + 27a3 + 21a2
27) ¡18n5 + 3n3 ¡ 21n + 3
30) ¡24x6 ¡ 4x4 + 12x3 + 4x2
29) ¡40x11 ¡ 20x12 + 50x13 ¡ 50x14
32) ¡10y 7 + 6y 10 ¡ 4y 10x ¡ 8y 8x
31) ¡32mn8 + 4m6n + 12mn4 + 16mn
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Section 1.1
1.1 Answers - Greatest Common Factor 1) prime
17) 5(6b9 + ab ¡ 3a2)
2) prime
18) 3y 2(9y 5 + 4x + 3)
3) 5(9x2 ¡ 5)
19) ¡8a2b (6b + 7a + 7a3)
4) prime
20) 5(6m6 + 3mn2 ¡ 5)
5) 7(8 ¡ 5p)
21) 5x3 y 2z(4x5z + 3x2 + 7y)
6) 10(5x ¡ 8y)
22) 3(p + 4q ¡ 5q 2r2)
7) 7ab(1 ¡ 5a)
23) 10(5x2 y + y 2 + 7xz 2)
8) 9x2 y 2(3y 3 ¡ 8x)
24) 10y 4z 3 (3x5 + 5z 2 ¡ x)
9) ¡3a2b(1 ¡ 2ab)
25) 5q(6pr ¡ p + 1)
10) 4x3(2y 2 + 1)
26) 7b(4 + 2b + 5b2 + b4)
11) ¡5x2(1 + x + 3x2)
27) ¡3(6n5 ¡ n3 + 7n ¡ 1)
12) ¡8n5(4n4 ¡ 4n ¡ 5)
28) 3a2(10a6 + 2a3 + 9a + 7) 29) ¡10x11(4 + 2x ¡ 5x2 + 5x3)
13) 10(2x4 ¡ 3x + 3) 14) 3(7p6 + 10p2 + 9)
30) ¡4x2(6x4 + x2 ¡ 3x ¡ 1)
15) 4(7m4 + 10m3 + 2)
31) ¡4mn(8n7 ¡ m5 ¡ 3n3 ¡ 4)
16) ¡2x(5x3 ¡ 10x ¡ 6)
32) ¡2y 7(5 ¡ 3y 3 + 2xy 3 + 4xy)
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CCBC ~ Math 083
Section 1.2
Section 1.2 - Factor by Grouping Objective: Factor polynomials with four terms using grouping. The rst thing we will always do when factoring is try to factor out a GCF. This GCF is often a monomial. For example, in the problem 5x y + 10xz, the GCF is the monomial 5x, so we would have 5x(y + 2z). However, a GCF does not have to be a monomial; it could be a polynomial. To see this, consider the following two examples. Example 1. Factor completely. 3ax ¡ 7bx =x(3a ¡ 7b)
Both terms have x in common; factor it out Our Answer
Now we have the same problem, but instead of x we have (2a + 5b) as the GCF. Example 2. Factor completely. 3a(2a + 5b) ¡ 7b(2a + 5b) =(2a + 5b)(3a ¡ 7b)
Both terms have (2a + 5b) in common; factor it out Our Answer
In the same way we factored out a GCF of x, we can factor out a GCF which is a binomial, (2a + 5b). This process can be extended to factor problems where either there is no GCF except 1, or after the GCF is factored out, there is more factoring that can be done. Here we will have to use another strategy to factor. We will use a process known as grouping. We will try grouping when factoring a polynomial with four terms. Remember, factoring is the reverse of multiplying, so rst we will look at a multiplication problem and then try to reverse the process. Example 3. Multiply. (2a + 3)(5b + 2) = 5b(2a + 3) + 2(2a + 3) = 10ab+15b + 4a + 6
Distribute (2a + 3) into second parentheses Distribute each monomial Our Answer
The product has four terms. We arrived at this answer by looking at the two parts, 5b(2a + 3) and 2(2a + 3). When we are factoring by grouping, we will always divide the expression into two groups: the rst two terms and the last two terms. Then we can factor the GCF out of each group of two terms. When we do this, our hope is what remains in the parentheses will match in both the left group and the right group. If they match, we can pull this matching GCF out front, putting the rest in parentheses and the expression will be factored. The next example is the same problem worked backwards, factoring instead of multiplying.
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CCBC ~ Math 083
Section 1.2
Example 4. Factor completely. 10ab + 15b + 4a + 6 = 10ab + 15b +4a + 6 = 5b(2a + 3) +2(2a + 3) = (2a + 3)(5b + 2)
Split expression into two groups Factor each group of two terms (2a + 3) is the same; Factor out this GCF Our Answer
The key for grouping to work is after the GCF is factored out of the left and right groups, the two binomials must match exactly. If there is any dierence between the two, we either have to do some adjusting or it can't be factored using the grouping method. When the third term of the four-term expression is subtracted, the GCF extracted should be a negative number. Example 5. Factor completely. 6x2 + 9xy ¡ 14x ¡ 21y = 6x2 + 9xy ¡14x ¡ 21y = 3x(2x + 3y) ¡7(2x + 3y) = (2x + 3y)(3x ¡ 7)
Split expression into two groups Factor each group of two terms (2x + 3y) is the same; Factor out this GCF Our Answer
Often, we can recognize early that we need to use the opposite of the GCF when factoring. If the rst term of the rst binomial is positive in the problem, we will also want the rst term of the second binomial to be positive. If the rst term of the second binomial is negative, then we will use the opposite of the GCF to be sure they match. Example 6. Factor completely. 5xy ¡ 8x ¡ 10y + 16 = 5xy ¡ 8x ¡10y + 16 = x(5y ¡ 8) ¡2(5y ¡ 8) = (5y ¡ 8)(x ¡ 2)
Split expression into two groups Factor each group of two terms (5y ¡ 8) is the same; Factor out this GCF Our Answer
Example 7. Factor completely. 12ab ¡ 14a ¡ 6b + 7 = 12ab ¡ 14a ¡6b + 7 = 2a(6b ¡ 7) ¡1(6b ¡ 7) = (6b ¡ 7)(2a ¡ 1)
Split expression into two groups Factor each group of two terms (6b ¡ 7) is the same; Factor out this GCF Our Answer
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CCBC ~ Math 083
Section 1.2
Example 8. Factor completely. 6x3 ¡ 15x2 + 2x ¡ 5 = 6x3 ¡ 15x2 +2x ¡ 5 = 3x2(2x ¡ 5) +1(2x ¡ 5) = (2x ¡ 5)(3x2 + 1)
Split expression into two groups Factor each group of two terms (2x ¡ 5) is the same; Factor out this GCF Our Answer
Example 9. Factor completely. 4a2 + 6ab ¡ 14ab2 ¡ 21b3 = 4a2 + 6ab ¡14ab2 ¡ 21b3 = 2a(2a + 3b) ¡7b2(2a + 3b) = (2a + 3b)(2a ¡ 7b2)
Split expression into two groups Factor each group of two terms (2a + 3b) is the same; Factor out this GCF Our Answer
Example 10. Factor completely. 8xy ¡ 12y ¡ 10x + 1 = 8xy ¡ 12y ¡10x + 15 = 4y(2x ¡ 3) ¡5(2x ¡ 3) = (2x ¡ 3)(4y ¡ 5)
Split expression into two groups Factor each group of two terms (2x ¡ 3) is this same; Factor out this GCF Our Answer
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CCBC ~ Math 083
Section 1.2
1.2 Practice - Factor by Grouping Factor completely. 1) 40r3 ¡ 8r2 ¡ 25r + 5
2) 35x3 ¡ 10x2 ¡ 56x + 16
5) 15b3 + 21b2 ¡ 35b ¡ 49
6) 6x3 ¡ 48x2 + 5x ¡ 40
4) 14v 3 + 10v 2 ¡ 7v ¡ 5
3) 3n3 ¡ 2n2 ¡ 9n + 6
7) 3x3 + 15x2 + 2x + 10
8) 9x3 + 3x2 + 4x + 8
9) 35x3 ¡ 28x2 ¡ 20x + 16
10) 7n3 + 21n2 ¡ 5n ¡ 15
13) 32xy + 40x2 + 12y + 15x
14) 15ab ¡ 6a + 5b3 ¡ 2b2
11) 7xy ¡ 49x + 5y ¡ 35
12) 42r3 ¡ 49r2 + 18r ¡ 21
15) 16xy ¡ 56x + 2y ¡ 7
16) 3mn ¡ 8m + 15n¡40
17) x3 ¡ 5x2 + 7x ¡ 21
18) 5mn + 2m ¡ 25n ¡ 10
21) 32uv ¡ 20u + 24v ¡ 15
22) 4uv + 14u2 + 12v + 42u
19) 40xy + 35x ¡ 8y 2 ¡ 7y
20) 6a2 + 3a ¡ 4b2 + 2b
23) 10xy + 30 + 25x + 12y
24) 24xy + 25y 2 ¡ 20x ¡ 30y 3
25) 3uv + 14u ¡ 6u2 ¡ 7v
26) 56ab + 14 ¡ 49a ¡ 16b
27) 2xy ¡ 8x2 + 7y 3 ¡ 28y 2x
28) 28p3 + 21p2 + 20p + 15
29) 16xy ¡ 3x ¡ 6x2 + 8y
30) 8xy + 56x ¡ y ¡ 7
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CCBC ~ Math 083
Section 1.2
1.2 Answers - Factor by Grouping
1) (8r2 ¡ 5)(5r ¡ 1)
2) (5x2 ¡ 8)(7x ¡ 2)
12) (7r2 + 3)(6r ¡ 7)
13) (8x + 3)(4y + 5x)
23) (5x + 6)(2y + 5) 24) (4x ¡ 5y 2)(6y ¡ 5)
3) (n2 ¡ 3)(3n ¡ 2)
14) (3a + b2)(5b ¡ 2)
5) (3b2 ¡ 7)(5b + 7)
16) (m + 5)(3n ¡ 8)
26) (7a ¡ 2)(8b ¡ 7)
18) (m ¡ 5)(5n + 2)
28) (7p2 + 5)(4p + 3)
4) (2v 2 ¡ 1)(7v + 5) 6) (6x2 + 5)(x ¡ 8) 7) (3x2 + 2)(x + 5) 8) prime 9) (7x2 ¡ 4)(5x ¡ 4) 10)(7n2 ¡ 5)(n + 3) 11) (7x + 5)(y ¡ 7)
15) (8x + 1)(2y ¡ 7) 17) prime
19) (5x ¡ y)(8y + 7)
25) (3u ¡ 7)(v ¡ 2u)
27) (2x + 7y 2)(y ¡ 4x)
20) prime
29) (2x + 1)(8y ¡ 3x)
21) (4u + 3)(8v ¡ 5)
30) (8x ¡ 1)(y + 7)
22) 2(u + 3)(2v + 7u)
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CCBC ~ Math 083
Section 1.3
Section 1.3 - Factoring Trinomials Whose Leading Coe cient Is 1 Objective: Factor trinomials where the coe cient of x2 is 1. Factoring polynomials with three terms (factoring trinomials) is the most important type of factoring to be mastered. Since factoring can be thought of as the reverse of multiplication, we will start with a multipication problem and look at how we can reverse the process. Example 1. Multiply and simplify. (x + 6)(x ¡ 4) = x(x + 6) ¡ 4(x + 6) = x2 + 6x ¡ 4x ¡ 24 = x2 + 2x ¡ 24
Distribute (x + 6) through second parentheses Distribute each monomial through parentheses Combine like terms Our Answer
You may notice that if you reverse the last three steps, the process looks like grouping. This is because it is grouping! The GCF of the left two terms is x and the GCF of the second two terms is ¡4. The way we will factor trinomials is to make them into a polynomial with four terms and then factor by grouping. This is shown in the following example, which is the same problem worked backwards: Example 2. Factor completely. x2 + 2x ¡ 24 = x2 + 6x ¡ 4x ¡ 24 = x(x + 6) ¡ 4(x + 6) = (x + 6)(x ¡ 4)
Split middle term into +6x ¡ 4x Grouping: Factor each group of two terms (x + 6) is the same; factor out this GCF Our Answer
The trick to making these problems work is in the way we split the middle term. Why did we pick +6x ¡ 4x and not +5x ¡ 3x? To nd the correct way to split the middle term, we will use what is called the ac method . The way the ac method works here is we nd a pair of numbers that multiply to obtain the last number in the trinomial and also add up to the coe cient of the middle term of the trinomial. In the previous example, that would mean we needed the numbers to multiply to ¡24 and add up to +2. The only numbers that can do this are 6 and ¡4 ( 6 ¡4 = ¡24 and 6 + ( ¡ 4) = 2). This process is shown in the next few examples.
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Section 1.3
Example 3. Factor completely. x2 + 9x + 18 = x2 + 6x + 3x + 18 = x(x + 6) + 3(x + 6) = (x + 6)(x + 3)
Want to multiply to 18; sum to 9 6 and 3; split the middle term Factor by grouping Our Answer
Example 4. Factor completely. x2 ¡ 4x + 3 = x2 ¡ 3x ¡ x + 3 = x(x ¡ 3) ¡ 1(x ¡ 3) = (x ¡ 3)(x ¡ 1)
Want to multiply to 3; sum to ¡4 ¡3 and ¡1; split the middle term Factor by grouping Our Answer
Example 5. Factor completely. x2 ¡ 8x ¡ 20 = x2 ¡ 10x + 2x ¡ 20 = x(x ¡ 10) + 2(x ¡ 10) = (x ¡ 10)(x + 2)
Want to multiply to ¡20; sum to ¡8 ¡10 and 2; split the middle term Factor by grouping Our Answer
Often when factoring we have two variables. These problems solve just like problems with one variable, using the coe cients to decide how to split the middle term. Example 6. Factor completely. a2 ¡ 9ab + 14b2 = a2 ¡ 7ab ¡ 2ab + 14b2 = a(a ¡ 7b) ¡ 2b(a ¡ 7b) = (a ¡ 7b)(a ¡ 2b)
Want to multiply to 14; sum to ¡9 ¡7 and ¡2; split the middle term Factor by grouping Our Answer
As the past few examples illustrate, it is very important to be aware of negatives as we nd the pair of numbers we will use to split the middle term. Consider the following example, done incorrectly, ignoring negative signs: Warning! x2 + 5x ¡ 6 = x2 + 2x + 3x ¡ 6 = x(x + 2) + 3(x ¡ 2) ???
Want to multiply to 6; sum to 5 2 and 3; split the middle term Factor by grouping Binomials do not match!
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Section 1.3
Because we did not use the negative sign with the 6 to nd our pair of numbers, the binomials did not match and grouping was not able to work at the end. Now the problem will be done correctly: Example 7. Factor completely. x2 + 5x ¡ 6 = x2 + 6x ¡ 1x ¡ 6 = x(x + 6) ¡ 1(x + 6) = (x + 6)(x ¡ 1)
Want to multiply to ¡6; sum to 5 6 and ¡1; split the middle term Factor by grouping Our Answer
You may have noticed a shortcut for factoring these problems. Once we identify the two numbers that are used to split the middle term, these are the two numbers in our factors! In the previous example, the numbers used to split the middle term were 6 and ¡1; our factors turned out to be (x + 6)(x ¡ 1). This pattern does not always work, so be careful getting in the habit of using it. We can use it, however, when the coe cient of the x2 term is 1. In all of the problems we have factored in this lesson, the coe cient of x2 is 1. If this is the case, then we can use this shortcut. This is shown in the next few examples. Example 8. Factor completely. x2 ¡ 7x ¡ 18 = (x ¡ 9)(x + 2)
Want to multiply to ¡18; sum to ¡7 ¡ 9 and 2; write the factors Our Answer
Example 9. Factor completely. m2 ¡ mn ¡ 30n2
= (m + 5n)(m ¡ 6n)
Want to multiply to ¡30; sum to ¡1 5 and ¡6; write the factors; don 0t forget the second variable Our Answer
It is possible to have an expression that does not factor. If there is no combination of numbers that multiply and add up to the correct numbers, then we say we cannot factor the polynomial or we say the polynomial is prime. This is shown in the following example. Example 10. Factor completely. x2 + 2x + 6 prime
Want to multiply to 6; sum to 2 1 6 and 2 3 are only possibilities to multiply to 6; none sum to 2 Our Answer
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Section 1.3
When factoring, it is important not to forget about the GCF. If all of the terms in a problem have a common factor, we will want to rst factor out the GCF before we attempt any other method. The next three examples illustrate this technique. Example 11. Factor completely. 3x2 ¡ 24x + 45 = 3(x2 ¡ 8x + 15) = 3(x ¡ 5)(x ¡ 3)
GCF of all terms is 3; factor this out Want to multiply to 15; sum to ¡8 ¡ 5 and ¡3; write the factors Our Answer
Example 12. Factor completely. 4x2 y ¡ 8xy ¡ 32y = 4y(x2 ¡ 2x ¡ 8) = 4y(x ¡ 4)(x + 2)
GCF of all terms is 4y; factor this out Want to multiply to ¡ 8; sum to ¡2 ¡ 4 and 2; write the factors Our Answer
Example 13. Factor completely. 7a4b2 + 28a3b2 ¡ 35a2b2 = 7a2b2(a2 + 4a ¡ 5) = 7a2b2(a ¡ 1)(a + 5)
GCF of all terms is 7a2b2; factor this out Want to multiply to ¡5; sum to 4 ¡ 1 and 5; write the factors Our Answer
Again it is important to comment on the shortcut of jumping right to the factors. This only works if the coe cient of x2 is 1. In the next lesson, we will look at how this process changes slightly when we have a number other than 1 as the coe cient of x2. Be careful not to use this shortcut on all factoring problems! World View Note: The rst person to use letters for unknown values was Francois Vieta in 1591 in France. He used vowels to represent variables we are solving for, just as codes used letters to represent an unknown message.
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CCBC ~ Math 083
Section 1.3
1.3 Practice - Factoring Trinomials Whose Leading Coe cient Is 1 Factor completely. 1) p2 + 17p + 72
2) x2 + x ¡ 72 4) x2 + x ¡ 30
3) n2 ¡ 9n + 8
6) x2 + 13x + 40
5) x2 ¡ 9x ¡ 10
8) b2 ¡ 17b + 70
7) b2 + 12b + 32
10) x2 + 3x ¡ 18
9) x2 + 3x ¡ 70
12) a2 ¡ 6a ¡ 27
11) n2 ¡ 8n + 15
14) p2 + 7p ¡ 30
13) p2 + 15p + 54
16) m2 ¡ 15mn + 50n2
15) c ¡ 4c + 9 2
18) m2 ¡ 3mn ¡ 40n2
17) u2 ¡ 8uv + 15v 2 19) m + 2mn ¡ 8n 2
21) x ¡ 11xy + 18y 2
23) x + xy ¡ 12y 2
20) x2 + 10xy + 16y 2
2
22) u2 ¡ 9uv + 14v 2
2
24) x2 + 14xy + 45y 2
2
26) 4x2 + 52x + 168
25) x2 + 4xy ¡ 12y 2
28) 7w2 + 5w ¡ 35
27) 5a2 + 60a + 100
30) 5v 2 + 20v ¡ 25
29) 6a2 + 24a ¡ 192
32) 5m2 + 30mn ¡ 80n2
31) 6x2 + 18xy + 12y 2
34) 6m2 ¡ 36mn ¡162n2
33) 6x2 + 96xy + 378y 2
36) 5n2 ¡ 45n + 40
35) n2 ¡ 15n + 56
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CCBC ~ Math 083
Section 1.3
1.3 Answers - Factoring Trinomials Whose Leading Coe cient Is 1
1) (p + 9)(p + 8) 2) (x ¡ 8)(x + 9)
14) (p + 10)(p ¡ 3) 15) prime
28) prime
3) (n ¡ 8)(n ¡ 1)
16) (m ¡ 5n)(m ¡ 10n)
5) (x + 1)(x ¡ 10)
18) (m + 5n)(m ¡ 8n)
4) (x ¡ 5)(x + 6)
27) 5(a + 10)(a + 2)
17) (u ¡ 5v)(u ¡ 3v)
29) 6(a ¡ 4)(a + 8) 30) 5(v ¡ 1)(v + 5)
19) (m + 4n)(m ¡ 2n)
31) 6(x + 2y)(x + y)
8) (b ¡ 10)(b ¡ 7)
21) (x ¡ 9y)(x ¡ 2y)
33) 6(x + 9y)(x + 7y)
10) (x ¡ 3)(x + 6)
23) (x ¡ 3y)(x + 4y)
6) (x + 5)(x + 8) 7) (b + 8)(b + 4)
9) (x ¡ 7)(x + 10) 11) (n ¡ 5)(n ¡ 3) 12) (a + 3)(a ¡ 9)
13) (p + 6)(p + 9)
20) (x + 8y)(x + 2y) 22) (u ¡ 7v)(u ¡ 2v)
32) 5(m ¡ 2n)(m + 8n) 34) 6(m ¡ 9n)(m + 3n)
24) (x + 5y)(x + 9y)
35) (n ¡ 8)(n ¡ 7)
25) (x + 6y)(x ¡ 2y)
36) 5(n ¡ 8)(n ¡ 1)
26) 4(x + 7)(x + 6)
17
CCBC ~ Math 083
Section 1.4
Section 1.4 - Factoring Trinomials Whose Leading Coe cient Is Not 1 Objective: Factor trinomials using the ac method when the leading coe cient of the polynomial is not 1. When factoring trinomials, we use the ac method to split the middle term and then factor by grouping. The ac method gets its name from the general trinomial expression, ax2 + bx + c, where a; b; and c are the numbers (coe cients) in front of x2 and x terms; and the constant at the end; respectively. World View Note: It was French philosopher Rene Descartes who rst used letters from the beginning of the alphabet to represent values we know (a; b; c) and letters from the end to represent values we don't know and are solving for (x; y; z). The ac method is named ac because we multiply a c to nd out what we want the two numbers to multiply to. In the previous lesson, we always multiplied to just c because there was no number written in front of x2. This meant the coe cient of x2 was 1 and we were multiplying 1 c, which is c. Now we will have a number other than 1 in front of x2; so, we will be looking for two numbers that multiply to ac and add up to b. Example 1. Factor completely. 3x2 + 11x + 6 = 3x2 + 9x + 2x + 6 = 3x(x + 3) + 2(x + 3) = (x + 3)(3x + 2)
Multiply to a c; or (3)(6) = 18; sum to 11 The numbers are 9 and 2; split the middle term Factor by grouping Our Answer
When a = 1, or no coefficient is written in front of x2, we are able to use a shortcut using the numbers that split the middle term into the factors. The previous example illustrates an important point the shortcut does not work when a= / 1. We must go through all the steps of grouping in order to factor the expression. Example 2. Factor completely. 8x2 ¡ 2x ¡ 15 = 8x2 ¡ 12x + 10x ¡ 15 = 4x(2x ¡ 3) + 5(2x ¡ 3) = (2x ¡ 3)(4x + 5)
Multiply to a c; or (8)(¡15) = ¡120; sum to ¡2 The numbers are ¡12 and 10; split the middle term Factor by grouping Our Answer
18
CCBC ~ Math 083
Section 1.4
Example 3. Factor completely. 10x2 ¡ 27x + 5 = 10x2 ¡ 25x ¡ 2x + 5 = 5x(2x ¡ 5) ¡ 1(2x ¡ 5) = (2x ¡ 5)(5x ¡ 1)
Multiply to a c; or (10)(5) = 50; sum to ¡27 The numbers are ¡25 and ¡2; split the middle term Factor by grouping Our Answer
The same process works with two variables in the expression. Example 4. Factor completely. 4x2 ¡ xy ¡ 5y 2 = 4x2 + 4xy ¡ 5xy ¡ 5y 2 = 4x(x + y) ¡ 5y(x + y) = (x + y)(4x ¡ 5y)
Multiply to a c; or (4)(¡5) = ¡20; sum to ¡1 The numbers are 4 and ¡5; split the middle term Factor by grouping Our Answer
As always, when factoring we will rst look for a GCF before using any other method, including the ac method . Factoring out the GCF rst also has the added bonus of making the numbers smaller so the ac method becomes easier. Example 5. Factor completely. 18x3 + 33x2 ¡ 30x = 3x[6x2 + 11x ¡ 10] = 3x[6x2 + 15x ¡ 4x ¡ 10] = 3x[3x(2x + 5) ¡ 2(2x + 5)] = 3x(2x + 5)(3x ¡ 2)
GCF = 3x; factor this out rst Multiply to a c; or (6)(¡10) = ¡60; sum to 11 The numbers are 15 and ¡4; split the middle term Factor by grouping Our Answer
As was the case with trinomials where a = 1, not all trinomials can be factored. If there is no combination that multiplies and adds up correctly, then we say we cannot factor the polynomial or we say the polynomial is prime. Example 6. Factor completely. 3x2 + 2x ¡ 7 prime
Multiply a c; or (3)(¡7) = ¡21; sum to 2 ¡3(7) and ¡7(3) are only two ways to multiply to ¡21; but neither sums to 2 Our Answer
19
CCBC ~ Math 083
Section 1.4
1.4 Practice - Factoring Trinomials Whose Leading Coe cient Is Not 1 Factor completely. 1) 7x2 ¡ 48x + 36
2) 7n2 ¡ 44n + 12
3) 7b2 + 15b + 2
4) 7v 2 ¡ 24v ¡ 16
5) 5a2 ¡ 13a ¡ 28
6) 5n2 ¡ 4n ¡ 20
7) 2x2 ¡ 5x + 2
8) 3r2 ¡ 4r ¡ 4
10) 7x2 + 29x ¡ 30
9) 2x2 + 19x + 35 11) 2b2 ¡ b ¡ 3
12) 5k 2 ¡ 26k + 24
13) 5k 2 + 13k + 6
14) 3r2 + 16r + 21
15) 3x2 ¡ 17x + 20
16) 3u2 + 13uv ¡ 10v 2
17) 3x2 + 17xy + 10y 2
18) 7x2 ¡ 2xy ¡ 5y 2
19) 5x2 + 28xy ¡ 49y 2
20) 5u2 + 31uv ¡ 28v 2 22) 10a2 ¡ 54a ¡ 36
21) 6x2 ¡ 39x ¡ 21
24) 21n2 + 45n ¡ 54
23) 21k 2 ¡ 87k ¡ 90 25) 14x2 ¡ 60x + 16
26) 4r2 + r ¡ 3
27) 6x2 + 29x + 20
28) 6p2 + 11p ¡ 7
29) 4k 2 ¡ 17k + 4
30) 4r2 + 3r ¡ 7
31) 4x2 + 9xy + 2y 2
32) 4m2 + 6mn + 6n2
33) 4m2 ¡ 9mn ¡ 9n2
34) 4x2 ¡ 6xy + 30y 2
37) 12x2 + 62xy + 70y 2
38) 16x2 + 60xy + 36y 2
36) 18u2 ¡ 3uv ¡ 36v 2
35) 4x2 + 13xy + 3y 2
39) 24x2 ¡ 52xy + 8y 2
40) 12x2 + 50xy + 28y 2
20
CCBC ~ Math 083
Section 1.4
1.4 Answers - Factoring Trinomials Whose Leading Coe cient Is Not 1
1) (7x ¡ 6)(x ¡ 6)
2) (7n ¡ 2)(n ¡ 6) 3) (7b + 1)(b + 2)
4) (7v + 4)(v ¡ 4)
5) (5a + 7)(a ¡ 4) 6) prime
7) (2x ¡ 1)(x ¡ 2) 8) (3r + 2)(r ¡ 2)
9) (2x + 5)(x + 7) 10) (7x ¡ 6)(x + 5) 11) (2b ¡ 3)(b + 1)
12) (5k ¡ 6)(k ¡ 4) 13) (5k + 3)(k + 2) 14) (3r + 7)(r + 3)
15) (3x ¡ 5)(x ¡ 4)
16) (3u ¡ 2v)(u + 5v)
17) (3x + 2y)(x + 5y) 18) (7x + 5y)(x ¡ y)
19) (5x ¡ 7y)(x + 7y) 20) (5u ¡ 4v)(u + 7v) 21) 3(2x + 1)(x ¡ 7) 22) 2(5a + 3)(a ¡ 6)
23) 3(7k + 6)(k ¡ 5)
24) 3(7n ¡ 6)(n + 3)
29) (k ¡ 4)(4k ¡ 1) 30) (r ¡ 1)(4r + 7) 31) (x + 2y)(4x + y) 32) 2(2m2 + 3mn + 3n2) 33) (m ¡ 3n)(4m + 3n) 34) 2(2x2 ¡ 3xy + 15y 2) 35) (x + 3y)(4x + y) 36) 3(3u + 4v)(2u ¡ 3v) 37) 2(2x + 7y)(3x + 5y)
25) 2(7x ¡ 2)(x ¡ 4)
38) 4(x + 3y)(4x + 3y)
27) (x + 4)(6x + 5)
40) 2(3x + 2y)(2x + 7y)
26) (r + 1)(4r ¡ 3)
28) (3p + 7)(2p ¡ 1)
21
39) 4(x ¡ 2y)(6x ¡ y)
CCBC ~ Math 083
Section 1.5
Section 1.5 - Factoring Special Products Objective: Identify and factor special products including a dierence of two perfect squares, perfect square trinomials, and sum and dierence of two perfect cubes. When factoring there are few special products that, if we can recognize them, help us factor polynomials. The rst one we have seen before. When multiplying special products, we found that a sum and a dierence could multiply to a dierence of two perfect squares. Here, we will use this special product to help us factor. Dierence of Two Perfect Squares: a2 ¡ b2 = (a + b)(a ¡ b) If we are subtracting two perfect squares, then the expression will always factor to the sum and dierence of the square roots. Example 1. Factor completely. x2 ¡ 16 = (x)2 ¡ (4)2 = (x + 4)(x ¡ 4)
Express each term as the square of a monomial Apply the dierence of two perfect squares formula with a = x and b = 4 Our Answer
Example 2. Factor completely. 36 ¡ y 2 = (6)2 ¡ (y)2 = (6 + y)(6 ¡ y)
Express each term as the square of a monomial Apply the dierence of two perfect squares formula with a = 6 and b = y Our Answer
Example 3. Factor completely. 9a2 ¡ 25b2 = (3a)2 ¡ (5b)2 = (3a + 5b)(3a ¡ 5b)
Express each term as the square of a monomial Apply the dierence of two perfect squares formula Our Answer
It is important to note that a sum of squares will never factor. It is always prime. This can be seen if we try to use the ac method to factor x2 + 36.
22
CCBC ~ Math 083
Section 1.5
Example 4. Factor completely. x2 + 36 = x2 + 0x + 36
No bx term; we use 0x: Multiply to 36; sum to 0 1 36; 2 18; 3 12; 4 9; 6 6 but no combination that multiplies to 36 and sums to 0 Our Answer
prime
It turns out that a sum of squares is always prime. Sum of Two Perfect Squares: a2 + b2 is prime Another factoring shortcut is the perfect square trinomial. We had a shortcut for squaring a binomial, which can be reversed to help us factor a perfect square trinomial. Perfect Square Trinomial: a2 + 2ab+b2 = (a + b)2 If we do not recognize a perfect square trinomial at rst glance, we use the ac method. If we get two of the same numbers, we know we have a perfect square trinomial. Then we can factor using the square roots of the rst and last terms, and the sign from the middle term. This is shown in the following examples. Example 5. Factor completely. x2 ¡ 6x + 9
= (x ¡ 3)2
Multiply to 9; sum to ¡6 Numbers are ¡3 and ¡3; the same; a perfect square trinomial Use square roots from rst and last terms and sign from middle Our Answer
Example 6. Factor completely. 4x2 + 20xy + 25y 2
= (2x + 5y)2
Multiply to 100; sum to 20 Numbers are 10 and 10; the same; perfect square trinomial Use square roots from rst and last terms and sign from middle Our Answer
World View Note: The rst known record of work with polynomials comes from the Chinese around 200 BC. For example, problems would be written as three sheafs of a good crop, two sheafs of a mediocre crop, and one sheaf of a bad crop sold for 29 dou. This would be the polynomial (trinomial) 3x + 2y + z = 29.
23
CCBC ~ Math 083
Section 1.5
Another factoring shortcut involves the sum or dierence of two perfect cubes. The sum and the dierence of two perfect cubes have very similar factoring formulas: Sum of Two Perfect Cubes: a3 + b3 = (a + b)(a2 ¡ ab+b2) Dierence of Two Perfect Cubes: a3 ¡ b3 = (a ¡ b)(a2 + ab+b2) Start by expressing each term as the cube of a monomial. Use these results to determine the factored form of the expression. Comparing the formulas, you may notice that the only dierence is the signs in between the terms. One way to keep these two formulas straight is to think of SOAP. S stands for Same sign as the problem. If we have a sum of cubes, we add rst; if we have a dierence of cubes we subtract rst. O stands for Opposite sign. If we have a sum, then subtraction is the second sign; a dierence would have addition for the second sign. Finally, AP stands for Always Positive. The last term for both formulas has an addition sign. The following examples demonstrate factoring the sum or dierence of two perfect cubes. Example 7. Factor completely. m3 ¡ 27 = (m)3 ¡ (3)3 = (m ¡ 3)(m2 + 3m + 9)
Express each term as the cube of a monomial Apply the dierence of two perfect cubes formula (m 3)(m2 3m 9); Use SOAP to ll in signs Our Answer
Example 8. Factor completely. 125p3 + 8r3 = (5p)3 + (2r)3
= (5p + 2r)(25p2 ¡ 10r + 4r2)
Express each term as the cube of a monomial Apply the sum of two perfect cubes formula (5p 2r)(25p2 10r 4r2); Use SOAP to ll in signs Our Answer
The previous example illustrates an important point. When we ll in the trinomial's rst and last terms, we square the monomials 5p and 2r. So, our squared terms in the second set of parentheses are 5p 5p = 25r2 and 2r 2r = 4r2 : Often students forget to square the number in addition to the variable. Notice that when done correctly, both the number and the variable are squared. Often after factoring a sum or dierence of cubes, students want to factor the second factor, the trinomial, further. As a general rule, this factor will always be prime (unless there is a GCF which should have been factored out before applying the appropriate cubes rule).
24
CCBC ~ Math 083
Section 1.5
The following table sumarizes all of the shortcuts that we can use to factor special products: Factoring Special Products Dierence of Squares Sum of Squares Perfect Square Trinomial Sum of Cubes Dierence of Cubes
a2 ¡ b2 = (a + b)(a ¡ b) a2 + b2 = prime a2 + 2ab + b2 = (a + b)2 a3 + b3 = (a + b)(a2 ¡ ab + b2) a3 ¡ b3 = (a ¡ b)(a2 + ab + b2)
As always, when factoring special products it is important to check for a GCF rst. Only after checking for a GCF should we be using the special products. This is shown in the following examples. Example 9. Factor completely. 72x2 ¡ 2 = 2(36x2 ¡ 1) = 2(6x + 1)(6x ¡ 1)
GCF is 2 Dierence of Squares: 36x2 = (6x)2 and 1 = (1)2 Our Answer
Example 10. Factor completely. 48x2 y ¡ 24xy + 3y = 3y(16x2 ¡ 8x + 1) = 3y(4x ¡ 1)2
GCF is 3y Multiply to 16; sum to 8 Numbers are 4 and 4; the same; perfect square trinomial Our Answer
Example 11. Factor completely. 128a4b2 + 54ab5 = 2ab2(64a3 + 27b3) = 2ab2(4a + 3b)(16a2 ¡ 12ab + 9b2)
GCF is 2ab2 Sum of cubes: 64a3 = (4a)3 and 27b3 = (3b)3 Our Answer
25
CCBC ~ Math 083
Section 1.5
1.5 Practice - Factoring Special Products Factor completely. 1) r2 ¡ 16
2) x2 ¡ 9
3) v 2 ¡ 25
4) x2 ¡ 1
5) p2 ¡ 4
6) 4v 2 ¡ 1
8) 9a2 ¡ 1
7) 9k 2 ¡ 4 9) 3x2 ¡ 27
10) 5n2 ¡ 20
13) 18a2 ¡ 50b2
14) 4m2 + 64n2
11) 16x2 ¡ 36
12) 125x2 + 45y 2
15) a2 ¡ 2a + 1
16) k 2 + 4k + 4
17) x2 + 6x + 9
18) n2 ¡ 8n + 16
19) x2 ¡ 6x + 9
20) k 2 ¡ 4k + 4
23) 25a2 + 30ab + 9b2
24) x2 + 8xy + 16y 2
25) 4a2 ¡ 20ab + 25b2
26) 49x2 + 36y 2
29) 8 ¡ m3
30) x3 + 64
21) 25p2 ¡ 10p + 1
22) x2 + 2x + 1
27) 8x2 ¡ 24xy + 18y 2
28) 20x2 + 20xy + 5y 2
31) x3 ¡ 64
32) x3 + 8
35) 125a3 ¡ 64
36) 64x3 ¡ 27
33) 216¡u3
34) 125x3 ¡ 216
37) 64x3 + 27y 3
38) 32m3 ¡ 108n3
39) 54x3 + 250y 3
40) 375m3 + 648n3
26
CCBC ~ Math 083
Section 1.5
1.5 Answers - Factoring Special Products
1) (r + 4)(r ¡ 4)
21) (5p ¡ 1)2
2) (x + 3)(x ¡ 3)
22) (x + 1)2
3) (v + 5)(v ¡ 5)
23) (5a + 3b)2
4) (x + 1)(x ¡ 1)
24) (x + 4y)2
5) (p + 2)(p ¡ 2)
25) (2a ¡ 5b)2
6) (2v + 1)(2v ¡ 1)
26) prime
7) (3k + 2)(3k ¡ 2)
27) 2(2x ¡ 3y)2
8) (3a + 1)(3a ¡ 1)
28) 5(2x + y)2
9) 3(x + 3)(x ¡ 3)
29) (2 ¡ m)(4 + 2m + m2)
10) 5(n + 2)(n ¡ 2)
30) (x + 4)(x2 ¡ 4x + 16)
11) 4(2x + 3)(2x ¡ 3)
31) (x ¡ 4)(x2 + 4x + 16)
12) 5(25x2 + 9y 2)
32) (x + 2)(x2 ¡ 2x + 4)
13) 2(3a + 5b)(3a ¡ 5b)
33) (6 ¡ u)(36 + 6u + u2)
14) 4(m2 + 16n2)
34) (5x ¡ 6)(25x2 + 30x + 36)
15) (a ¡ 1)2
35) (5a ¡ 4)(25a2 + 20a + 16)
16) (k + 2)2
36) (4x ¡ 3)(16x2 + 12x + 9)
17) (x + 3)2
37) (4x + 3y)(16x2 ¡ 12xy + 9y 2)
18) (n ¡ 4)2
38) 4(2m ¡ 3n)(4m2 + 6mn + 9n2)
19) (x ¡ 3)2
39) 2(3x + 5y)(9x2 ¡ 15xy + 25y 2)
20) (k ¡ 2)2
40) 3(5m + 6n)(25m2 ¡ 30mn + 36n2)
27
CCBC ~ Math 083
Section 1.6
Section 1.6 - Factoring Strategy Objective: Identify and use the correct method to factor various polynomials. With so many dierent tools used to factor, it is easy to get lost as to which strategy to use when. Here, we will attempt to organize all the dierent factoring methods we have seen. A large part of deciding how to solve a problem is based on how many terms are in the problem. For all problem types, we will always try to factor out the GCF rst. Factoring Strategy (GCF First!!!!!)
2 terms: sum or dierence of perfect squares or cubes: a2 ¡ b2 = (a + b)(a ¡ b) a2 + b2 = prime
a3 + b3 = (a + b)(a2 ¡ ab + b2) a3 ¡ b3 = (a ¡ b)(a2 + ab+b2)
3 terms: ac method ; watch for perfect square trinomials, trinomials with a leading coe cient of 1, and trinomials with a leading coe cient other than 1: a2 + 2ab + b2 = (a + b)2 Multiply to ac; and sum to b
4 terms: grouping.
We will use the above strategy to factor each of the following examples. Here, the emphasis will be on which strategy to use rather than the steps used in that method. Example 1. Factor completely. x2 ¡ 23x+42 = (x ¡ 2)(x ¡ 21)
GCF = 1; so nothing to factor out of all three terms Three terms; multiply to 42; sum to ¡23 ¡ 2 and ¡21 Our Answer
Example 2. Factor completely. z 2 + 6z ¡ 9 prime
GCF = 1; so nothing to factor out of all three terms Three terms; multiply to ¡ 9; sum to 6 Try (¡1)(9); (1)(¡9); (3)(¡3); none sum to 6 Our Answer
28
CCBC ~ Math 083
Section 1.6
Example 3. Factor completely. 4x2 + 56xy + 196y 2 = 4(x2 + 14xy + 49y 2) = 4(x + 7y)2
GCF rst; factor out 4 from each term Three terms; try ac method, multiply to 49; sum to 14 7 and 7; a perfect square trinomial Our Answer
Example 4. Factor completely. 5x 2 y + 15xy ¡ 35x2 ¡ 105x = 5x(xy + 3y ¡ 7x ¡ 21) = 5x[y(x + 3) ¡ 7(x + 3)] = 5x(x + 3)(y ¡ 7)
GCF rst; factor out 5x from each term Four terms; try grouping (x + 3) match Our Answer
Example 5. Factor completely. 100x2 ¡ 400 = 100(x2 ¡ 4) = 100(x + 4)(x ¡ 4)
GCF rst; factor out 100 from both terms Two terms; dierence of perfect squares Our Answer
Example 6. Factor completely. 108x3 y 2 ¡ 39x2 y 2 + 3xy 2 = 3xy 2(36x2 ¡ 13x + 1) = 3xy 2(36x2 ¡ 9x ¡ 4x + 1) = 3xy 2[9x(4x ¡ 1) ¡ 1(4x ¡ 1)] = 3xy 2(4x ¡ 1)(9x ¡ 1)
GCF rst; factor out 3xy 2 from each term Three terms; ac method; multiply to 36; sum to ¡13 ¡9 and ¡4; split middle term Factor by grouping Our Answer
World View Note: Variables originated in ancient Greece where Aristotle would use a single capital letter to represent a number. Example 7. Factor completely. 5 + 625y 3 = 5(1 + 125y 3) = 5(1 + 5y)(1 ¡ 5y + 25y 2)
GCF rst; factor out 5 from each term Two terms; sum of perfect cubes Our Answer
It is important to be comfortable and con dent not just with using all the factoring methods, but also with deciding on which method to use. This is why your practice with these problems is very important!
29
CCBC ~ Math 083
Section 1.6
1.6 Practice - Factoring Strategy Factor completely. 1) 24az ¡ 18ah + 60yz ¡ 45yh
2) 2x2 ¡ 11x + 15
5) ¡2x3 + 128y 3
6) 20uv ¡ 60u3 ¡ 5xv + 15xu2
7) 5n3 + 7n2 ¡ 6n
8) 2x3 + 5x2 y + 3y 2x
9) 54u3 ¡ 16
10) 54 ¡ 128x3
11) n2 ¡ n
12) 5x2 ¡ 22x ¡ 15
3) 5u2 ¡ 9uv + 4v 2
4) 16x2 + 48xy + 36y 2
13) x2 ¡ 4xy + 3y 2
14) 45u2 ¡ 150uv + 125v 2
17) m2 ¡ 4n2
18) 12ab ¡ 18a + 6nb ¡ 9n
21) 128 + 54x3
22) 64m3 + 27n3
15) 64x2 + 49y 2
16) x3 ¡ 27y 3
19) 36b2c ¡ 16xd ¡ 24b2d + 24xc
20) 3m3 ¡ 6m2n ¡ 24n2m
24) 3ac + 15ad2 + x2c + 5x2d2
23) 2x3 + 6x2 y ¡ 20y 2x
26) 64m3 ¡ n3
25) n3 + 7n2 + 10n
28) 16a2 ¡ 9b2
27) 27x3 ¡ 64
30) 2x2 ¡ 10x + 12
29) 5x2 + 2x 31) 3k 3 ¡ 27k 2 + 60k
32) 32x2 ¡ 18y 2
35) 16x2 ¡ 8xy + y 2
36) v 2 + v
33) mn ¡ 12x + 3m ¡ 4xn
34) 2k 2 + k ¡ 10
37) 27m2 ¡ 48n2
38) x3 + 4x2
39) 9x3 + 21x2 y ¡ 60y 2x
40) 9n3 ¡ 3n2
43) 5x2-6x+7
44) 9x2 ¡ 25y 2
41) 2m2 + 6mn ¡ 20n2
42) 2u2v 2 ¡ 11uv 3 + 15v 4
30
CCBC ~ Math 083
Section 1.6
1.6 Answers - Factoring Strategy
1) 3(2a + 5y)(4z ¡ 3h)
23) 2x(x + 5y)(x ¡ 2y) 24) (3a + x2)(c + 5d2)
2) (2x ¡ 5)(x ¡ 3)
25) n(n + 2)(n + 5)
3) (5u ¡ 4v)(u ¡ v)
26) (4m ¡ n)(16m2 + 4mn + n2)
4) 4(2x + 3y)2
27) (3x ¡ 4)(9x2 + 12x + 16)
5) 2(¡x + 4y)(x2 + 4xy + 16y 2)
28) (4a + 3b)(4a ¡ 3b)
6) 5(4u ¡ x)(v ¡ 3u2)
29) x(5x + 2)
7) n(5n ¡ 3)(n + 2)
30) 2(x ¡ 2)(x ¡ 3)
8) x(2x + 3y)(x + y)
31) 3k(k ¡ 5)(k ¡ 4)
9) 2(3u ¡ 2)(9u2 + 6u + 4)
32) 2(4x + 3y)(4x ¡ 3y)
10) 2(3 ¡ 4x)(9 + 12x + 16x2)
33) (m ¡ 4x)(n + 3)
11) n(n ¡ 1)
34) (2k + 5)(k ¡ 2)
12) (5x + 3)(x ¡ 5)
35) (4x ¡ y)2
13) (x ¡ 3y)(x ¡ y)
36) v(v + 1)
14) 5(3u ¡ 5v)
2
37) 3(3m + 4n)(3m ¡ 4n)
15) prime
38) x2(x + 4)
16) (x ¡ 3y)(x2 + 3xy + 9y 2)
39) 3x(3x ¡ 5y)(x + 4y)
17) (m + 2n)(m ¡ 2n)
40) 3n2(3n ¡ 1)
18) 3(2a + n)(2b ¡ 3)
41) 2(m ¡ 2n)(m + 5n)
19) 4(3b + 2x)(3c ¡ 2d) 2
42) v 2(2u ¡ 5v)(u ¡ 3v)
20) 3m(m + 2n)(m ¡ 4n)
43) prime
21) 2(4 + 3x)(16 ¡ 12x + 9x2)
44) (3x + 5y)(3x ¡ 5y)
22) (4m + 3n)(16m2 ¡ 12mn + 9n2)
31
CCBC ~ Math 083
Section 1.7
Section 1.7 - Solve Equations by Factoring Objective: Solve equations by factoring and using the zero product rule. When solving linear equations such as 2x ¡ 5 = 21, we can solve for the variable directly by adding 5 and dividing by 2 to get 13. However, when we have x2 (or a higher power of x), we cannot just isolate the variable as we did with the linear equations. One method that we can use to solve for the variable is known as the zero product rule. Zero Product Rule: If ab = 0; then either a = 0 or b = 0: The zero product rule tells us that if two factors are multiplied together and the answer is zero, then one of the factors must be zero. We can use this rule to help us solve equations having zero on one side and a factored expression on the other side as in the following example. Example 1. Solve the equation. One factor must be zero Set each factor equal to zero Solve each equation
(2x ¡ 3)(5x + 1) = 0 2x ¡ 3 = 0 or 5x + 1 = 0 ¡1 ¡ 1 +3 + 3 2x = 3 or 5x = ¡1 2 2 5 5 3 1 x = or ¡ 2 5
Our Solutions
For the zero product rule to work, we must have factors to set equal to zero. This means if the problem is not already factored, we will need to factor it rst, if at all possible. Example 2. Solve the equation. 4x2 + x ¡ 3 = 0 4x2 ¡ 3x + 4x ¡ 3 = 0 x(4x ¡ 3) + 1(4x ¡ 3) = 0 (4x ¡ 3)(x + 1) = 0 4x ¡ 3 = 0 or x + 1 = 0 +3 + 3 ¡1 ¡ 1 4x = 3 or x = ¡1 4 4 3 x = or ¡1 4
Factor using the ac method; multiply to ¡12; sum to 1 The numbers are ¡3 and 4; split the middle term Factor by grouping Since one factor must be zero; set each factor equal to zero Solve each equation
Our Solutions
32
CCBC ~ Math 083
Section 1.7
Another important part of the zero product rule is that before we factor, one side of the equation must be zero. If one side of the equation is not zero, we must move terms around so that one side of the equation is zero. Generally, we like the coe cient of the x2 term to be positive. Example 3. Solve the equation. x2 = 8x ¡ 15 ¡8x + 15 ¡8x + 15 x2 ¡ 8x + 15 = 0 (x ¡ 5)(x ¡ 3) = 0 x ¡ 5 = 0 or x ¡ 3 = 0 +5 + 5 +3 + 3 x = 5 or x = 3
Set equal to zero by moving terms to the left Factor using the ac method; multiply to 15; sum to ¡8 The numbers are ¡5 and ¡3 Set each factor equal to zero Solve each equation Our Solutions
Example 4. Solve the equation. (x ¡ 7)(x + 3) = ¡9 x ¡ 7x + 3x ¡ 21 = ¡9 x2 ¡ 4x ¡ 21 = ¡9 +9 +9 x2 ¡ 4x ¡ 12 = 0 (x ¡ 6)(x + 2) = 0 x ¡ 6 = 0 or x + 2 = 0 +6 + 6 ¡2 ¡ 2 x = 6 or ¡2 2
Not equal to zero; multiply rst; use FOIL Combine like terms Move ¡9 to other side so right side of equation equals zero Factor using the ac method; multiply to ¡12; sum to ¡4 The numbers are 6 and ¡2 Set each factor equal to zero Solve each equation Our Solutions
Example 5. Solve the equation. 3x2 + 4x ¡ 5 = 7x2 + 4x ¡ 14 ¡3x2 ¡ 4x + 5 ¡3x2 ¡ 4x + 5 0 = 4x2 ¡ 9 0 = (2x + 3)(2x ¡ 3) 2x + 3 = 0 or 2x ¡ 3 = 0 ¡3 ¡ 3 +3 + 3 2x = ¡3 or 2x = 3 2 2 2 2 3 3 x = ¡ or 2 2
Set equal to zero by moving terms to the right side of the equal sign Factor using the dierence of squares One factor must be zero Set each factor equal to zero Solve each equation
Our Solutions
33
CCBC ~ Math 083
Section 1.7
Most problems with x2 will have two unique solutions. However, it is possible to have only one solution as the next example illustrates. Example 6. Solve the equation. 4x2 = 12x ¡ 9 ¡12x + 9 ¡12x + 9 4x2 ¡ 12x + 9 = 0 (2x ¡ 3)2 = 0 2x ¡ 3 = 0 +3 + 3 2x = 3 2 2 3 x= 2
Set equal to zero by moving terms to left Factor using the ac method; multiply to 36; sum to ¡12 ¡6 and ¡6 the same; a perfect square trinomial Set this factor equal to zero Solve the equation
Our Solution
As always, it will be important to factor out the GCF rst if we have one. This GCF is also a factor and must also be set equal to zero using the zero product rule. This may give us more than just two solutions. The next few examples illustrate this. Example 7. Solve the equation. 4x2 = 8x ¡8x ¡ 8x 4x2 ¡ 8x = 0 4x(x ¡ 2) = 0 4x = 0 or x ¡ 2 = 0 4 4 +2 + 2 x = 0 or 2
Set equal to zero by moving the 8x to left side of the equal sign Factor out the GCF of 4x One factor must be zero Set each factor equal to zero Solve each equation Our Solutions
Example 8. Solve the equation. 2x3 ¡ 14x2 + 24x = 0 2x(x2 ¡ 7x + 12) = 0 2x(x ¡ 3)(x ¡ 4) = 0 2x = 0 or x ¡ 3 = 0 or x ¡ 4 = 0 2 2 +3 + 3 +4 + 4 x = 0 or 3 or 4
Factor out the GCF of 2x Factor with ac method; multiply to 12; sum to ¡ 7 The numbers are ¡ 3 and ¡4 Set each factor equal to zero Solve each equation Our Solutions
34
CCBC ~ Math 083
Section 1.7
Example 9. Solve the equation. 6x2 + 21x ¡ 27 = 0 3(2x2 + 7x ¡ 9) = 0 3(2x2 + 9x ¡ 2x ¡ 9) = 0 3[x(2x + 9) ¡ 1(2x + 9)] = 0 3(2x + 9)(x ¡ 1) = 0 3 = 0 or 2x + 9 = 0 or x ¡ 1 = 0 +1 + 1 ¡9 ¡ 9 3= /0 2x = ¡9 or x = 1 2 2 9 x=¡ 2 9 x = ¡ or 1 2
Factor out the GCF of 3 Factor with ac method; multiply to ¡18; sum to 7 The numbers are 9 and ¡2 Factor by grouping One factor must be zero Set each factor equal to zero Solve each equation
Our Solutions
In the previous example, the GCF did not have a variable in it. When we set this factor equal to zero, we get a false statement. No solution comes from this factor. Often a student will skip setting the GCF factor equal to zero if there is no variables in the GCF. Just as not all polynomials can be factored, not all equations can be solved by factoring. If an equation cannot be solved by factoring, we will have to use another method. These other methods are saved for another lesson. World View Note: While factoring works great to solve problems with x2, Tartaglia, in 16th century Italy, developed a method to solve problems with x3. He kept his method a secret until another mathematician, Cardan, talked him out of his secret and published the results. To this day, the formula is known as Cardan's Formula. A question often asked is if it is possible to get rid of the square on the variable by taking the square root of both sides. While it is possible, there are a few properties of square roots that we have not covered yet, and thus it is common to break a rule of roots that we are not aware of at this point. The short reason we want to avoid this for now is because taking a square root will only allow us to consider the principal or positive square root. When we talk about roots, we will come back to problems like these and see how we can solve them using square roots in a method called completing the square. For now, never take the square root of both sides!
35
CCBC ~ Math 083
Section 1.7
1.7 Practice - Solve Equations by Factoring Solve each equation by factoring. 1) (k ¡ 7)(k + 2) = 0
2) (a + 4)(a ¡ 3) = 0
5) 6x2 ¡ 150 = 0
6) p2 + 4p ¡ 32 = 0
3) (x ¡ 1)(x + 4) = 0
4) 0 = (2x + 5)(x ¡ 7) 8) m2 ¡ m ¡ 30 = 0
7) 2n2 + 10n ¡ 28 = 0
10) 40r2 ¡ 285r ¡ 280 = 0
9) 7x2 + 26x + 15 = 0 11) 5n2 ¡ 9n ¡ 2 = 0
12) 2b2 ¡ 3b ¡ 2 = 0
13) x2 ¡ 4x ¡ 8 = ¡8
14) v 2 ¡ 8v ¡ 3 = ¡3
15) x2 ¡ 5x ¡ 1 = ¡5
16) a2 ¡ 6a + 6 = ¡2
19) 7x2 + 17x ¡ 20 = ¡8
20) 4n2 ¡ 13n + 8 = 5
17) 49p2 + 371p ¡ 163 = 5
18) 7k 2 + 57k + 13 = 5
21) 7r2 + 84 = ¡49r
22) 7m2 ¡ 224 = 28m
23) x2 ¡ 6x = 16
24) 7n2 ¡ 28n = 0
25) 3v 2 + 7v = 40
26) 6b2 = 5 + 7b
27) 35x2 + 120x = ¡45
28) 9n2 + 39n = ¡36
29) 4k 2 + 18k ¡ 23 = 6k ¡ 7
30) a2 + 7a ¡ 9 = ¡3 + 6a
31) 9x2 ¡ 46 + 7x = 7x + 8x2 + 3
32) x2 + 10x + 30 = 6
33) 2m2 + 19m + 40 = ¡2m
34) 5n2 + 41n + 40 = ¡2
35) 5x(3x ¡ 6) ¡ 5x2 = x2 ¡ 15x + 24
36) 24x2 + 11x ¡ 80 = 3x
37) 40p2 + 183p ¡ 168 = p + 5p2
38) 121w2 + 8w ¡ 7 = 8w ¡ 6
36
CCBC ~ Math 083
Section 1.7
1.7 Answers - Solve Equations by Factoring
1) 7; ¡2 2) ¡4; 3 3) 1; ¡4 4)
5 ¡2; 7
5) ¡5; 5 6) 4; ¡8 7) 2; ¡7 8) ¡5; 6 5
9) ¡ 7 ; ¡3 7
10) ¡ 8 ; 8 1
11) ¡ 5 ; 2 1
12) ¡ 2 ; 2
15) 1; 4
4
28) ¡ 3 ; ¡3
16) 4; 2 17)
3 ; ¡8 7
18)
1 ¡ 7 ; ¡8
19)
4 ; ¡3 7
20)
1 ;3 4
29) ¡4; 1 30) 2; ¡3 31) ¡7; 7 32) ¡4; ¡6 5
33) ¡ 2 ; ¡8
21) ¡4; ¡3
6
22) 8; ¡4
34) ¡ 5 ; ¡7
24) 4; 0
35) ¡ 1,
23) 8; ¡2 25)
8 3
5
36) 3 ; ¡2
8 ; ¡5 3
4
1 5
37) 5 ; ¡6
3
38) ¡ 11 ; 11
13) 4; 0
26) ¡ 2 ; 3
14) 8; 0
27) ¡ 7 ; ¡3
1
37
1
CCBC ~ Math 083
Chapter 2
Chapter 2: Rational Expressions and Equations Section 2.1 - Reduce Rational Expressions.......................................................40 Section 2.2 - Multiply and Divide Rational Expressions..................................47 Section 2.3 - Least Common Denominator.......................................................52 Section 2.4 - Add and Subtract Rational Expressions......................................58 Section 2.5 - Proportions .................................................................................64 Section 2.6 - Solving Rational Equations..........................................................72 Section 2.7 - Motion and Work Applications....................................................79 Section 2.8 - Variation......................................................................................86
39
CCBC ~ Math 083
Section 2.1
Section 2.1 - Reduce Rational Expressions Objective: Reduce rational expressions by dividing out common factors. A rational expression is a quotient of polynomials. Examples of rational expressions include: x2 ¡ x ¡ 12 x2 ¡ 9x + 20
and
3 x¡2
and
a¡b b¡a
and
3 2
As rational expressions often contain a variable in the denominator, it is important to remember that the denominator of a fraction cannot have a value of zero. For this reason, rational expressions may have one or more excluded x -values; that is, values that the variable cannot be or the expression would be unde ned. We determine the values that will give us zero in the denominator, and then exclude those values from the list (domain) of possible values the variable could be. Example 1. State the excluded value(s). x2 ¡ 1 3x2 + 5x 2 3x + 5x = 0 x(3x + 5) = 0 x = 0 or 3x + 5 = 0 ¡5 ¡ 5 3x = ¡5 3 3 5 x=¡ 3 5 Excluded values: x = 0 or ¡ 3
Find values of x for which denominator is 0 Factor Set each factor equal to zero Subtract 5 from second equation Divide by 3 Second equation is solved Our Answer
5
This means we can use any value for x in the equation except for 0 and ¡ 3 . We can, however, evaluate using any other value of x in the expression. World View Note: The number zero was not widely accepted in mathematical thought around the world for many years. It was the Mayans of Central America who rst used zero to aid in the use of their base 20 system as a place holder. Rational expressions are easily evaluated by substituting the given value for the variable and then simplifying using the Order of Operations.
40
CCBC ~ Math 083
Section 2.1
Example 2. Evaluate the expression. x2 ¡ 4 when x = ¡6 x2 + 6x + 8
Substitute ¡6 in for each variable
(¡6)2 ¡ 4 (¡6)2 + 6(¡6) + 8
Exponents rst
=
36 ¡ 4 36 + 6(¡6) + 8
Multiply
=
36 ¡ 4 36 ¡ 36 + 8
Add and subtract
=
32 8
Reduce Our Answer
=4
Just as we reduced the previous example, often a rational expression can be reduced, even without knowing the value of the variable. When we reduce, we divide out common factors. If the expression only has monomials, we can reduce the coe cients, and subtract exponents on the variables. Example 3. Simplify the expression. 15x4 y 2 25x2 y 6
Reduce 15 and 25 by dividing each by 5; subtract the exponents (x4¡2 ) = x2; (y 2¡6) = y ¡4: Negative exponents move to denominator; so y ¡4 becomes
=
3x2 5y 4
1 : y4
Our Answer
However, if there is more than just one term in either the numerator or denominator, we can't divide out common factors unless we rst factor the numerator and denominator. Example 4. Simplify the expression. 28 8x2 ¡ 16 =
=
28 8(x2 ¡ 2) 7 2(x2 ¡ 2)
Denominator has a common factor of 8; factor: Reduce by dividing 28 and 8 by 4: Our Answer
41
CCBC ~ Math 083
Section 2.1
Example 5. Simplify the expression. 9x ¡ 3 18x ¡ 6
Numerator has a common factor of 3; denominator has a
=
3(3x ¡ 1) 6(3x ¡ 1)
Divide out common factor (3x ¡ 1); and divide both 3 and 6 by 3
=
1 2
Our Answer
common factor of 6; factor both the numerator and denominator
Example 6. Simplify the expression. x2 ¡ 25 x2 + 8x + 15
Numerator is the dierence of two perfect squares;
=
(x + 5)(x ¡ 5) (x + 5)(x + 3)
Divide out the common factor (x + 5)
=
x¡5 x+3
Our Answer
denominator can be factored using the ac method; factor both the numerator and the denominator
It is important to remember that we cannot reduce terms, only factors. This means if there are any + or ¡ signs between the parts we want to reduce, we x¡5 cannot reduce. In the previous example, we had the solution x + 3 ; we cannot divide out the xs because they are terms (separated by + or ¡), not factors (separated by multiplication). Example 7. Simplify the expression. 5¡x x¡5
Rewrite the numerator
=
¡x + 5 x¡5
Factor ¡ 1 from the numerator
=
¡1(x ¡ 5) x¡5
Divide out common factor (x ¡ 5)
= ¡1
Our Answer
42
CCBC ~ Math 083
Section 2.1
Example 8. Simplify the expression. 7¡x x2 ¡ 49
Rewrite the numerator
¡x + 7 x2 ¡ 49
Factor ¡ 1 from the numerator; Factor denominator as
=
¡1(x ¡ 7) (x + 7)(x ¡ 7)
Divide out common factor (x ¡ 7)
=
¡1 x+7
Our Answer
=
dierence of two perfect squares
43
CCBC ~ Math 083
Section 2.1
2.1 Practice - Reduce Rational Expressions Evaluate. 1)
4v + 2 6
3)
x¡3 x2 ¡ 4x + 3
5)
b+2 b2 + 4b + 4
2)
b¡3 3b ¡ 9
when x = ¡4
4)
a+2 a2 + 3a + 2
when b = 0
6)
n2 ¡ n ¡ 6 n¡3 27p 18p2 ¡ 36p
when v = 4
when b = ¡2 when a = ¡1 when n = 4
State the excluded value(s). 7)
3k 2 + 30k k + 10
8)
9)
15n2 10n + 25
10)
x + 10 8x2 + 80x
11)
10m2 + 8m 10m
12)
10x + 16 6x + 20
13)
r 2 + 3r + 2 5r + 10
14)
6n2 ¡ 21n 6n2 + 3n
15)
b2 + 12b + 32 b2 + 4b ¡ 32
16)
10v2 + 30v 35v 2 ¡ 5v
Simplify each expression. 17)
21x2 18x
18)
24a 40a2
19)
32x3 8x4
20)
90x2 20x
21)
18m ¡ 24 60
22)
20 4p + 2
23)
x+3 3+x
24)
5¡x x¡5
25)
9¡n 9n ¡ 81
26)
x+7 7¡x
27)
x+1 x2 + 8x + 7
28)
28m + 12 36
29)
32x2 28x2 + 28x
30)
49r + 56 56r
31)
n2 + 4n ¡ 12 n2 ¡ 7n + 10
32)
b2 + 14b + 48 b2 + 15b + 56
9v + 54 ¡ 60
34)
30x ¡ 90 50x + 40
33)
v 2 ¡ 4v
35)
12x2 ¡ 42x 30x2 ¡ 42x
36)
k2 ¡ 12k + 32 k2 ¡ 64
37)
6a ¡ 10 10a + 4
38)
9p + 18 p2 + 4p + 4
39)
2n2 + 19n ¡ 10 9n + 90
40)
3x2 ¡ 29x + 40 5x2 ¡ 30x ¡ 80
44
CCBC ~ Math 083
41)
Section 2.1
3¡x 2
5x ¡ 45
42)
9r 2 + 63r 5r 2 + 40r + 35
43)
2x2 ¡ 10x + 8 3x2 ¡ 7x + 4
44)
50b ¡ 80 50b + 20
45)
7n2 ¡ 32n + 16 4n ¡ 16
46)
35v + 35 21v + 7
47)
n2 ¡ 2n + 1 6n + 6
48)
56x ¡ 48 24x2 + 56x + 32
49)
7a2 ¡ 26a ¡ 45 6a2 ¡ 34a + 20
50)
4k3 ¡ 2k 2 ¡ 2k 9k 3 ¡ 18k 2 + 9k
45
CCBC ~ Math 083
Section 2.1
2.1 Answers - Reduce Rational Expressions
1) 3
20)
9x 2
37)
3a ¡ 5 5a + 2
1 3
21)
3m ¡ 4 10
38)
9 p+2
22)
10 2p + 1
39)
2n ¡ 1 9
40)
3x ¡ 5 5(x + 2)
41)
¡1 5(x + 3)
42)
9r 5(r + 1)
43)
2(x ¡ 4) 3x ¡ 4
2)
1
3) ¡ 5
4) unde ned 5)
1 2
24) ¡1
6) 6
1
7) ¡10 8) 0; 2 5
9) ¡ 2 10) 0; ¡10 11) 0
10
12) ¡ 3 13) ¡2 14)
23) 1 25) ¡ 9
26) cannot be simpli ed 27)
1 x+7
28)
7m + 3 9
29)
8x 7(x + 1)
44)
5b ¡ 8 5b + 2
30)
7r + 8 8r
45)
7n ¡ 4 4
46)
5(v + 1) 3v + 1
47)
(n ¡ 1)2 6(n + 1)
48)
7x ¡ 6 (3x + 4)(x + 1)
49)
7a + 9 2(3a ¡ 2)
50)
2(2k + 1) 9(k ¡ 1)
31)
1 0; ¡ 2
n+6 n¡5
32)
b+6 b+7
1
33)
9 v ¡ 10
15) ¡8; 4 16) 0; 7 17)
7x 6
34)
3(x ¡ 3) 5x + 4
18)
3 5a
35)
2x ¡ 7 5x ¡ 7
19)
4 x
36)
k ¡4 k+8
46
CCBC ~ Math 083
Section 2.2
Section 2.2 - Multiply and Divide Rational Expressions Objective: Multiply and divide rational expressions. Multiplying and dividing rational expressions is the same process we use to multiply and divide fractions. Example 1. Multiply, expressing the resulting fraction in its lowest terms. 15 14 49 45
First; reduce by dividing out the common factors from numerator
=
1 2 7 3
Multiply the numerators together and the denominators together
=
2 21
Our Answer
and denominator (15 and 7)
When multiplying rational expressions, we rst divide out common factors, and then multiply straight across. Example 2. Multiply, expressing the resulting fraction in its lowest terms. 25x 24y 9y 8 55x7 2
4
Reduce coe cients by dividing out common factors from the numerator and the denominator (3 and 5) Reduce the variables by subtracting exponents (x2¡7) = x¡5; (y 4¡8) = y ¡4 1 Negative exponent (x¡5) moves to the denominator as 5 ; x 1 ¡4 likewise; (y ) to the denominator as 4 y
=
5 8 4 3y 11x5
Multiply across (numerators together; denominators together)
=
40 33x5 y 4
Our Answer
The process for division has the extra step of multiplying by the reciprocal ( ipped fraction) of the divisor. Example 3. Divide, expressing the resulting fraction in its lowest terms.
=
a4b2 b4 a 4
Multiply the rst expression by the reciprocal of the second expression
a4b2 4 a b4
Subtract exponents ; negative exponents move to denominator
47
CCBC ~ Math 083
Section 2.2
=
a3 4 1 b2
Multiply across (numerators together; denominators together)
=
4a3 b2
Our Answer
If the rational expression in either the numerator or the denominator is factorable, it must be factored rst. That way, any common factors can be divided out before multiplying. Example 4. Multiply, expressing the resulting fraction in its lowest terms. x2 ¡ 8x + 16 x2 ¡ 9 3x + 9 x2 + x ¡ 20
Factor each numerator and denominator
=
(x + 3)(x ¡ 3) (x ¡ 4)(x ¡ 4) (x ¡ 4)(x + 5) 3(x + 3)
Divide out the common factors (x + 3) and (x ¡ 4)
=
x¡3 x¡4 x+5 3
Multiply across
=
(x ¡ 3)(x ¡ 4) 3(x + 5)
Our Answer
Again we follow the same pattern with division with the extra rst step of multiplying by the reciprocal of the divisor. Example 5. Divide, expressing the resulting fraction in its lowest terms. x2 ¡ x ¡ 12 5x2 + 15x x2 ¡ 2x ¡ 8 x2 + x ¡ 2
Multiply by the reciprocal
=
x2 ¡ x ¡ 12 x2 + x ¡ 2 x2 ¡ 2x ¡ 8 5x2 + 15x
Factor each numerator and denominator
=
(x ¡ 4)(x + 3) (x + 2)(x ¡ 1) (x + 2)(x ¡ 4) 5x(x + 3)
Divide out the common factors:
=
1 x¡1 1 5x
Multiply across
=
x¡1 5x
Our Answer
(x ¡ 4) and (x + 3) and (x + 2)
48
CCBC ~ Math 083
Section 2.2
We can combine multiplying and dividing of rational expressions into one problem as shown below. To perform these operations, we change any divisions to multiplication by the reciprocal, factor wherever possible, reduce if possible, and then multiply the remaining factors. Example 6. Multiply and/or divide as indicated, expressing the resulting fraction in its lowest terms. a¡1 a2 + 7a + 10 a + 1 2 a + 6a + 5 4a + 8 a + 2
Factor each expression
=
(a ¡ 1) (a + 5)(a + 2) (a + 1) (a + 5)(a + 1) 4(a + 2) (a + 2)
Multiply by the reciprocal of last fraction
=
(a + 5)(a + 2) (a + 1) (a + 2) (a + 5)(a + 1) 4(a + 2) (a ¡ 1)
Divide out the common factors
a+2 4(a ¡ 1)
Our Answer
=
(a + 5); (a + 2); (a + 1)
World View Note: Indian mathematician Aryabhata, in the 6th century, pubn(n + 1)(n + 2) for the sum of lished a work which included the rational expression 6 1 2 2 2 the rst n squares (1 + 2 + 3 + :::: + n ):
49
CCBC ~ Math 083
Section 2.2
2.2 Practice - Multiply and Divide Rational Expressions Multiply or divide, expressing the result in its lowest terms. 2
9
2)
8x 3x
7
4)
9m 5m2
6
6)
10p 5
8)
7 10(n + 3)
1)
8x2 9
3)
9n 2n
5)
5x2 4
7)
7 (m ¡ 6) m¡6
9)
7r 7r(r + 10)
5n 5 5m(7m ¡ 5) 7(7m ¡ 5)
4
7 7
2 8
10 n¡2
(n + 3)(n ¡ 2)
10)
6x(x + 4) x¡3
(x ¡ 3)(x ¡ 6) 6x(x ¡ 6)
12)
9 b2 ¡ b ¡ 12
b2 ¡ b ¡ 12
14)
v ¡1 4
v2 ¡ 11v + 10
16)
1 a¡6
18)
p¡8 p2 ¡ 12p + 32
6
20)
x2 ¡ 7x + 10 x¡2
m¡5 5m2
22)
2r r+6
r¡6
(r ¡ 6)2
11)
25n + 25 5
13)
x ¡ 10 35x + 21
15)
x2 ¡ 6x ¡ 7 x+5
17)
8k 24k2 ¡ 40k
4
30n + 30 7
35x + 21 x+5
x¡7 1
15k ¡ 25
19) (n ¡ 8) 10n ¡ 80
b¡5
4
8a + 80 8 1
p ¡ 10 x + 10
x2 ¡ x ¡ 20
2r
21)
4m + 36 m+9
23)
3x ¡ 6 (x + 3) 12x ¡ 24
24)
2n2 ¡ 12n ¡ 54 n+7
25)
b+2 (5b ¡ 3) 40b2 ¡ 24b
26)
21v2 + 16v ¡ 16 3v + 4
27)
n¡7 6n ¡ 12
12 ¡ 6n
28)
x2 + 11x + 24 6x3 + 18x2
29)
27a + 36 9a + 63
6a + 8 2
30)
31)
x2 ¡ 12x + 32 x2 ¡ 6x ¡ 16
n2 ¡ 13n + 42
7x2 + 14x
7x2 + 21x
33) (10m2 + 100m) 35)
7p2 + 25p + 12 6p + 48
37)
10b2 30b + 20
39)
7r 2 ¡ 53r ¡ 24 7r + 2
41)
x2 ¡ 1 2x ¡ 4
18m3 ¡ 36m2 20m2 ¡ 40m
3p ¡ 8
21p2 ¡ 44p ¡ 32
30b + 20
2b2 + 10b 49r + 21
49r + 14
x2 ¡ 4
x2 ¡ x ¡ 2
x2 + x ¡ 2 3x ¡ 6
k ¡7 ¡ 12
k2 ¡ k
(2n + 6)
35v ¡ 20 v ¡9
6x3 + 6x2
x2 + 5x ¡ 24 7k 2 ¡ 28k
8k2 ¡ 56k
32)
9x3 + 54x2 x2 + 5x ¡ 14
34)
n¡7 n2 ¡ 2n ¡ 35
10n + 50
36)
7x2 ¡ 66x + 80 49x2 + 7x ¡ 72
38)
35n2 ¡ 12n ¡ 32 49n2 ¡ 91n + 40
40)
12x + 24 10x2 + 34x + 28
42)
50
7r + 42
a3 + b3 a2 + 3ab + 2b2
x2 + 5x ¡ 14 10x2 9n + 54
7x2 + 39x ¡ 70
49x2 + 7x ¡ 72
7n2 + 16n ¡ 15 5n + 4 15x + 21 5
3a ¡ 6b
3a2 ¡ 3ab + 3b2
a2 ¡ 4b2 a + 2b
CCBC ~ Math 083
Section 2.2
2.2 Answers - Multipy and Divide Rational Expressions
1) 4x2
16)
a + 10 a¡6
2)
14 3
17) 5
3)
63 10n
18)
p ¡ 10 p¡4
4)
63 10m
19)
3 5
5)
3x2
20)
x + 10 x+4
6)
5p 2
21)
4(m ¡ 5) 5m2
2
7) 5m
x+3 4
24)
n¡9 n+7
10) x + 4
25)
b+2 8b
11)
2 3
26)
v ¡9 5
12)
9 b¡5
27) ¡ n ¡ 6
13)
x ¡ 10 7
14)
1 v ¡ 10
9)
7 10 r¡6 r + 10
15) x + 1
x¡4 x+3
32)
9(x + 6) 10
33) 9m2(m + 10)
22) 7 23)
8)
31)
34)
10 9(n + 6)
35)
p+3 6(p + 8)
36)
x¡8 x+7
37)
5b b+5
38) n + 3 39) r ¡ 8
1
40)
18 5
28)
x+1 x¡3
41)
3 2
29)
1 a+7
42)
1 a + 2b
30)
7 8(k + 3)
51
CCBC ~ Math 083
Section 2.3
Section 2.3 - Least Common Denominator Objective: Identify the least common denominator and build equivalent fractions using that common denominator. Finding the least common denominator, LCD, and expressing equivalent fractions using that LCD is very important when working with rational expressions. The process we use to nd the LCD is based on the process used to nd the LCD of integers. Example 1. Find the Least Common Denominator. Find the LCD of
5 1 and 6 8
LCD is 24
Consider multiples of the larger denominator: 8; 16; 24:::: 24 is the rst multiple of 8 that is also divisible by 6 Our Answer
When nding the LCD of several monomials, we rst nd the LCD of the coe cients, then use all unique variables and attach the highest exponent for each variable. Example 2. Find the Least Common Denominator. 1
7
Find the LCD of 4x2 y5 and 6x4 y3z6 :
LCD is 12x4 y 5z 6
First nd the LCD of coe cients 4 and 6: 12 is the LCD of 4 and 6 because it is the smallest number that both 4 and 6 divide into without a remainder Use all variables with highest exponent for each variable: x4 y 5z 6 Our Answer
The same pattern can be used on polynomials that have more than one term. However, we must rst factor each polynomial so that we can identify all the factors to be used (attaching highest exponent if necessary). Example 3. Find the Least Common Denominator. Find the LCD of
4 x2 + 2x ¡ 3
and
9 x2 ¡ x ¡ 12
x2 + 2x ¡ 3 = (x ¡ 1)(x + 3) x2 ¡ x ¡ 12 = (x ¡ 4)(x + 3) (x ¡ 1)(x + 3) and (x ¡ 4)(x + 3) LCD is (x ¡ 1)(x + 3)(x ¡ 4)
52
Factor each polynomial LCD uses all unique factors; here; (x + 3) is common to both expressions; we only express it once Our Answer
CCBC ~ Math 083
Section 2.3
Notice we only used (x + 3) once in our LCD. This is because it only appears as a factor once in either polynomial. The only time we need to repeat a factor or use an exponent on a factor is if there are multiple like factors associated with one of the polynomials in completely factored form. Example 4. Find the Least Common Denominator. 3x
1
Find the LCD of x2 ¡ 10x + 25 and x2 ¡ 14x + 45 . x2 ¡ 10x + 25 = (x ¡ 5)(x ¡ 5) = (x ¡ 5)2 x2 ¡ 14x + 45 = (x ¡ 5)(x ¡ 9) (x ¡ 5)2 and (x ¡ 5)(x ¡ 9) LCD is (x ¡ 5)2(x ¡ 9)
Factor each polynomial
LCD uses all unique factors with highest exponent Our Answer
The previous example could have also been done with factoring the rst polynomial to (x ¡ 5)(x ¡ 5) and not expressing it as (x ¡ 5)2. In that case, we would have included (x ¡ 5) twice in the LCD because it showed up twice in one of the polynomials. However, expressing the factors using exponents allows us to use the same pattern (including the highest exponent in the LCD) that we previously used with monomials. Once we know the LCD, our goal will be to build up fractions so that they have matching denominators. Whenever we alter the denominator of a fraction by multiplying to get the LCD, we must multiply by the same factor in the numerator of that fraction in order to keep the fraction equivalent to its original value. We can build up a fraction to an equivalent one with a speci ed denominator by multiplying the numerator and denominator by any factors that are part of the LCD, but not part of the original denominator. Example 5. Build the rational expression to an equivalent expression with the speci ed denominator. State the value of the missing numerator. 5a ? = 5 3 2 3a b 6a b
=
Identify what factors we need to match denominators: The missing factor is 2a3b2 because 3 2 = 6 and we need three more as and two more bs
5a 2a3b2 3a2b 2a3b2
Multiply both numerator and denominator by missing factor
10a4b2 6a5b3
Find the value of the numerator
10a4b2
Our Answer
53
CCBC ~ Math 083
Section 2.3
Example 6. Build the rational expression to an equivalent expression with the speci ed denominator. State the value of the missing numerator. ? x¡2 = 2 x + 4 x + 7x + 12 (x + 4)(x + 3)
=
Factor to identify factors we need to match denominators: The missing factor is (x + 3)
x¡2 x+3 x+4 x+3
Multiply numerator and denominator by missing factor
(x ¡ 2) (x + 3) (x + 4) (x + 3)
Multiply (FOIL) the numerator
x2 + x ¡ 6 (x + 4)(x + 3)
Find the value of the numerator
=
x2 + x ¡ 6
Our Answer
As the above example illustrates, we multiply out our numerators, but keep our denominators factored. The reason for multiplying out only the numerators is that in order to add and subtract fractions, we need to combine like terms in the numerator. However, once the like terms have been added/subtracted in the numerator, we factor so that we can see if the expression can be reduced further. Since we can only reduce factors, both the numerator and the denominator should be in completely factored format, and only common factors can be divided out. Once we know how to nd the LCD and how to build up fractions using the desired common denominator, we can use this knowledge to nd a common denominator and build up those fractions. In this lesson, we will not be adding and subtracting fractions, just building them up to a common denominator. Example 7. Build up each rational expression so they have a common denominator. 3c 5a and 3 6a2b 4b c
First; identify the LCD; in this case the least common denominator is 12a2b3c Determine what factors each fraction 0s denominator is missing: First is missing 3a2 and second is missing 2b2c
54
CCBC ~ Math 083
Section 2.3
5a 3a2 3c 2b2c and 4b3c 3a2 6a2b 2b2c
Multiply each corresponding fraction 0s numerator
15a3 6b2c2 and 12a2b3c 12a2b3c
Our Answers
and denominator by that denominator 0s missing factors to get equivalent fractions
Example 8. Build up each rational expression so they have a common denominator. 5x x2 ¡ 5x ¡ 6
and
(x ¡ 6)(x + 1)
x¡2
Factor deminators to nd LCD
x2 + 4x + 3 (x + 1)(x + 3)
x+3 x¡2 x¡6 5x and (x + 1)(x + 3) x ¡ 6 (x ¡ 6)(x + 1) x + 3
5x2 + 15x x2 ¡ 8x + 12 and (x ¡ 6)(x + 1)(x + 3) (x ¡ 6)(x + 1)(x + 3)
Use factors to nd LCD: LCD is (x ¡ 6)(x + 1)(x + 3) Determine what factors each fraction 0s denominator is missing: First: (x + 3) Second: (x ¡ 6) Multiply each corresponding
fraction 0s numerator and denominator by that denominator 0s missing factors to get equivalent fractions Our Answers
World View Note: When the Egyptians began working with fractions, they 4 expressed all fractions as a sum of unit fractions. Rather than 5 , they would write 1 1 1 the fraction as the sum, 2 + 4 + 20 . An interesting problem with this system is this is not a unique solution,
4 5
is also equal to the sum
55
1 3
1
1
1
+ 5 + 6 + 10 .
CCBC ~ Math 083
Section 2.3
2.3 Practice - Least Common Denominator Build the rational expression to an equivalent expression with the speci ed denominator. State the value of the missing numerator. 1)
3 8
= 48
?
2)
a 5
3)
a x
= xy
?
4)
5 2x2
5)
2 3a3b2c
= 9a5b2c4
6)
4 3a5b2c4
7)
2 x+4
= x2 ¡ 16
?
8)
x+1 x¡3
9)
x¡4 x+2
= x2 + 5x + 6
?
?
10)
?
= 5a ?
= 8x3y ?
= 9a5b2c4 ?
= x2 ¡ 6x + 9
x¡6 x+3
?
= x2 ¡ 2x ¡ 15
Find the Least Common Denominator. 11)
1 7 3 ; ; 2a3 6a4b2 4a3b5
12)
4 9 ; 5x2 y 25x3 y5z
13)
5 1 2 ; ; x2 ¡ 3x x ¡ 3 x
14)
7 3 1 ; ; 4x ¡ 8 x ¡ 2 4
15)
2 8 ; x+2 x¡4
16) x ; x ¡ 7 ; x + 1
17)
4 6 ; x2 ¡ 25 x + 5
18)
19)
10 x2 + 3x + 2
5
1
20)
; x2 + 5x + 6
1
9x
7x 3 ; x2 ¡ 9 x2 ¡ 6x + 9 2 x2 ¡ 7x + 10
20
3x2
; x2 ¡ 2x ¡ 15 ; x2 + x ¡ 6
Find the LCD and build up each rational expression so they have a common denominator. 21)
2 3a ; 5b2 10a3b
22)
3x 2 ; x¡4 x+2
23)
x+2 x¡3 ; x¡3 x+2
24)
5 2 ¡3 ; ; x2 ¡ 6x x x ¡ 6
25)
x 3x ; x2 ¡ 16 x2 ¡ 8x + 16
26)
5x + 1 4 ; x2 ¡ 3x ¡ 10 x ¡ 5
27)
x+1 2x + 3 ; x2 ¡ 36 x2 + 12x + 36
28)
3x + 1 2x ; x2 ¡ x ¡ 12 x2 + 4x + 3
29)
4x x+2 ; x2 ¡ x ¡ 6 x ¡ 3
30)
3x x¡2 5 ; ; x2 ¡ 6x + 8 x2 + x ¡ 20 x2 + 3x ¡ 10
56
CCBC ~ Math 083
Section 2.3
2.3 Answers - Least Common Denominator 1) 18 2) a2 3) ay 4) 20xy 5) 6a2c3 6) 12 7) 2x ¡ 8
8) x2 ¡ 2x ¡ 3 9) x2 ¡ x ¡ 12
10) x2 ¡ 11x + 30 11) 12a4b5
12) 25x3 y 5z 13) x (x ¡ 3) 14) 4(x ¡ 2)
15) (x + 2)(x ¡ 4)
16) x(x ¡ 7)(x + 1) 17) (x + 5)(x ¡ 5)
18) (x ¡ 3)2(x + 3)
19) (x + 1)(x + 2)(x + 3) 20) (x ¡ 2)(x ¡ 5)(x + 3) 21)
2b 6a4 ; 10a3b2 10a3b2
22)
3x2 + 6x 2x ¡ 8 ; (x ¡ 4)(x + 2) (x ¡ 4)(x + 2)
23)
x2 + 4x + 4 x2 ¡ 6x + 9 ; (x ¡ 3)(x + 2) (x ¡ 3)(x + 2)
24)
5 2x ¡ 12 ¡3x ; ; x(x ¡ 6) x(x ¡ 6) x(x ¡ 6)
25)
x2 ¡ 4x 3x2 + 12x ; (x ¡ 4)2(x + 4) (x ¡ 4)2(x + 4)
26)
5x + 1 4x + 8 ; (x ¡ 5)(x + 2) (x ¡ 5) (x + 2)
27)
x2 + 7x + 6 2x2 ¡ 9x ¡ 18 ; (x ¡ 6)(x + 6)2 (x ¡ 6)(x + 6)2
28)
3x2 + 4x + 1 2x2 ¡ 8x ; (x ¡ 4)(x + 3)(x + 1) (x ¡ 4)(x + 3)(x + 1)
29)
4x x2 + 4x + 4 ; (x ¡ 3)(x + 2) (x ¡ 3)(x + 2)
30)
3x2 + 15x x2 ¡ 4x + 4 5x ¡ 20 ; ; (x ¡ 4)(x ¡ 2)(x + 5) (x ¡ 4)(x ¡ 2)(x + 5) (x ¡ 4)(x ¡ 2)(x + 5)
57
CCBC ~ Math 083
Section 2.4
Section 2.4 - Add and Subtract Rational Expressions Objective: Add and subtract rational expressions with and without common denominators. You will recall that when adding fractions with a common denominator, we add the numerators and keep the denominator. This same process is used with rational expressions. Remember to reduce your sum or dierence, if possible, to obtain your nal answer. Example 1. Add the rational expressions, and simplify if possible. x¡4
x2 ¡ 2x ¡ 8 = =
+
x+8 x2 ¡ 2x ¡ 8
2x + 4 x2 ¡ 2x ¡ 8 2(x + 2) (x + 2)(x ¡ 4)
2 = x¡4
Same denominator; add numerators; combine like terms Factor numerator and denominator Divide out (x + 2) to reduce the fraction to its lowest terms Our Answer
Subtraction with common denominators follows the same pattern. However, subtraction can cause problems if we are not careful to avoid sign errors. Consequently, we will rst distribute the subtraction sign to every term in the numerator of the fraction that follows the subtraction sign. Then we can treat it like an addition problem. Example 2. Subtract the rational expressions, and simplify if possible. 6x ¡ 12 15x ¡ 6 ¡ 3x ¡ 6 3x ¡ 6
Add the opposite of the second fraction
=
6x ¡ 12 ¡15x + 6 + 3x ¡ 6 3x ¡ 6
Add the numerators; combine like terms
=
¡9x ¡ 6 3x ¡ 6
Factor numerator and denominator
=
¡3(3x + 2) 3(x ¡ 2)
Divide out common factor of 3
=
¡(3x + 2) x¡2
Our Answer
(distribute the negative sign to each term in the second fraction)
58
CCBC ~ Math 083
Section 2.4
World View Note: The Rhind papyrus of Egypt from 1650 BC gives some of the earliest known symbols for addition and subtraction: a pair of legs walking in the direction one reads for addition, and a pair of legs walking in the opposite direction for subtraction. When we don't have a common denominator, we will have to nd the least common denominator (LCD) and build up each fraction so that the denominators match. The following example shows this process with fractions. Example 3. Add the fractions, and simplify if possible. 5 1 + 6 4 5 2 1 3 = + 6 2 4 3
The LCD is 12: Build up; multiply 6 by 2 and 4 by 3 to get the common denominator; do the same in each numerator 5 Multiply 6
2 1 3 and 2 4 3
=
10 3 + 12 12
Add the numerators
=
13 12
Our Answer
The same process is used with rational expressions containing variables. Example 4. Add the rational expressions, and simplify if possible. 4b 7a + 2 3a b 6ab4 7a 2b3 4b a = 2 + 3a b 2b3 6ab4 a
The LCD is 6a2b4: We will then build up each fraction: (3a2b)(2b3) = 6a2b4 and (6ab4)(a) = 6a2b4
2b3 a Multiply rst fraction by and second by 3 2b a
=
14ab3 4ab + 6a2b4 6a2b4
Add the numerators; there are no like terms to combine
=
14ab3 + 4ab 6a2b4
Factor the numerator
=
2ab(7b3 + 2) 6a2b4
Reduce; dividing out factors 2; a; and b
=
7b3 + 2 3ab3
Our Answer
The same process can be used for subtraction; we will simply include the rst step of changing the subtraction to addition of the opposite value.
59
CCBC ~ Math 083
Section 2.4
Example 5. Subtract the rational expressions, and simplify if possible.
=
7b 4 ¡ 5a 4a2
Change subtraction to addition of the opposite
4 ¡7b + 5a 4a2
LCD is 20a2: Build up each fraction:
4 4a ¡7b 5 = + 2 5a 4a 4a 5 =
16a ¡35b + 20a2 20a2
(5a)(4a) = 20a2 and (4a2)(5) = 20a2 4a 5 Multiply rst fraction by and second by 4a 5 Add numerators Cannot be simpli ed
=
16a ¡ 35b 20a2
Our Answer
If our denominators have more than one term in them, then we will need to factor rst to nd the LCD. Next, we build up each fraction using the factors that are missing from each denominator. Example 6. Add the rational expressions, and simplify if possible.
=
6 3a + 8a + 4 8
Factor denominators to nd LCD
6 3a + 4(2a + 1) 8
LCD is 8(2a + 1); build up each fraction: 4(2a + 1) 2 = 8(2a + 1) and (8)(2a + 1) = 8(2a + 1)
2 3a 2a + 1 6 = + 8 2a + 1 4(2a + 1) 2
=
12 6a2 + 3a + 8(2a + 1) 8(2a + 1)
2 Multiply rst fraction by 2 ; 2a + 1 second by 2a + 1 Add numerators Cannot be simpli ed
=
6a2 + 3a + 12 8(2a + 1)
Our Answer
60
CCBC ~ Math 083
Section 2.4
Whenever you encounter a subtraction problem, remember to rewrite the problem using addition of the opposite. Example 7. Subtract the rational expressions, and simplify if possible. x+1 x+1 ¡ x ¡ 4 x2 ¡ 7x + 12
Add the opposite: Distribute the negative to
=
x+1 ¡x ¡ 1 + x ¡ 4 x2 ¡ 7x + 12
Factor denominators to nd LCD
=
¡x ¡ 1 x+1 + x ¡ 4 (x ¡ 3)(x ¡ 4)
LCD is (x ¡ 4)(x ¡ 3); build up each fraction
x+1 x¡3 ¡x ¡ 1 = + x¡4 x¡3 (x ¡ 3)(x ¡ 4)
each term in the numerator of the second fraction
Only rst fraction needs to be multiplied by (x ¡ 3)
=
x2 ¡ 2x ¡ 3 ¡x ¡ 1 + (x ¡ 3)(x ¡ 4) (x ¡ 3)(x ¡ 4)
Add the numerators; combine like terms
=
x2 ¡ 3x ¡ 4 (x ¡ 3)(x ¡ 4)
Factor the numerator
=
(x ¡ 4)(x + 1) (x ¡ 3)(x ¡ 4)
Divide out the common factor of (x ¡ 4)
=
x+1 x¡3
Our Answer
61
CCBC ~ Math 083
Section 2.4
2.4 Practice - Add and Subtract Rational Expressions Add or subtract . Simplify your answers whenever possible. 1)
2 a+3
3)
t2 + 4t t¡1
5)
4
2)
2t ¡ 7 t¡1
4)
+ a+3 +
2x2 + 3 x2 ¡ 6x + 5
7)
5 6r
9)
8 9t3
x2 ¡ 5x + 9
¡ x2 ¡ 6x + 5
5
¡ 8r 5
+ 6t2
x2 x¡2
6x ¡ 8 x¡2
¡
a2 + 3a
4
a2 + 5a ¡ 6
6)
3 x
8)
7 xy2
¡ a2 + 5a ¡ 6
4
+ x2 3
+ x2 y
10)
x+5 8
+
x¡3 12
11)
a+2 2
¡
a¡4 4
12)
2a ¡ 1 3a2
+
5a + 1 9a
13)
x¡1 4x
¡
2x + 3 x
14)
2c ¡ d c2 d
¡
c+d cd2
15)
5x + 3y 2x2 y
16)
2 x¡1
+ x+1
17)
2z z ¡1
3z
18)
2 x¡5
+ 4x
19)
8 x2 ¡ 4
3
20)
4x x2 ¡ 25
21)
t t¡3
5
22)
2 x+3
23)
2 5x2 + 5x
24)
3a 4a ¡ 20
25)
t y ¡t
26)
x x¡5
27)
x x2 + 5x + 6
28)
2x x2 ¡ 1
¡ x2 + 5x + 4
29)
x x2 + 15x + 56
30)
2x x2 ¡ 9
+ x2 + x ¡ 6
31)
5x x2 ¡ x ¡ 6
¡ x2 ¡ 9
32)
4x x2 ¡ 2x ¡ 3
¡ x2 ¡ 5x + 6
33)
2x x2 ¡ 1
4
34)
x¡1 x2 + 3x + 2
+ x2 + 4x + 3
35)
x+1 x2 ¡ 2x ¡ 35
36)
3x + 2 3x + 6
+ 4 ¡ x2
37)
4 ¡ a2 a2 ¡ 9
¡ 3¡a
38)
4y y2 ¡ 1
¡ y ¡ y+1
39)
2z 1 ¡ 2z
+ 2z + 1 ¡ 4z 2 ¡ 1
40)
2r r 2 ¡ s2
41)
2x ¡ 3 x2 + 3x + 2
+ x2 + 5x + 6
3x ¡ 1
42)
x+2 x2 ¡ 4x + 3
+ x2 + 4x ¡ 5
43)
2x + 7 x2 ¡ 2x ¡ 3
¡ x2 + 6x + 5
3x ¡ 2
44)
3x ¡ 8 x2 + 6x + 8
+ x2 + 3x + 2
¡
3x + 4y xy2
¡ z+1 ¡ x+2
¡ 4t ¡ 12 4
¡ 3x + 3 y
¡ y+t 2
¡ x2 + 3x + 2 7
¡ x2 + 13x + 42 18
¡ x2 + 2x ¡ 3 x+6
+ x2 + 7x + 10
a ¡2 3z
3
62
2
3
x
+ x+5 4
+ (x + 3)2 9a
+ 6a ¡ 30
+
x¡5 x 3
5
3
x+5
x
2
2
1
1
+ r+s ¡ r¡s 4x + 5
2x ¡ 3
CCBC ~ Math 083
Section 2.4
2.4 Answers - Add and Subtract Rational Expressions
1)
6 a+3
2) x ¡ 4 3) t + 7 4)
a+4 a+6
5)
x+6 x¡5
6)
3x + 4 x2
7)
5 24r
8)
7x + 3y x2 y 2
9)
15t + 16 18t3
17)
¡z 2 + 5z z2 ¡ 1
32)
4x + 1 (x + 1)(x ¡ 2)
33)
2x + 4 x2 + 4x + 3
18)
11x ¡ 15 4x(x ¡ 5)
19)
14 ¡ 3x x2 ¡ 4
34)
20)
x2 ¡ x x2 ¡ 25
2x + 7 x2 + 5x + 6
21)
4t ¡ 5 4(t ¡ 3)
35)
2x ¡ 8 x2 ¡ 5x ¡ 14
22)
2x + 10 (x + 3)2
36)
¡3x2 + 7x + 4 3(x + 2)(2 ¡ x)
23)
6 ¡ 20x 15x(x + 1)
37)
a¡2 a2 ¡ 9
24)
9a 4(a ¡ 5)
38)
2 y2 ¡ y
25)
t2 + 2ty ¡ y 2 y 2 ¡ t2
39)
z ¡3 2z ¡ 1
40)
2 r+s
41)
5(x ¡ 1) (x + 1)(x + 3)
42)
5x + 5 x2 + 2x ¡ 15
10)
5x + 9 24
11)
a+8 4
26)
2x2 ¡ 10x + 25 x(x ¡ 5)
12)
5a2 + 7a ¡ 3 9a2
27)
x¡3 (x + 3)(x + 1)
13)
¡7x ¡ 13 4x
28)
2x + 3 (x ¡ 1)(x + 4)
14)
¡c2 + cd ¡ d2 c2d2
29)
x¡8 (x + 8)(x + 6)
15)
3y 2 ¡ 3xy ¡ 6x2 2x2 y 2
30)
2x ¡ 5 (x ¡ 3)(x ¡ 2)
43)
¡(x ¡ 29) (x ¡ 3)(x + 5)
16)
4x x2 ¡ 1
31)
5x + 12 x2 + 5x + 6
44)
5x ¡ 10 x2 + 5x + 4
63
CCBC ~ Math 083
Section 2.5
Section 2.5 - Proportions Objective: Solve proportions using the cross product and use proportions to solve application problems. When two fractions are equal, they are said to be in proportion. This de nition can be generalized for two equal rational expressions. Solving proportions follows a procedure called the cross product that prescribes how to calculate unknown values. a c Cross Product: If = ; then ad=bc: d b The cross product tells us we can multiply diagonally (cross multiply) to express an equation in a more familiar format one that we can solve more readily. Example 1. Solve the proportion for x. 20 x = 9 6 (20)(9) = 6x 180 = 6x 6 6 30 = x
Calculate the cross product Multiply Divide both sides by 6
Our Solution
World View Note: The rst clear de nition of a proportion and the notation for a proportion came from the German mathematician Leibniz who wrote, I write dy: x = dt: a; for dy is to x as dt is to a, is indeed the same as, dy divided by x is equal to d t divided by a. From this equation follow then all the rules of proportion. If the proportion has more than one term in either the numerator or denominator, we distribute when calculating the cross product. Example 2. Solve the proportion for x. x+3 2 = 4 5 5(x + 3) = (4)(2) 5x + 15 = 8 ¡15 ¡ 15
Calculate the cross product Multiply and distribute Solve Subtract 15 from both sides
64
CCBC ~ Math 083
Section 2.5
5x = ¡7 5 5 x=¡
7 5
Divide both sides by 5
Our Solution
This same idea can be seen when the variable appears in several parts of the proportion. Example 3. Solve the proportion for x. 4 6 = x 3x + 2
Calculate the cross product
4(3x + 2) = 6x
Distribute
12x + 8 = 6x ¡12x ¡12x
Move variables to one side of the equation Subtract 12x from both sides
8 = ¡6x ¡6 ¡6 ¡
4 =x 3
Divide both sides by ¡6
Our Solution
Example 4. Solve the proportion for x. 2x ¡ 3 2 = 7x + 4 5
Calculate the cross product
5(2x ¡ 3) = 2(7x + 4)
Distribute
10x ¡ 15 = 14x + 8 ¡10x ¡10x
Move variables to one side of the equation Subtract 10x from both sides
¡15 = 4x + 8 ¡8 ¡8 ¡23 = 4x 4 4 ¡
23 =x 4
Subtract 8 from both sides
Divide both sides by 4
Our Solution
65
CCBC ~ Math 083
Section 2.5
As we solve proportions, we may end up with a quadratic that we will have to solve. In this section, we will solve the quadratic equations in the same way we solved quadratics previously by factoring. Other methods, such as completing the square or utilizing the quadratic formula, will be discussed in a later chapter. As before, we will generally end up with two solutions. Example 5. Solve the proportion for k. 8 k+3 = k ¡2 3
Calculate the cross product FOIL and multiply
(k + 3)(k ¡ 2) = (8)(3) k 2 + k ¡ 6 = 24 ¡24 ¡ 24
Set the equation equal to zero Subtract 24 from both sides
k 2 + k ¡ 30 = 0
Factor completely Set each factor equal to zero
(k + 6)(k ¡ 5) = 0 k+6 ¡6
= 0 or k ¡ 5 ¡6 +5
Solve each equation Add or subtract
=0 +5
k = ¡6 or k = 5
Our Solutions
Proportions are very useful in that they can be used in many dierent types of applications. We can use them to compare dierent quantities and make conclusions about how quantities are related. As we set up these problems, it is important to remember to stay organized. For example, if we are comparing dogs and cats and the number of dogs is in the numerator of the rst fraction, then the numerator of the second fraction should also refer to the dogs. This consistency of the numerator and denominator is essential in setting up proportions. Example 6. Solve. A six foot tall man casts a shadow that is 3.5 feet long. If the shadow of a ag pole is 8 feet long, how tall is the ag pole? shadow height
We will put shadows in numerator; heights in denominator The man has a shadow of 3.5 feet and a height of 6 feet: Write
66
3.5 6
CCBC ~ Math 083
Section 2.5
The agpole has a shadow of 8 feet; but we don 0t know the height: 8 Write x Set up the proportion 3.5 8 = 6 x 3.5x = (8)(6)
Calculate the cross product Multiply
3.5x = 48 3.5 3.5
Divide both sides by 3.5
x = 13.7ft
Our Solution
Example 7. Solve. In a basketball game, the home team was down by 9 points at the end of the game. They only scored 6 points for every 7 points the visiting team scored. What was the nal score of the game? home visitor
We will put the home team in numerator; visitors in the denominator Visitor 0s score is unknown; and the home team scored 9 points less: x¡9 Write x Home team scored 6 points for every 7 points the visiting team scored: 6 Write 7 Set up the propotion:
x¡9 6 = x 7
Calculate the cross product
7(x ¡ 9) = 6x
Distribute
7x ¡ 63 = 6x ¡7x ¡7x
Move variables to one side Subtract 7x from both sides
67
CCBC ~ Math 083
¡63 = ¡x ¡1 ¡1
Section 2.5
Divide both sides by ¡1
63 = x
We used x for the visiting team 0s score:
63 ¡ 9 = 54
Subtract 9 to get the home team 0s score
63 to 54
Our Solution ¡ The visiting team scored 63 points; whereas the home team scored 54 points
68
CCBC ~ Math 083
Section 2.5
2.5 Practice - Proportions Solve each proportion. 1)
10 a
3)
7 6
5)
6 x
=2
7)
m¡1 5
9)
2 9
=8
6
2)
7 9
=6
=k
2
4)
8 x
=8
8
6)
n ¡ 10 8
=2
8)
8 5
10
10)
9 n+2
12)
9 4
= r¡4
14)
n 8
=
16)
x+1 9
=
x+2 2
18)
n+8 10
=
n¡9 4
4
20)
k+5 k¡6
=5
8
= p¡4
n 4 9
=3 3
= x¡8 3
=9
11)
b ¡ 10 7
13)
x 5
15)
3 10
17)
v ¡5 v+6
=9
19)
7 x¡1
= x¡6
21)
x+5 5
= x¡2
6
22)
4 x¡3
=
23)
m+3 4
= m¡4
11
24)
x¡5 8
= x¡1
25)
2 p+4
=
p+5 3
26)
5 n+1
=
n¡4 10
27)
n+4 3
= n¡2
¡3
28)
1 n+3
=
n+2 2
29)
3 x+4
=
x+2 5
30)
x¡5 4
= x+3
=
b
=4
x+2 9 a
= a+2 4
r
n¡4 3
8
x+5 5 4
¡3
Answer each question. Round your answer to the nearest tenth. Round dollar amounts to the nearest cent. 31) The currency in Western Samoa is the Tala. The exchange rate is approximately S0.70 to 1 Tala. At this rate, how many dollars would you get if you exchanged 13.3 Tala? 32) If you can buy one plantain for S0.49 then how many can you buy with S7.84? 33) Kali reduced the size of a painting to a height of 1.3 in. What is the new width if it was originally 5.2 in. tall and 10 in. wide? 34) A model train has a scale of 1.2 in : 2.9 ft. If the model train is 5 in tall then how tall is the real train?
69
CCBC ~ Math 083
Section 2.5
35) A bird bath that is 5.3 ft tall casts a shadow that is 25.4 ft long. Find the length of the shadow that a 8.2 ft adult elephant casts. 36) Victoria and Georgetown are 36.2 mi from each other. How far apart would the cities be on a map that has a scale of 0.9 in : 10.5 mi? 37) The Vikings led the Timberwolves by 19 points at the half. If the Vikings scored 3 points for every 2 points the Timberwolves scored, what was the half time score? 38) Sarah worked 10 more hours than Josh. If Sarah worked 7 hr for every 2 hr Josh worked, how long did they each work? 39) Computer Services Inc. charges S8 more for a repair than Low Cost Computer Repair. If the ratio of the costs is 3 : 6, what will it cost for the repair at Low Cost Computer Repair? 40) Kelsey's commute is 15 minutes longer than Christina's. If Christina drives 12 minutes for every 17 minutes Kelsey drives, how long is each commute?
70
CCBC ~ Math 083
Section 2.5
2.5 Answers - Proportions 40
15) a = 7
2) n =
14 3
3) k =
12 7
4) x = 16 5) x =
3 2
6) n = 34 7) m = 21 8) x =
79 8
9) p = 49 10) n = 25 11)
29) x = ¡7; 1
6
1) a= 3
40 b=¡ 3
12) r =
36 5
13) x =
5 2
14) n =
32 5
16
30) x = ¡1; 3
16) v = ¡ 7 17) v =
69 5
31) S9.31
18) n =
61 3
32) 16
19) x =
38 3
33) 2.5 in
20) k =
73 3
34) 12.1 ft
21) x = ¡8; 5 22) x = ¡7; 5
23) m = ¡7; 8 24) x = ¡3; 9
25) p = ¡7; ¡2 26) n = ¡6; 9 27) n = ¡1
28) n = ¡4; ¡1
71
35) 39.4 ft 36) 3.1 in 37) T: 38, V: 57 38) J: 4 hr, S: 14 hr 39) S8 40) C: 36 minutes, K: 51 minutes
CCBC ~ Math 083
Section 2.6
Section 2.6 - Solving Rational Equations Objective: Solve rational equations by identifying and multiplying by the least common denominator. When solving equations that are made up of rational expressions, we will solve them using the same strategy we used to solve linear equations with fractions. When we solved problems like the next example, we cleared the fraction by multiplying the entire equation by the least common denominator (LCD). Example 1. Solve the equation. 5 3 2 x¡ = 6 4 3 5(12) 3(12) 2(12) x¡ = 6 4 3
8x ¡10 =9 +10 + 10 8x = 19 8 8 x=
19 8
Multiply each term by the LCD; 12 Reduce fractions; which clears the denominators 2(12) 5(12) 3(12) x = 8x; ¡ = ¡10; =9 3 6 4 Solve for x First; add 10 to both sides Then; divide both sides by 8
Our Solution
We will use the same process to solve rational equations; the only dierence is that our LCD will be more involved. We will also have to be aware of domain issues whenever there is a variable in the denominator. You will recall that whenever the denominator is zero, the fraction is unde ned; so, any solution that would make the denominator's value zero is not a part of the solution set. For this reason, we will always check our solutions in the denominator to make sure that the solution is possible. Example 2. Solve the equation. 5x + 5 x2 + 3x = x+2 x+2 (5x + 5)(x + 2) x2(x + 2) + 3x(x + 2) = x+2 x+2
Multiply each term by the LCD; (x + 2) Reduce fractions
72
CCBC ~ Math 083
Section 2.6
5x + 5 + 3x(x + 2) = x2 5x + 5 + 3x2 + 6x = x2 3x2 + 11x + 5 = x2 ¡x2 ¡x2 2x2 + 11x + 5 = 0 (2x + 1)(x + 5) = 0 2x + 1 = 0 or x + 5 = 0 ¡1 ¡ 1 ¡5 ¡ 5 2x = ¡1 or x = ¡5 2 2 1 x = ¡ or ¡5 2 3 1 +2= x+2= ¡ 2 2 x + 2 = (¡5) + 2 = ¡3
Distribute Combine like terms Make equation equal zero by subtracting x2 from both sides Factor completely Set each factor equal to zero Solve each equation
Check solutions; LCD can 0t be zero 1 Check ¡ in (x + 2); it works 2 Check ¡ 5 in (x + 2); it works Neither make LCD zero; both are solutions
x=¡
1 or ¡5 2
Our Solutions
The LCD can contain several factors in these problems. As the LCD gets more complex, it is important to remember the process we are using to solve is still the same. Example 3. Solve the equation. 1 5 x + = x + 2 x + 1 (x + 1)(x + 2)
Multiply each term by the LCD; (x + 1)(x + 2)
x(x + 1)(x + 2) 1(x + 1)(x + 2) 5(x + 1)(x + 2) + = x+2 x+1 (x + 1)(x + 2)
Reduce fractions Distribute
x(x + 1) + 1(x + 2) = 5
73
CCBC ~ Math 083
Section 2.6
Combine like terms
x2 + x + x + 2 = 5 x2 + 2x + 2 = 5 ¡5 ¡ 5
Make equation equal to zero by subtracting 5 from both sides
x2 + 2x ¡ 3 = 0
Factor Set each factor equal to zero
(x + 3)(x ¡ 1) = 0 x + 3 = 0 or x ¡ 1 = 0 ¡3 ¡ 3 +1 + 1
Solve each equation
x = ¡3 or x = 1
Check solutions; LCD can 0t be zero
(¡3 + 1)(¡3 + 2) = (¡2)(¡1) = 2
Check ¡3 in (x + 1)(x + 2); it works Check 1 in (x + 1)(x + 2); it works
(1 + 1)(1 + 2) = (2)(3) = 6 x = ¡3 or 1
Our Solutions
In the previous example the denominators were factored for us. More often, we will need to factor before nding the LCD. Example 4. Solve the equation. x 1 11 ¡ = x¡1 x¡2 x2 ¡ 3x + 2 (x ¡ 1)(x ¡ 2) x(x ¡ 1)(x ¡ 2) 1(x ¡ 1)(x ¡ 2) 11(x ¡ 1)(x ¡ 2) ¡ = x¡2 (x ¡ 1)(x ¡ 2) x¡1
x(x ¡ 2) ¡ 1(x ¡ 1) = 11 x2 ¡ 2x ¡ x + 1 = 11 x2 ¡ 3x + 1 = 11 ¡11 ¡ 11
74
Factor denominator Identify LCD; (x ¡ 1)(x ¡ 2) Multiply by the LCD; reduce Distribute Combine like terms Make equation equal zero Subtract 11 from both sides
CCBC ~ Math 083
Section 2.6
x2 ¡ 3x ¡ 10 = 0 (x ¡ 5)(x + 2) = 0 x ¡ 5 = 0 or x + 2 = 0 ¡2 ¡ 2 +5 + 5 x = 5 or x = ¡2 (5 ¡ 1)(5 ¡ 2) = (4)(3) = 12 (¡2 ¡ 1)(¡2 ¡ 2) = (¡3)(¡4) = 12 x = 5 or ¡2
Factor completely Set each factor equal to zero Solve each equation
Check answers; LCD can 0t be 0 Check 5 in (x ¡ 1)(x ¡ 2); it works Check ¡2 in (x ¡ 1)(x ¡ 2); it works Our Solutions
World View Note: Maria Agnesi was the rst women to publish a math textbook in 1748. It took her over 10 years to write her book! This textbook covered everything from arithmetic through dierential equations, and was over 1,000 pages in length! If we are subtracting a fraction in the problem, it may be easier to avoid future sign errors by rst distributing the negative throughout the numerator in the subtrahend (rational expresssion immediately following the subtraction sign). Example 5. Solve the equation. x¡2 x+2 5 ¡ = x¡3 x+2 8 x ¡ 2 ¡x ¡ 2 5 + = x¡3 x+2 8
Distribute the negative through numerator Identify LCD; 8(x ¡ 3)(x + 2); multiply each term
(x ¡ 2)8(x ¡ 3)(x + 2) (¡x ¡ 2)8(x ¡ 3)(x + 2) 5 8(x ¡ 3)(x + 2) + = Reduce x+2 8 x¡3 8(x ¡ 2)(x + 2) + 8(¡x ¡ 2)(x ¡ 3) = 5(x ¡ 3)(x + 2)
8(x2 ¡ 4) + 8(¡x2 + x + 6) = 5(x2 ¡ x ¡ 6) 8x2 ¡ 32 ¡ 8x2 + 8x + 48 = 5x2 ¡ 5x ¡ 30 8x + 16 = 5x2 ¡ 5x ¡ 30
75
Multiply (FOIL) Distribute Combine like terms Make equation equal zero
CCBC ~ Math 083
Section 2.6
¡8x ¡ 16
¡8x ¡ 16
0 = 5x2 ¡ 13x ¡ 46 0 = (5x ¡ 23)(x + 2) 5x ¡ 23 = 0 or x + 2 = 0 +23 + 23 ¡2 ¡ 2
Subtract 8x and 16 Factor using ac method Set each factor equal to zero Solve each equation
5x = 23 or x = ¡2 5 5 x=
23 or ¡2 5
23 23 8 33 2112 8 ¡3 +2 =8 = 5 5 5 5 25
8(¡2 ¡ 3)(¡2 + 2) = 8(¡5)(0) = 0 x=
23 5
Check solutions; LCD can 0t be 0 Check
23 in 8(x ¡ 3)(x + 2); it works 5
Check ¡2 in 8(x ¡ 3)(x + 2); can 0t be 0 Our Solution
In the previous example, one of the solutions we found made the LCD zero. When this happens, we ignore this result and only use the solution(s) that do not result with zero in the denominator.
76
CCBC ~ Math 083
Section 2.6
2.6 Practice - Solving Rational Equations Solve the following equations for the given variable: 1
1
4
1) 3x ¡ 2 ¡ x = 0
2) x + 1 = x + 1
20
5x
4)
x2 + 6 x¡1
6
2x
6)
x¡4 x¡1
8)
6x + 5 2x2 ¡ 2x
3) x + x ¡ 4 = x ¡ 4 ¡ 2 5) x + x ¡ 3 = x ¡ 3 7)
2x 3x ¡ 4
9)
3m 2m ¡ 5
4x + 5
3
= 6x ¡ 1 ¡ 3x ¡ 4 7
3
7 3¡x
+ 2 = 4¡x
14)
2 3¡x
¡ 8¡x = 1
3x + 8
16)
x+2 3x ¡ 1
5
18)
x¡1 x¡3
20)
3x ¡ 5 5x ¡ 5
22)
x¡1 x¡2
+ 2x + 1 = 2x2 ¡ 3x ¡ 2
= 3¡x
13)
7 y ¡3
¡ 2 = y¡4
15)
1 x+2
¡ 2 ¡ x = x2 ¡ 4
17)
x+1 x¡1
¡ x+1 = 6
19)
3 2x + 1
21)
x¡2 x+3
¡ x ¡ 2 = x2 + x ¡ 6
23)
3 x+2
+ x + 5 = 6x + 24
25)
x x¡1
¡ x + 1 = x2 ¡ 1
27)
2x x+1
¡ x + 5 = x2 + 6x + 5
29)
x¡5 x¡9
+ x ¡ 3 = x2 ¡ 12x + 27
31)
x¡3 x¡6
+ x + 3 = x2 ¡ 3x ¡ 18
33)
4x + 1 x+3
y¡2
8x2
2x + 1
+ 1 ¡ 2x = 1 ¡ 4x2 ¡ 1 1
3x
12)
12
4¡x 1¡x
x¡1
2
¡ 1 ¡ x2 = x2 ¡ 1
4x 2x ¡ 6
11)
1
12
= 3¡x + 1
10)
¡ 3m + 1 = 2
1
x¡2
+ x ¡ 1 = 2x
1
4
1
¡ 5x ¡ 15 = 2 1
3
6
1
3x ¡ 3
¡ x = 3x2 ¡ x x+2
3
+ x+3 = 4 5x ¡ 1
x¡4
+ 7x ¡ 7 ¡ 1 ¡ x = 2 x+4
1
x¡1
5x + 20
24)
x x+3
¡ x ¡ 2 = x2 + x ¡ 6
2
4x2
26)
2x x+2
+ x ¡ 4 = x2 ¡ 2x ¡ 8
28)
x x+1
¡ x + 3 = x2 + 4x + 3
3
¡8x2
4
¡5x2
2
3x
3
¡2x2
x¡2
x2
x¡2
9x2
x+3
¡4x2
30)
x¡3 x+6
+ x ¡ 3 = x2 + 3x ¡ 18
x+5
¡2x2
32)
x+3 x¡2
+ x + 1 = x2 ¡ x ¡ 2
34)
3x ¡ 1 x+6
+
5x ¡ 3 x¡1
8x2
= x2 + 2x ¡ 3
77
¡
2x ¡ 3 x¡3
¡3x2
= x2 + 3x ¡ 18
CCBC ~ Math 083
Section 2.6
2.6 Answers - Solving Rational Equations 1 2
1) ¡ 2 ; 3
13)
16 ;5 3
2) ¡3; 1
14) 2, 13
4) ¡1; 4
16) 2
3) 3
15) ¡8
8)
9) ¡5
3 2
30) ¡1
2
31)
21) 0, 5
11) ¡5; 0 12) 5; 10
3 10
29) ¡ 3
20) 10 7
27)
9
19)
10) ¡ 15
1 2
18) ¡ 5 ; 1
1 3
1 ¡3
26)
28) 1
17) ¡ 5 ; 5
7) ¡1
2 3
1
5) 2 6)
25)
13 4
32) 1
5
22) ¡2; 3
33) ¡10
23) 4; 7
34)
24) ¡1
78
7 4
CCBC ~ Math 083
Section 2.7
Section 2.7 - Motion and Work Applications Objective: Solve application problems by creating a rational equation to model the problem. Uniform Motion Problems: Solving motion problems where time traveled is the main focus usually involves rational equations. The distance formula, d= rt, is still utilized. However, since the focus is time, d the formula is rewritten as t = . The rational expressions that form the times are set r equal to each other, giving us a proportion. Example 1. Use a rational equation to solve. In the time it takes for a car to travel 120 miles, a train can travel 180 miles. If the train's rate is 20 miles per hour faster than that of the car's rate, what is the average rate for each? distance Train
180
Car
120
rate
time 180 r + 20 r + 20 120 r r
180 120 = r + 20 r
We do not know rate; r; or time; t; traveled by either the train or the car: But we do know the distances traveled; and that they traveled for the same amount of time: Since the times traveled by the train and car are the same; we can set them equal to each other : We have a proportion; cross multiply
180r = (120)(r + 20)
Distribute on the right side
180r = 120r + 2400 ¡120r ¡120r
Add (¡120r) to both sides
60r = 2400
Divide both sides by 60
60r 2400 = 60 60
Solve for r
r = 40 r + 20 = (40) + 20 = 60 car: 40 mph; train: 60 mph
Speed of the car Speed of the train Our Solutions
79
CCBC ~ Math 083
Section 2.7
World View Note: The world's fastest man (at the time of printing) is Jamaican Usain Bolt who set the record of running 100 meters in 9.58 seconds on August 16, 2009 in Berlin, Germany. That is a speed of over 23 miles per hour! Another type of simultaneous product distance problem concerns a boat traveling in a river either with the current or against the current (or an airplane ying with the wind or against the wind). If a boat is traveling downstream, the current will push it, increasing the rate by the speed of the current. If a boat is traveling upstream, the current will pull against it, decreasing the rate by the speed of the current. This is demonstrated in the following example. Example 2. Use a rational equation to solve. A man rows downstream for 30 miles, then turns around. He travels 20 miles upstream in the same amount of time. In still water, his boat averages 15 miles per hour. What is the speed of the water's current? distance downstream
30
upstream
20
rate
time 30 15 + x 15 + x 20 15 ¡ x 15 ¡ x 20 30 = 15 + x 15 ¡ x
(30)(15 ¡ x) = (20)(15 + x)
Let x represent the speed of the water 0s current: We know the time traveled is the same in both directions: We can set the times equal to each other: We have a proportion; cross multiply Simplify by distributing on both sides
450 ¡ 30x = 300 + 20x +30x + 30x
Add (+30x) to both sides
450 = 300 + 50x ¡300 ¡ 300 150 = 50x
Add (¡300) to both sides Divide both sides by 50
150 50x = 50 50 x=3 3 mph
This is the speed of the current Our Solution
80
CCBC ~ Math 083
Section 2.7
Example 3. Use a rational equation to solve. Two planes leave an airport at the same time and traveled at the same speed. The rst plane traveled to a remote island, a distance of 450 miles, in 1.2 hours with a tailwind. The other plane ew to a mountain resort, a distance of 250 miles, in the same amount of time with a head wind. The wind current's speed for both planes was 55 miles per hour. What would be each plane's average speed in still air? distance with a tailwind
450
with a head wind
250
rate
time 450 x + 55 x + 55 250 x ¡ 55 x ¡ 55 450 250 = x + 55 x ¡ 55
450(x ¡ 55) = 250(x + 55) 450x ¡ 24750 = 250x + 13750 ¡250x ¡250x 200x ¡ 24750 = 13750 +24750 +24750 200x = 38500
Let x represent the speed of the planes in still air: We know the travel time is the same for both planes: We can set the times equal to each other: We have a proportion; cross multiply Simplify by distributing on both sides Add (¡250x) to both sides of the equation
Add 24750 to both sides Divide both sides by 200
200x 38500 = 200 200 x = 192.5 192.5 mph
This is the rate of the planes in still air Our Solution
Work Problems: Work problems typically have two individuals working together to complete a job. The rst person does a job in A hours, and a second person does that job in B hours. So, each person is performing part of the work. The time spent working together is denoted by x . x The rst person's portion of the work is A and the second person's portion of the work is x : Working together the job is completed; so, we add the two parts together and set the B equation equal to 1 to signify that the whole job is nished. We use the rational equation below for this type of problem. Work Equation:
81
x x + =1 A B
CCBC ~ Math 083
Section 2.7
The equation above is used to solve the work problems included in this section. Example 4. Use a rational equation to solve. Pat can paint a room in 3 hours. Les can paint the same room in 6 hours. How long will it take them to do the job together? Pat: 3 hours; Les: 6 hours
x x + =1 3 6 x(6) x(6) + = 1(6) 6 3 2x + x = 6 3x = 6
The time working together is unknown; or x Use the teamwork equation to form the rational equation: Multiply each term by LCD; 6 Reduce; clear fractions Combine like terms Divide by 3 on both sides of the equation
3x 6 = 3 3 x=2 The job takes them 2 hours working together
Our solution for x Our Solution
Example 5. Use a rational equation to solve. Adam can clean a room in 3 hours. His sister Maria can clean it in 12 hours. How long will it them to do the job together? Adan: 3 hours; Maria: 12 hours
x x + =1 3 12 x(12) x(12) + = 1(12) 3 12 4x + x = 12
The time working together is unknown; or x Use the teamwork equation to form the equation: Multiply each term by LCD; 12 Reduce; clear fractions Combine like terms
82
CCBC ~ Math 083
Section 2.7
5x = 12
Divide by 5 on both sides of the equation
5x 12 = 5 5 x=
12 2 =2 5 5
2 The job takes them 2 hours 5 working together
Our solution for x Our Solution
83
CCBC ~ Math 083
Section 2.7
2.7 Practice - Motion and Work Applications Use a rational equation to solve. 1) A sailboat travels upstream for 50 miles in the same amount of time that another sailboat travels 80 miles downstream. The rate of still water for both is 65 mph. What is the rate of the current? 2) Mary drove 840 miles to Texas in the same amount of time that Sue drove 770 miles to Louisiana. Mary was traveling 5 miles per hour faster than Sue. How fast was each traveling? 3) Steve runs uphill 3 miles in the time that it takes Mark to run 5 miles downhill. If Steve is traveling 4 miles per hour slower than Mark, how fast is each one running? 4) Alberta went on a kayaking trip. She traveled 2 miles upstream in the same amount of time that she spent kayaking 6 miles downstream. The water's current was 2 miles per hour. What was her average rate of travel in still water? 5) A car travels 240 miles in the same amount of time that a motorcyclist travels 360 miles. The car is traveling an average of 20 miles per hour slower. How fast is each traveling? 6) If A can do a piece of work alone in 6 days and B can do it alone in 4 days, how long will it take the two working together to complete the job? 7) A can do a piece of work in 4 days and B can do it in half the time. How long will it take them to do the work together? 8) A cistern can be lled by one pipe in 20 minutes and by another in 30 minutes. How long will it take both pipes together to ll the tank? 9) If Sam can do a certain job in 3 days, while it takes Fred 6 days to do the same job, how long will it take them, working together, to complete the job? 10) Tim can nish a certain job in 10 hours. It take his wife JoAnn only 8 hours to do the same job. If they work together, how long will it take them to complete the job?
84
CCBC ~ Math 083
Section 2.7
2.7 Answers - Motion and Work Applications '
1) 15 mph current
5) 60 mph
2) Mary, 60 mph; Sue, 55 mph
6) 2.4
3) Steve, 6 mph; Mark, 10 mph
8) 12 min
4) 4 mph
4
10) 4 9 days
1
7) 1 3 days 9) 2 days
85
CCBC ~ Math 083
Section 2.8
Section 2.8 - Variation Objective: Model and solve direct and inverse variation problems. Variation problems are used to show the relationship between quantities. In this section, we will examine how quantities can vary directly and inversely, as well as explore what happens when more than one relationship must be considered within a problem. Each type of variation problem will require that we rst nd the constant of variation, k . Once that constant has been established, the relationship is de ned and speci c problems can be solved. Direct Variation: When quantities vary directly, we say that y varies directly as x or that y varies directly in proportion to x . Direct variation problems are modeled using equations in the form of y = kx. Direct Variation Equations: y=kx We must rst nd the value of the constant of variation; then, we solve for a speci c quantity. Example 1. Solve the variation problem. Given: y varies directly as x , and y = 80 when x = 20. Find y when x = 65. 80 = k(20)
80 k(20) = 20 20 4=k y = 4(65) y = 260
Use y = kx as the variation model: Substitute the given values for x and y ; and solve for k: Divide both sides by 20 The constant of variation; k; is 4: Substitute k = 4 and x = 65 into y = kx ; multiply to nd the value of y Our Solution
An example of an application involving direct variation follows.
86
CCBC ~ Math 083
Section 2.8
Example 2. Solve the variation problem. The dosage of a medication that is prescribed by a doctor varies directly as the weight of the patient. A doctor prescribes 2.5 ml of a medication for a 200-pound patient. How many milliliters (ml) of this medication can be prescribed for someone who weighs 220 pounds? 2.5 = k(200)
The dosage y varies directly as the weight x of the patient: Use y = kx as the variation model: Substitute the given values y = 2.5 ml and x = 200 pounds; and solve for k:
2.5 k(200) = 200 200
Divide both sides by 200
0.0125 = k
The constant of variation; k; is 0.0125
y = 0.0125(220)
Substitute k = 0.0125 and x = 220 into y = kx; multiply to nd the value of y
y = 2.75ml
Our Solution
Inverse Variation: When quantities vary inversely, we say that y varies inversely as x or that y varies inversely in proportion to x . Inverse variation problems are modeled using k equations in the form of y= . x Inverse Variation Equations: y =
k x
Example 3. Solve the variation problem. Given: y varies inversely as x , and y = 95 when x = 3. Find y when x = 15. 95 =
k 3
k 3 (95)(3) = 3 1 k = 285
k as the variation model: Substitute the x given values for x and y; and solve for k: Use y =
Multiply both sides by 3 The constant of variation; k; is 285
87
CCBC ~ Math 083
Section 2.8
y=
285 15
k Substitute k = 285 and x = 15 into y = ; x divide to nd the value of y
y = 19
Our Solution
An example of an application involving inverse variation follows. Example 4. Solve the variation problem. The number of imported toys stocked by ZZ Playthings Company is inversely proportional to the number of toys stocked domestically. If 2000 imported toys are stocked, the number of domestic toys stocked is 50. How many imported toys are stocked if the company has 250 domestic toys in stock? 2000 =
k 50
The number of toys y varies inversely as the number of k as the variation model: x Substitue the given values for x and y; and solve for k: domestic toys x: Use y =
k (2000)(50) = 50
50 1
10000 = k y=
10000 250
y = 40 imported toys
Multiply both sides by 50 The constant of variation; k; is 10000 k Substitute k = 10000 and x = 250 into y = ; x divide to nd the value of y Our Solution
88
CCBC ~ Math 083
Section 2.8
2.8 Practice - Variation Solve the following variation problems. 1) y varies directly as x . When x = 15, y = 9. What is the value of y when x = 6.3? 2) y varies directly as x . When x = 80, y = 65. What is the value of y when x = 72? 3) y varies directly as x . When x = 15, y = 540. What is the value of y when x = 9.75? 4) y varies directly as x . When x = 4.8, y = 25.2. What is the value of y when x = 7.4? 5) y varies inversely as x . When x = 65, y = 9. What is the value of y when x = 50? 6) y varies inversely as x . When x = 12, y = 5. What is the value of y when x = 4? 7) y varies inversely as x . When x = 9.2, y = 2.5. What is the value of y when x = 4.6? 8) y varies inversely as x . When x = 3.2, y = 70. What is the value of y when x = 8? 9) The force, F, needed to stretch a spring varies directly as a certain distance, x , where k is the spring constant measured in Newtons/cm. If a 24 Newton force stretches a certain spring by 20 cm, determing the force needed to stretch the spring 36 cm. 10) Jean wants to invest in property in order to build and sell houses. In the area where she would like to make her purchase, she has found that zoning regulations dictate that the amount of land is directly proportional to the number of houses that can be built on that land. She was able to determine that on 3.42 acres of land, 6 houses can be built. She wishes to build 10 houses. How much land does she need? 11) The time that it takes to ll a swimming pool with water varies directly as the depth of the pool. A swimming pool manufacturer claims that a backyard swimming pool that is 4 feet deep can be lled with water in 3 hours. If this is true, how long would it take to ll a pool that is 8 feet deep? 12) There is a direct relationship between the number of hours spent working on a project and the grade a student receives for that project. Students who spend 2.5 hours on a project earn an average of 75 points on that project. How many points should Greg earn if he spends 3 hours on his project?
89
CCBC ~ Math 083
Section 2.8
13) A company's manager has noticed that as the temperature rises, the number of people attending the monthly stockholder's meeting falls. Last month, 16 people attended the meeting and the temperature was 70 degrees. This month, the temperature was 80 degrees. How many people attended the meeting? 14) The frequency of a vibrating piano string varies inversely to its length. An 18inch piano string vibrates at 336 cycles/second. What is the frequency of a vibrating 21-inch piano string? 15) Entrepreneurs know that product production varies inversely with the price of their product. Alice produces natural chew toys for dogs. She knows that she can sell 250 toys per month when she charges S6 per toy. How many toys should she expect to sell if she charges S4 per toy? 16) The cost per person to rent an airport limousine for one ride is inversely proportional to the number of passengers in the vehicle. When ve people rent the airport limousine, each person pays S50. How much would each person pay if eight people rent the airport limousine?
90
CCBC ~ Math 083
Section 2.8
2.8 Answers - Variation 1) 3.78
7) 5
13) 14 people
2) 58.5
8) 28
3) 351
9) 43.2 Newtons
4) 38.85
10) 5.7 acres
5) 11.7
11) 6 hours
6) 15
12) 90 points
91
14) 288 cycles/second 15) 375 toys 16) S31.25
CCBC ~ Math 083
Chapter 3
Chapter 3: Radical Expressions and Equations Section 3.1 - Square Roots................................................................................94 Section 3.2 - Higher Roots................................................................................99 Section 3.3 - Add and Subtract Radical Expressions......................................104 Section 3.4 - Multiply and Divide Radical Expressions..................................108 Section 3.5 - Rationalize Denominators..........................................................113 Section 3.6 - Rational Exponents ..................................................................117 Section 3.7 - Solving Radical Equations.........................................................123 Section 3.8 - Complex Numbers.....................................................................129
93
CCBC ~ Math 083
Section 3.1
Section 3.1 - Square Roots Objective: Simplify expressions with square roots. To reverse the process of squaring a number, we nd the square root of a number. In other words, a square root un-squares a number. De nition of the Principal Square Root: If a is a nonnegative real number, then the principal square root of a is the nonnegative number b such that b 2=a. p We write b= a . p is called the radical sign and the number a, under the radical The symbol sign, is called the radicand . An expression containing a radical sign is called a radical expression. Square roots are the most common type of radical expressions used. 2 For example, we say the principal p square root of 25 is 5 because 5 = 25. The square root of 25 is written as 25 .
World View Note: When rst used, the radical sign was an R with a line through the tail, similar to our prescription symbol today. The R came from the Latin word, radix, which can be translated as source or foundation. It wasn't until the 1500s that our current symbol was rst used in Germany (but even then it was just a check mark with no bar over the numbers)! The following example gives several square roots: Example 1. Evaluate. p p 1 = 1 because 12 = 1 121 = 11 because 112 = 121 p p 4 = 2 because 22 = 4 625 = 25 because 252 = 625 p p 9 = 3 because 32 = 9 81 = 9 because 92 = 81 p p 16 = 4 because 42 = 16 ¡81 is not a real number p Notice that ¡81 is not a real number because there is no real number whose square is -81. If we square a positive number or a negative number, the result will always be positive. Thus we can only take square roots of positive numbers. In another lesson we will de ne a method we can use to work with and evaluate square roots of negative numbers, but for now we will say they are not real numbers. We call the numbers like 1, 4, 9, 16, 81, 121, and 625 perfect squares because they are squares of integers. Not all numbers are perfect squares. For example, 8 is p not a perfect square so if we used our calculator to nd 8 , the answer would be 2.828427124746190097603377448419... and even this number is a rounded approximation of the square root. To be as accurate as possible, we will never use the calculator to nd decimal approximations of square roots. Instead, we will express roots in simplest radical form. We will do this using a property known as the product rule of radicals: p p p Product Rule of Square Roots: a b = a b
94
CCBC ~ Math 083
Section 3.1
We will use the product rule to evaluate a square root expression by rst nding a factor of the radicand that is a perfect square. For example, we want p to evaulate p p 180 by factoring 180 as 36 5: Then use the product rule to split 180 = p p p 36 5 into two roots, 36 5 . Finally, simplify the rst root so we have 6 5 . This process is shown in the next few examples. Example 2. Simplify. p 75 p = 25 3 p p = 25 3 p =5 3
75 is divisible by the perfect square 25 Split radicand into factors Product rule; take the square root of 25 Our Answer
If there is a coe cient in front of the radical to begin with, the problem becomes a big multiplication problem. Example 3. Simplify. p 5 63 p =5 97 p p =5 9 7 p =53 7 p = 15 7
63 is divisible by the perfect square 9 Split radicand into factors Product rule; take the square root of 9 Multiply coe cients Our Answer
As we simplify radicals using this method, it is important to be sure our nal answer can be simpli ed no more. Example 4. Simplify. p 72 p = 98 p p = 9 8 p =3 8 p =3 42 p p =3 4 2 p =32 2 p =6 2
72 is divisible by the perfect square 9 Split radicand into factors Product rule; take the square root of 9 But 8 is also divisible by the perfect square 4 Split into factors Product rule; take the square root of 4 Multiply Our Answer
The previous example could have been done in fewer steps if we had noticed that 72 = 36 2, but often the time it takes to discover the larger perfect square is more than it would take to simplify the radicand in several steps.
95
CCBC ~ Math 083
Section 3.1
Variables are often part of the radicand as well. To simplify radical expressions involving variables, use the property below: p Simplifying a2 : p For any nonnegative real number a, a2 = a: Note this property only holds if a is nonnegative. For this reason, we will assume that all variables involved in a radical expression are nonnegative. When simplifying with variables, variables with exponents that are divisible by 2 are perfect squares. For example, bypthe power of a power rule of exponents, p (x4)2 = x8. So x8 is a perfect square and x8 = (x4)2 = x4. A shortcut for taking the p square roots of variables is to divide the exponent by 2. In our example, x8 = x4, because we divide the exponent of 8 by 2. When squaring, we multiply the exponent by two, so when taking a square root we divide the exponent by 2. This process is shown in the following example. Example 5. Simplify. p ¡ 5 18x4 y 6z 10 p = ¡5 9 2x4 y 6z 10 p p p p p = ¡5 9 2 x4 y 6 z 10 p = ¡5 3x2 y 3z 5 2 p = ¡15x2 y 3z 5 2
18 is divisible by the perfect square 9 Split radicand into factors Product rule; simplify roots; divide exponents by 2 Multiply coe cients Our Answer
We can't always evenly divide the exponent of a variable by 2. Sometimes we have a remainder. If there is a remainder, this means the variable with an exponent equal to the remainder will remain inside the radical sign. On the outside of the radical, the exponent of the variable will be equal to the whole number part. This process is shown in the following example. Example 6. Simplify. p 20x5 y 9z 6 p = 4 5x5 y 9z 6 p p p p p = 4 5 x5 y 9 z 6 p = 2x2 y 4z 3 5xy
20 is divisible by the perfect square 4 Split radicand into factors Product rule; divide exponents by 2; remainder is left inside Our Answer 5
In the previous example, for the variable x, we divided 2 = 2 R 1, so x2 came out of the radicand and x1=x remained inside the radicand. For the variable y, we 9 divided 2 = 4 R 1, so y 4 came out of the radicand and y 1=y remained 6 inside. For the variable z , we divided 2 = 3 R 0, so z 3 came out of the radicand and no z s remained inside.
96
CCBC ~ Math 083
Section 3.1
3.1 Practice - Square Roots Simplify. Assume that all variables represent positive real numbers. p p 2) ¡100 1) 36 p p 3) ¡ 196 4) 12 p p 5) 125 6) 72 p p 7) 245 8) 3 24 p p 10) 6 128 9) 5 48 p p 12) ¡7 63 11) ¡8 392 p p 14) 343b 13) 192n p p 16) 100n3 15) 169v 2 p p 18) 200a3 17) 252x2 p p 20) ¡4 175p4 19) ¡ 100k 4 p p 21) ¡7 64x4 22) ¡2 128n p p 23) ¡5 36m 24) 8 112p2 p p 25) 45x2 y 2 26) 72a3b4 p p 27) 16x3 y 3 28) 98a4b2 p p 29) 320x4 y 4 30) 512m4n3 p p 31) 6 80xy 2 32) 8 98mn p p 33) 5 245x2 y 3 34) 2 72x2 y 2 p p 35) ¡2 180u3v 36) ¡5 28x3 y 4 p p 37) ¡8 108x4 y 2z 4 38) 6 50a4bc2 p p 39) 2 80hj 4k 40) ¡ 32xy 2z 3 p p 41) ¡4 54mnp2 42) ¡8 56m2 p4 q
97
CCBC ~ Math 083
1) 6 2) not a real number 3) ¡14 p 4) 2 3 p 5) 5 5 p 6) 6 2 p 7) 7 5 p 8) 6 6 p 9) 20 3 p 10) 48 2 p 11) ¡112 2 p 12) ¡21 7 p 13) 8 3n p 14) 7 7b 15) 13v
Section 3.1
3.1 Answers - Square Roots p p 16) 10n n 31) 24y 5x p p 17) 6x 7 32) 56 2mn p 18) 10a 2a p 33) 35xy 5y 19) ¡10k 2 p p 34) 12xy 2 20) ¡20p2 7 p 21) ¡56x2 35) ¡12u 5uv p 22) ¡16 2n p 36) ¡10xy 2 7x p 23) ¡30 m p p 37) ¡48x 2 yz 2 3 24) 32p 7 p p 25) 3xy 5 38) 30a2c 2b p p 26) 6ab 2 2a 39) 8j 2 5hk p 27) 4xy xy p 40) ¡4yz 2xz p 28) 7a2b 2 p p 41) ¡12p 6mn 29) 8x2 y 2 5 p p 42) ¡16mp 2 14q 30) 16m2n 2n
98
CCBC ~ Math 083
Section 3.2
Section 3.2 - Higher Roots Objective: Simplify radicals with an index greater than two. While square roots are the most common type of radical we work with, we can take higher roots of numbers as well: cube roots, fourth roots, fth roots, etc. Following is a de nition of higher roots. De nition of the n th Root: p n
a = b if bn = a
We call b the n th root of a. The small letter n inside the radical is called the index. It tells us which root we are taking, or which power we are un-doing. For square roots, the index is 2. As this is the most common root, the two is not usually written. World View Note: The word for root comes from the French mathematician Franciscus Vieta in the late 16th century. The following example includes several higher roots. Example 1. Evaluate. p p 3 125 = 5 because 53 = 125 3 ¡64 = ¡4 because (¡4)3 = ¡64 p p 4 81 = 3 because 34 = 81 7 ¡128 = ¡2 because (¡2)7 = ¡128 p p 5 4 32 = 2 because 25 = 32 ¡16 is not a real number We must be careful of a few things as we work with higher roots.pFirst, it is important not to forget to check the index on the root. For example 81 = 9 but p 4 81 = 3 because 92 = 81 and 34 = 81. Another thing to watch out for is negative numbers in the radicand. We can take an odd root of a negative number because a negative number raised to an odd power is still negative. However, the even root of a negative number is not a real number. In a later section we will discuss how to work with even roots of negative numbers, but for now we will say they are not real numbers. We can simplify higher roots in much the same way we simpli ed square roots, using the product rule of radicals. Product Rule of Radicals:
p n
p p a n b =n ab
Often we are not as familiar with perfect nth powers as we are with perfect squares. It is important to remember what index we are working with as we express higher roots in simplest radical form.
99
CCBC ~ Math 083
Section 3.2
Example 2. Simplify. p 3 We are working with a cube root; want perfect third powers 54 Test 2: 23 = 8; 54 is not divisible by 8: Test 3: 33 = 27; 54 is divisible by 27! p = 3 27 2 Split radicand into factors p p 3 3 = 27 2 Product rule; take the cube root of 27 p 3 =3 2 Our Answer Just as with square roots, if we have a coe cient, then we multiply the new coef cients together. Example 3. Simplify. p 3 4 48 We are working with a fourth root; want perfect fourth powers Test 2: 24 = 16; 48 is divisible by 16! p Split radicand into factors = 3 4 16 3 p p 4 4 = 3 16 3 Product rule; take the fourth root of 16 p 4 =32 3 Multiply coe cients p 4 =6 3 Our Answer We can also take higher roots of variables. To simplify radical expressions involving variables, use the property below: p Simplifying n an: p For any nonnegative real number a, n an = a: Note this property only holds if a is nonnegative. For this reason, we will assume that all variables involved in a radical expression are nonnegative. As with square roots, when simplifying with variables, we will divide the variable's exponent by the index. The whole number part of the division is how many factors of that varible will come out of the n th root. Any remainder is how many factors of the variable are left behind in the radicand. This process is shown in the following examples. Example 4. Simplify. p 5 25 17 3 Divide each exponent by 5; whole number outside; remainder inside x y z p = x5 y 3 5 y 2z 3 Our Answer 25
In the previous example, for the variable x, we divided 5 = 5R 0, so x5 came out 17 and no xs remained inside. For the y, we divided 5 = 3R 2, so y 3 came out, and y 2 remained inside. For the z, when we divided inside.
100
3 5
= 0R3, all three or z 3 remained
CCBC ~ Math 083
Section 3.2
Example 5. Simplify. 2
p 3
40a4b8
p p p p 3 3 = 2 3 8 3 5 a4 b8 p 3 = 2 2ab2 5ab2 p 3 = 4ab2 5ab2
40 is divisible by the perfect cube 8; split radicand into factors Product rule; divide exponents by 3; remainders left inside Multiply coe cients Our Answer
101
CCBC ~ Math 083
Section 3.2
3.2 Practice - Higher Roots Simplify. p 1) 3 64 p 3) 4 16 p 5) 5 ¡1 p 7) 3 625 p 9) 3 750 p 11) 3 875 p 13) ¡44 96 p 15) ¡4 112 p 4 17) 648a2 p 5 19) 224n3 p 21) 5 224p5 p 23) ¡37 896r p 25) ¡23 ¡48v 7 p 3 27) ¡7 320n6 p 29) 3 ¡135x5 y 3 p 31) 3 ¡32x4 y 4 p 33) 3 256x4 y 6 p 35) 73 ¡81x3 y 7 p 3 37) 2 375u2v 8 p 3 39) ¡3 192ab2 p 41) 63 ¡54m8n3 p7 p 4 43) ¡2 405a5b8c p 45)54 324x7 yz 7
p 3 ¡27 p 4) 4 ¡16 p 6) 8 ¡1 p 8) 3 375 p 10) 3 250 p 12) 3 24 p 14) 34 48 p 16) 54 243 p 4 18) 64n3 p 20) 5 ¡96x3 p 6 22) 256x6 p 7 24) ¡8 384b8 p 3 26) 4 250a6 p 3 28) ¡ 512n6 p 3 30) 64u5v 3 p 3 32) 1000a4b5 p 34) 3 189x3 y 6 p 36) ¡43 56x2 y 8 p 38) 83 ¡750xy p 40) 33 135xy 3 p 42) ¡84 80m4 p7 q 4 p 44) 74 128h6 j 8 k 8 2)
102
CCBC ~ Math 083
Section 3.2
3.2 Answers - Higher Roots 1) 4 2) ¡3 3) 2
4) not a real number 5) ¡1
6) not a real number p 7) 53 5 p 8) 53 3 p 9) 53 6 p 10) 53 2 p 11) 53 7 p 12) 23 3 p 13) ¡84 6 p 14) 64 3 p 15) ¡24 7 p 16) 154 3
p 4 17) 3 8a2 p 4 18) 2 4n3 p 5 19) 2 7n3 p 5 20) ¡2 3x3 p 21) 2p5 7 p 22) 2x6 4 p 23) ¡67 7r p 24) ¡16b7 3b p 25) 4v 2 3 6v p 26) 20a2 3 2 p 27) ¡28n2 3 5 28) ¡8n2
p 3 29) ¡3xy 5x2 p 3 30) 4uv u2 p 31) ¡2xy 3 4xy p 3 32) 10ab ab2
103
p 33) 4xy 2 3 4x p 34) 3xy 2 3 7 p 35) ¡21xy 2 3 3y
p 36) ¡8y 2 3 7x2 y 2 p 3 37) 10v 2 3u2v 2 p 38) ¡403 6xy p 3 39) ¡12 3ab2 p 40) 9y 3 5x
p 41) ¡18m2np2 3 2m2 p p 42) ¡16mpq 4 5p3 p 43) ¡6ab2 4 5ac p 4 44) 14hj 2k 2 8h2 p 45) 15xz 4 4x3 yz 3
CCBC ~ Math 083
Section 3.3
Section 3.3 - Add and Subtract Radical Expressions Objective: Add and subtract like radicals by rst simplifying each radical. Adding and subtracting radical expressions is very similar to adding and subtracting with variables. If two or more radical expressions have the same indices and the same radicands, they are called like radicals. Consider the following example. Example 1. Perform the indicated operation. 5x + 3x ¡ 2x = 6x
Combine like terms Our Answer
p p p 5 11 + 3 11 ¡ 2 11 p = 6 11
Combine like radicals Our Answer
p Notice that when we combined the terms with 11 it was just like combining terms with x. When adding and subtracting like radicals, we add and subtract the coe cients in front of the radical, and the radical stays the same. This process is shown in the following example. Example 2. Perform the indicated operation. p p p p 7 5 6 +4 5 3 ¡9 5 3 +5 6 p p =8 5 6¡5 5 3
p p p p Combine like radicals 7 5 6 + 5 6 and 4 5 3 ¡ 9 5 3 Our Answer
We cannot simplify this expression any more because the radicals are not alike. Often radical expressions do not look like at rst. However, if we simplify the radicals, we may nd we do in fact have like radicals. Example 3. Perform the indicated operation. p p p p 5 45 + 6 18 ¡ 2 98 + 20 p p p p =5 9 5 + 6 9 2 ¡ 2 49 2 + 4 5 p p p p =5 3 5 + 6 3 2 ¡ 2 7 2 + 2 5 p p p p =15 5 + 18 2 ¡ 14 2 + 2 5 p p =17 5 + 4 2
Simplify radicals; nd perfect square factors Take square roots where possible Multiply coe cients Combine like radicals Our Answer
104
CCBC ~ Math 083
Section 3.3
Example 4. Perform the indicated operation. p p p 4 3 54 ¡ 9 3 16 + 5 3 9 p p p = 4 3 27 2 ¡ 9 3 8 2 + 5 3 9 p p p =43 3 2 ¡92 3 2 +5 3 9 p p p = 12 3 2 ¡ 18 3 2 + 5 3 9 p p = ¡6 3 2 + 5 3 9
Simplify each radical; nding perfect cube factors Take cube roots where possible Multiply coe cients p p Combine like radicals 12 3 2 ¡ 18 3 2 Our Answer
World View Note: The Arab writers of the 16th century used the symbol similar to the greater than symbol with a dot underneath for radicals.
105
CCBC ~ Math 083
Section 3.3
3.3 Practice - Add and Subtract Radical Expressions Perform the indicated operation. p p p 1) 2 5 + 2 5 + 2 5 p p p 3) ¡3 2 + 3 5 + 3 5 p p p 5) ¡2 6 ¡ 2 6 ¡ 6 p p p 7) 3 6 + 3 5 + 2 5 p p p 9) 2 2 ¡ 3 18 ¡ 2 p p p 11) ¡3 6 ¡ 12 + 3 3 p p p 13) 3 2 + 2 8 ¡ 3 18 p p p 15) 3 18 ¡ 2 ¡ 3 2 p p p p 17) ¡3 6 ¡ 3 6 ¡ 3 + 3 6 p p p p 19) ¡2 18 ¡ 3 8 ¡ 20 + 2 20 p p p p 21) ¡2 24 ¡ 2 6 + 2 6 + 2 20 p p p p 23) 3 24 ¡ 3 27 + 2 6 + 2 8 p p p 25) ¡23 16 + 23 16 + 23 2 p p p 27) 24 243 ¡ 24 243 ¡ 4 3 p p p 29) 34 2 ¡ 24 2 ¡ 4 243 p p p 31) ¡4 324 + 34 324 ¡ 34 4 p p p p 33) 24 2 + 24 3 + 34 64 ¡ 4 3 p p p p 35) ¡35 6 ¡ 5 64 + 25 192 ¡ 25 64
p p p 2) ¡3 6 ¡ 3 3 ¡ 2 3 p p p 4) ¡2 6 ¡ 3 + 3 6 p p p 6) ¡3 3 + 5 3 + 2 3 p p p 8) ¡ 5 + 2 3 ¡ 2 3 p p p 10) ¡ 54 ¡ 3 6 + 3 27 p p p 12) 4 5 ¡ 5 ¡ 2 48 p p p 14) 2 20 + 2 20 ¡ 3 p p p 16) 3 27 + 2 3 ¡ 12 p p p p 18) ¡2 2 ¡ 2 + 3 8 + 3 6 p p p p 20) ¡3 18 ¡ 8 + 5 8 + 2 8 p p p p 22) ¡3 8 ¡ 5 ¡ 3 6 + 2 18 p p p p 24) 2 6 ¡ 54 ¡ 3 27 ¡ 3 p p p 26) 33 135 ¡ 3 81 ¡ 3 135 p p p 28) ¡34 4 + 34 324 + 24 64 p p p 30) 24 6 + 24 4 + 34 6 p p p 32) ¡24 243 ¡ 4 96 + 24 96 p p p p 34) 24 48 ¡ 34 405 ¡ 34 48 ¡ 4 162
106
CCBC ~ Math 083
Section 3.3
3.3 Answers - Add and Subtract Radical Expressions p p p 1) 6 5 19) ¡12 2 + 2 5 p p p 2) ¡3 6 ¡ 5 3 20) 3 2 p p p p 3) ¡3 2 + 6 5 21) ¡4 6 + 4 5 p p p p 22) ¡ 5 ¡ 3 6 4) 6 ¡ 3 p p p p 5) ¡5 6 23) 8 6 ¡ 9 3 + 4 2 p p p 24) ¡ 6 ¡ 10 3 6) 4 3 p p p 25) 23 2 7) 3 6 + 5 5 p p p 8) ¡ 5 26) 63 5 ¡ 33 3 p p 9) ¡8 2 27) ¡4 3 p p p 10) ¡6 6 + 9 3 28) 104 4 p p p p 11) ¡3 6 + 3 29) 4 2 ¡ 34 3 p p p p 12) 3 5 ¡ 8 3 30) 54 6 + 24 4 p p 13) ¡2 2 31) 34 4 p p p p 14) 8 5 ¡ 3 32) ¡64 3 + 24 6 p p p p 15) 5 2 33) 24 2 + 4 3 + 64 4 p p p p 16) 9 3 34) ¡24 3 ¡ 94 5 ¡ 34 2 p p p p 17) ¡3 6 ¡ 3 35) 5 6 ¡ 65 2 p p 18) 3 2 + 3 6
107
CCBC ~ Math 083
Section 3.4
Section 3.4 - Multiply and Divide Radical Expressions Objective: Multiply and divide radical expressions using the product and quotient rules for radicals. Multiplying radicals is very simple if the index on all the radicals match. The product rule of radicals which we have already been using can be generalized as follows: p p p n n n Product Rule for Radicals: a b c d = a c b d Another way of stating this rule is we are allowed to multiply the factors outside the radical and we are allowed to multiply the factors inside the radicals, as long as the index matches. This is shown in the following example. Example 1. Multiply. p p ¡ 5 14 4 6 p = ¡20 84 p = ¡20 4 21 p = ¡20 2 21 p = ¡40 21
Multiply outside and inside the radical Simplify the radical; divisible by 4 Take the square root where possible Multiply coe cients Our Answer
The same process works with higher roots. Example 2. Multiply. p p 2 3 18 6 3 15 p = 12 3 270 p = 12 3 27 10 p = 12 3 3 10 p = 36 3 10
Multiply outside and inside the radical Simplify the radical; divisible by 27 Take the cube root where possible Multiply coe cients Our Answer
When multiplying with radicals we can still use the distributive property or FOIL just as we could when multiplying polynomials. Example 3. Multiply. p p p 7 6 (3 10 ¡ 5 15 ) p p = 21 60 ¡ 35 90 p p = 21 4 15 ¡ 35 9 10 p p = 21 2 15 ¡ 35 3 10 p p = 42 15 ¡ 105 10
Distribute; following rules for multiplying radicals Simplify each radical; nding perfect square factors Take the square root where possible Multiply coe cients Our Answer
108
CCBC ~ Math 083
Section 3.4
Example 4. Multiply. p p p p ( 5 ¡ 2 3 )(4 10 + 6 6 ) p p p p = 4 50 + 6 30 ¡ 8 30 ¡ 12 18 p p p p = 4 25 2 + 6 30 ¡ 8 30 ¡ 12 9 2 p p p p = 4 5 2 + 6 30 ¡ 8 30 ¡ 12 3 2 p p p p = 20 2 + 6 30 ¡ 8 30 ¡ 36 2 p p = ¡16 2 ¡ 2 30
FOIL; following rules for multiplying radicals Simplify radicals; nding perfect square factors Take the square root where possible Multiply coe cients Combine like radicals Our Answer
Example 5. Multiply. p p p p (2 5 ¡ 3 6 )(7 2 ¡ 8 7 ) p p p p =14 10 ¡ 16 35 ¡ 21 12 ¡ 24 42 p p p p =14 10 ¡ 16 35 ¡ 21 4 3 ¡ 24 42 p p p p =14 10 ¡ 16 35 ¡ 21 2 3 ¡ 24 42 p p p p =14 10 ¡ 16 35 ¡ 42 3 ¡ 24 42
FOIL; following rules for multiplying radicals Simplify radicals; nding perfect square factors Take the square root where possible Multiply coe cients Our Answer
Remember when squaring a binomial we either have to FOIL or use our shortcut to square the rst, twice the product and square the last. The next example uses the shortcut. Example 6. Multiply. p p ( 2 + 3 )2 p p p p = ( 2 + 3 )( 2 + 3 ) p p p p = 4+ 6+ 6+ 9 p =2+2 6 +3 p =5+2 6
Write as a product of two same binomials FOIL; following rules for multiplying radicals Take the square root where possible Combine like terms Our Answer
As we are multiplying we always look at our nal answer to check if all the radicals are simpli ed and all like radicals have been combined. Division with radicals is very similar to multiplication. If we think about division as reducing fractions, we can reduce the coe cients outside the radicals and reduce the values inside the radicals to get our nal solution. r p an b a n b Quotient Rule for Radicals: p = c d cn d
109
CCBC ~ Math 083
Section 3.4
Example 7. Divide. p 15 3 108 p 20 3 2
p 3 15 108 Reduce by dividing by 5 and reduce p by dividing by 2 3 20 2
p 3 3 54 = 4
Simplify radical; 54 is divisible by 27
p 3 3 27 2 = 4
Take the cube root of 27
p 33 3 2 = 4
Multiply coe cients
p 93 2 = 4
Our Answer
Example 8. Divide. p =
50x3 y 7 p 2xy 2
p
25x2 y 5
p = 5xy 2 y
Divide
50 x3 y7 ; ; and 2 2 x y
Simplify radical; 25 is a perfect square; divide exponents by 2 Our Answer
There is one catch to dividing with radicals. It is considered bad practice to have a radical in the denominator of our nal answer. We will see how to handle this situation in the next section. World View Note: Clay tablets have been discovered revealing much about Babylonian mathematics dating p back from 1800 to 1600 BC. In one of the tablets there is an approximation of 2 accurate to ve decimal places (1.41421).
110
CCBC ~ Math 083
Section 3.4
3.4 Practice - Multiply and Divide Radical Expressions Perform the indicated operation. p p 1) 3 5 ¡4 16 p p 3) 12m 15m p p 3 3 5) 4x3 2x4 p p 7) 6 ( 2 + 2) p p 9) ¡5 15 (3 3 + 2) p p 11) 6 10 (5n + 2 ) p p 13) (2 + 2 2 )(¡3 + 2 ) p p 15) ( 5 ¡ 5)(2 5 ¡ 1) p p p p 17) ( 2a + 2 3a )(3 2a + 5a ) p p 19) (¡5 ¡ 4 3 )(¡3 ¡ 4 3 ) 21) 23)
p p 2) ¡5 10 15 p p 4) 5r3 ¡5 10r2 p p 6) 3 (4 ¡ 6 ) p p p 8) 10 ( 5 + 2 ) p p p 10) 7 ( 3 + 5 14 ) p p p 12) 15 ( 5 ¡ 3 3v ) p p 14) (¡2 + 3 )(¡5 + 2 3 ) p p p p 16) (2 3 + 5 )(2 3 ¡ 5 ) p p p p 18) (2 2p + q )(3 2p + q ) p 20) (7 ¡ 3 )2
p
12 p 5 100 p 5 p 4 125
22) 24)
111
p 15 p 2 4 p 12 p 3
CCBC ~ Math 083
Section 3.4
3.4 Answers - Multiply and Divide Radical Expressions p p 1) ¡48 5 13) ¡2 ¡ 4 2 p p 14) 16 ¡ 9 3 2) ¡25 6 p p 15) 15 ¡ 11 5 3) 6m 5 p 16) 7 4) ¡25r2 2r p p p p 17) 6a + a 10 + 6a 6 + 2a 15 5) 2x2 3 x p 18) 12p + 5 2pq + q p p 6) 4 3 ¡ 3 2 p 19) 63 + 32 3 p p p 7) 2 3 + 2 6 20) 52 ¡ 14 3 p p p 8) 5 2 + 2 5 3 21) 25 p p 9) ¡45 5 ¡ 10 15 p 15 p p 22) 4 10) 21 + 35 2 p p 1 11) 30n 10 + 12 5 23) 20 p p 12) 5 3 ¡ 9 5v 24) 2
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CCBC ~ Math 083
Section 3.5
Section 3.5 - Rationalize Denominators Objective: Rationalize the denominators of radical expressions. It is considered bad practice to have a radical in the denominator of a fraction in nal form. If there is a radical in the denominator, we will rationalize it or clear out any radicals in the denominator. In this section, we will focus only on rationalizing denominators with a monomial that is a square root in the denominator. Multiply both the numerator and denominator by the same square root that will produce a perfect square in the ¡p 2 p p denominator. Use the property that ( a )( a ) = a = a. This idea is shown in the following examples. Example 1. Rationalize the denominator. p 6 p 5 p p 6 5 =p p 5 5 p 30 = 5
p Multiply numerator and denominator by 5
Multiply the numerators; p p p multiply 5 5 = 25 = 5 in the denominator Our Answer
Example 2. Rationalize the denominator. 12 p 3x p 12 3x =p p 3x 3x
p Multiply numerator and denominator by 3x
Multiply the numerators; p p p multiply 3x 3x = 9x2 = 3x in the denominator
p 12 3x = 3x
Reduce the fraction
p 4 3x = x
Our Answer
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CCBC ~ Math 083
Section 3.5
Example 3. Rationalize the denominator. p p 2+ 3 p Multiply numerator and denominator by 7 7 p p 2+ 3 7 p = p 7 7
Distribute in the numerator; multiply
p p 2 7 + 21 = 7
Our Answer
¡
p
p p 7 7 = 49 = 7
in the denominator
Example 4. Rationalize the denominator. p p 3 ¡9 p Multiply numerator and denominator by 6 2 6 p p ( 3 ¡ 9) 6 p = p 2 6 6 p p 18 ¡ 9 6 = 26 p p 92 ¡9 6 = 12 p p 3 2¡9 6 = 12 p p 2¡3 6 = 4
Distribute in the numerator; p p p multiply 6 6 = 36 = 6in the denominator Simplify radicals in numerator; multiply out denominator Take square root where possible Reduce by dividing each term by 3 Our Answer
It is important to remember that when reducing the fraction we cannot reduce with just the 3 and 12 or just the 9 and 12. When we have addition or subtraction in the numerator or denominator, we must divide all terms by the same number. As we are rationalizing, it will always be important to constantly check our answer to see if it can be simpli ed more. We ask ourselves, can the fraction be reduced? Can the radicals be simpli ed? These steps may happen several times on our way to the solution. World View Note: During the 5th century BC in India, Aryabhata published a treatise on astronomy. His work included a method for nding the square root of numbers that have many digits.
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CCBC ~ Math 083
Section 3.5
3.5 Practice - Rationalize Denominators Rationalize the denominator. 1
5
1) p
2) p
3)
4)
5)
2 p 10 p 6 p 2 4 p 3 3
6)
8) p
5x2
7) p 4
2p2 3p
9) p 11) 13)
4
5 3xy 4 p 8n2 10) p 10n p 5+4 12) p 4 17 p p 5¡ 2 p 14) 3 6
3x3 y 3
p
3 p 2 p 3 5 p 4 3 p 15
p 2¡5 5 p 4 13 p p 2¡3 3 p 3
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CCBC ~ Math 083
1) 2) 3) 4) 5) 6) 7)
p 2 2 p 5 3 3 p 15 3 p 10 15 p 4 3 9 p 4 5 5 p 5 3xy 12y2
Section 3.5
3.5 Answers - Rationalize Denominators p 4 3x 8) 15xy2 p 6p 9) 3 p 2 5n 10) 5 p p 2 13 ¡ 5 65 11) 52 p p 85 +4 17 12) 68 p 6¡9 13) 3 p p 30 ¡ 2 3 14) 18
116
CCBC ~ Math 083
Section 3.6
Section 3.6 - Rational Exponents Objectives: Convert between radical notation and exponential notation and simplify expressions with rational exponents using the properties of exponents. Multiply and divide radical expressions with dierent indices. We de ne rational exponents as follows: m p De nition of Rational Exponents: a n = (n a )m
The denominator of a rational exponent is the same as the index of our radical while the numerator serves as an exponent. We can use this property to change any radical expression into an exponential expression. Notice the special case of n th roots that follows from the de nition: 1 p p a n = (n a )1 = n a
Example 1. Rewrite with rational exponents. 3 5 p p (5 x )3 = x 5 (6 3x )5 = (3x) 6 3 2 1 1 ¡7 ¡3 = a = (xy) p p 2 7 3 ( xy ) ( a )3
Index is denominator; exponent is numerator Negative exponents from reciprocals
We can also change any rational exponent into a radical expression by using the denominator as the index. Example 2. Rewrite using radical notation. 5 2 p p a 3 = (3 a )5 (2mn) 7 = (7 2mn )2 4 2 1 1 ¡ ¡9 x 5= p (xy) = p 5 9 4 ( xy )2 ( x)
Exponent is numerator; index is denominator Negative exponent means reciprocals
World View Note: Nicole Oresme, a Mathematician born in Normandy, was the rst to use rational exponents. He used the notation However, his notation went largely unnoticed.
1 3
1
9p to represent 9 3 .
The ability to change between exponential expressions and radical expressions allows us to evaluate expressions we had no means of evaluating previously by changing to a radical.
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CCBC ~ Math 083
Section 3.6
Example 3. Use radical notation to rewrite and evaluate. 3
16 2
=
Change to radical format; numerator is exponent; denominator is index
¡ p 3 16
Evaluate radical
= (4)3
Evaluate exponent
= 64
Our Answer
Example 4. Use radical notation to rewrite and evaluate. 27 =
4
¡3
1 4
27 3
1 = p 3 ( 27 )4
Negative exponent is reciprocal Change to radical format; numerator is exponent; denominator is index Evaluate radical
=
1 (3)4
Evaluate exponent
=
1 81
Our Answer
The largest advantage of being able to change a radical expression into an exponential expression is we are now allowed to use all our exponent properties to simplify. The following table reviews all of our exponent properties. Properties of Exponents aman = am+n
(ab)m = ambm
a¡m =
am = am¡n an
m a am = m b b
a¡m
(am)n = amn
a0 = 1
1
1 am
= am
¡m a bm = m b a
When adding and subtracting with fractions we need to be sure to have a common denominator. When multiplying we only need to multiply the numerators together and denominators together. The following examples show several dierent problems, using dierent properties to simplify the rational exponents.
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CCBC ~ Math 083
Section 3.6
Example 5. Simplify. 2
1
1
1
Need common denominator for as (6) and for bs (10)
a3 b2 a6 b5 4 6
5 10
5 6
7 10
1 6
=a b a b =a b
2 10
Add exponents on as and bs Our Answer
Example 6. Simplify.
1
2
x3 y 5 1
3
Multiply each exponent by
4
3
3 4
Our Answer
= x 4 y 10 Example 7. Simplify. 2
x2 y 3
Need common denominator for xs (2) to subtract exponents
7
x 2 y0 4
=
2
x2 y 3 x
7 2
3
=x
Subtract exponents on x; in denominator; y 0 = 1
y0
¡2
2
Negative exponent moves down to denominator
y3
2
=
y3
Our Answer
3
x2 We will use rational exponents to multiply or divide radical expressions with different indices. We will write our answer as a single radical expression. Example 8. Write the expression using a single radical. p p 5 x x 1
1
Need common denominator of 10 to add exponents
= x5 x2 2
5
= x 10 x 10 7
= x 10 =
p x7
10
Rewrite radical expressions using rational exponents
Add exponents Rewrite as a radical expression Our Answer
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CCBC ~ Math 083
Section 3.6
Example 9. Write the expression using a single radical. p 3
y2
p 5
y2
Rewrite radical expressions using rational exponents
2
=
y3 2
Need common denominator of 15 to subtract exponents
y5 10
=
y 15 6
Subtract exponents
y 15 4
= y 15 p = 15 y 4
Rewrite as a radical expression Our Answer
It is important to remember that as we simplify with rational exponents, we are using the exact same properties we used when simplifying integer exponents. The only dierence is we need to follow our rules for fractions as well. It may be worth reviewing your notes on exponent properties to be sure you are comfortable with using the properties.
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CCBC ~ Math 083
Section 3.6
3.6 Practice - Rational Exponents Write each expression in radical form. 3
1) m 5
2) (10r) 3
3) (7x) 2
4) (6b)
3
¡4 4
¡3
Write each expression in exponential form. p 1 6) v 5) p 3 ( 6x ) p 1 7) (p 8) 5a 4 n )7 Evaluate. 2
1
9) 8 3
10) 16 4 3
11) 4 2
12) 100
3
¡2
Simplify. Your answer should contain only positive exponents. 1
3
2
13) x 3 y xy 2 1
14) 4v 3 v ¡1
1
5
15) (a 2 b 2 )¡1 1
16) (x 3 y ¡2)0
3
18) u
17) (x0 y 3 ) 2 x0
21)
a 4 b¡1 b 4 3b¡1 3y
¡
x
5 4
¡
5 5 ¡ 4y 3
1
3 m 2 n¡2 4
(mn 3 )¡1
22)
ab 3 2b 4a
!7 4
24)
(y
¡
1 3 2)2
3
1
x2 y2
(m2n 2 )0
26)
3
27)
¡
4 1 ¡ 3y 3
y0 3
1
(x 4 y ¡1) 3
n4 (x
5 4
1 2 ¡ 2b 3
¡
1
25)
1
xy 2
¡
1
y)¡1
28)
1 x 3 y ¡2
(x 2 y 0)
¡
y 4 x¡2 y
Write the expression using a single radical.
p p 29) x 4 x
30)
121
p 5 2 x p 6 x
4 3
¡
2 3
3
¡2
2x¡2 y 3
20)
1 ¡ y ¡1 2y 3
23)
3
v (u 2 ) 5
7
3
19)
5
¡4 2
CCBC ~ Math 083
Section 3.6
3.6 Answers - Rational Exponents p 1) (5 m )3 1
2) p 4
( 10r )3
p 3) ( 7x )3 4) p 3
5) (6x) 6) v
3 ¡2
1 2
7) n
15)
35
1 v3
23)
1 1 1 a2b2
24)
1
17) y 2 18)
7
¡4 1 2
9) 4
19) 20)
10) 2 11) 8
21)
1 1000 4
5
13) x 3 y 2
22)
m8 7
n6
25)
1 3 5 x2 y4
1 3
n4
v2 7
1
u2
8) (5a)
12)
4
16) 1
1
6b )4
(
14)
26)
3 7 a4b4
y3 1
x4
3
4
27) xy 3
17 2y 6 7
4
x4
x3
28)
1 3y 12
y
2
29)
3 a2 1
30)
2b 4
122
10 3
p 4 x3 p
30
x7
CCBC ~ Math 083
Section 3.7
Section 3.7 - Solving Radical Equations Objective: Solve equations with radicals and check for extraneous solutions. Here we look at equations that have roots in the problem. As you might expect, to clear a root we can raise both sides to an exponent. So to clear a square root, we can raise both sides to the second power. To clear a cube root, we can raise both sides to the third power. There is one catch to solving radical equations. Sometimes we end up with solutions that do not actually work in the original equation. This will only happen if the index on the root is even, and it will not happen all the time for those roots. So, for radical equations solved by raising both sides to an even power, we must check our answers by substituting each result into the original equation. If a value does not work, it is called an extraneous solution, and is not included in the nal solution. Note: When solving a radical equation with an even index, always check your answers! Example 1. Solve the equation. p 7x + 2 = 4 Even index; we will have to check all results p ( 7x + 2 )2 = 42 Square both sides; simplify exponents 7x + 2 = 16 Solve ¡2 ¡2 Subtract 2 from both sides 7x = 14 Divide both sides by 7 7 7 x = 2 Need to check this result in the original equation p
7(2) + 2 = 4 p 14 + 2 = 4 p 16 = 4 4=4 x=2
Multiply Add Square root True; it works Our Solution
Example 2. Solve the equation. p 3 x ¡ 1 = ¡4 Odd index; we don 0t need to check our results p (3 x ¡ 1 )3 = (¡4)3 Cube both sides; simplify exponents x ¡ 1 = ¡64 Solve +1 +1 Add 1 to both sides x = ¡63
Our Solution
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CCBC ~ Math 083
Section 3.7
Example 3. Solve the equation. p 4 3x + 6 = ¡3 p 4 ( 3x + 6 )4 = (¡3)4 3x + 6 = 81 ¡6 ¡6 3x = 75 3 3 x = 25 p 4
3(25) + 6 = ¡3 p 4 75 + 6 = ¡3 p 4 81 = ¡3 3 = ¡3 No Solution
Even index; we will have to check all results Raise both sides to the fourth power Solve Subtract 6 from both sides Divide both sides by 3 Need to check result in the original equation Multiply Add Simplify the radical False; extraneous solution; thus; x = 25 is not a solution Our Solution
If the radical is not alone on one side of the equation, we will have to isolate the radical before we raise it to an exponent. Example 4. Solve the equation. p x + 4x + 1 = 5 ¡x ¡x p 4x + 1 = 5 ¡ x p ( 4x + 1 )2 = (5 ¡ x)2 4x + 1 = 25 ¡ 10x + x2 4x + 1 = x2 ¡ 10x + 25 ¡4x ¡ 1 ¡4x ¡1 2 0 = x ¡ 14x + 24 0 = (x ¡ 12)(x ¡ 2) x ¡ 12 = 0 or x ¡ 2 = 0 +12 + 12 +2 + 2 x = 12 or x = 2 p (12) + 4(12) + 1 = 5 p 12 + 48 + 1 = 5 p 12 + 49 = 5 12 + 7 = 5 19 = 5
Even index; we will have to check all results Isolate radical by subtracting x from both sides Square both sides Evaluate exponents; recall (a ¡ b)2 = a2 ¡ 2ab + b2 Reorder terms Rewrite equation equal to zero Subtract 4x and 1 from both sides Factor Set each factor equal to zero Solve each equation Need to check both results by substituting each into the original equation Check x = 12 rst; multiply inside the root sign Add inside the root sign Take the square root Add False; extraneous solution; thus; x = 12 is not a solution
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CCBC ~ Math 083
Section 3.7
p (2) + 4(2) + 1 = 5 p 2+ 8+1 =5 p 2+ 9 =5 2+3=5 5=5 x=2
Check x = 2 second; multiply inside the root sign Add inside the root sign Take the square root Add True; it works Our Solution
The above example illustrates that as we square both sides of the equation we could end up with an x2 term, which means that the equation is a quadratic. In this case, we must set the equation to zero and solve by factoring. We will have to check both solutions if the root in the problem was even (for example, a square root or a fourth root). Sometimes both values work, sometimes only one value works, and sometimes neither value works. World View Note: The Babylonians were the rst known culture to solve quadratics in radicals as early as 2000 BC! If there is more than one square root in a problem we will clear all of the roots at the same time. This means we must rst make sure that one root is isolated on one side of the equal sign before squaring both sides. Example 5. Solve the equation. p
p 3x ¡ 8 ¡ x = 0 p p + x + x p p 3x ¡ 8 = x p p ( 3x ¡ 8 )2 = ( x )2 3x ¡ 8 = x ¡3x ¡3x ¡8 = ¡2x ¡2 ¡2 4=x p
p 3(4) ¡ 8 ¡ 4 = 0 p p 12 ¡ 8 ¡ 4 = 0 p p 4¡ 4=0 2¡2=0 0=0 x=4
Even index; we will have to check all results p Isolate rst root by adding x to both sides Square both sides Evaluate exponents Solve the equation Subtract 3x from both sides Divide both sides by ¡2 Need to check result in original equation Multiply inside the root sign Subtract inside the root sign Take roots Subtract True; it works Our Solution
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CCBC ~ Math 083
Section 3.7
When the index of the roots is not 2, we need to raise both sides of the equation to the power that corresponds to that index before solving the equation. Example 6. Solve the equation. p p 4 x¡1=4 8 ¡p ¡p 4 x¡1 4= 4 8 4 x¡1=8 +1 +1 x=9 p 4
p (9) ¡ 1 = 4 8 p p 4 8=4 8 x=9
Even index; we will have to check all results Raise both sides to the fourth power Evaluate exponents Add 1 to both sides of the equation Need to check result in original equation Subtract True; it works Our Solution
Example 7. Solve the equation. p 3
p x2 + 5 = 3 x2 ¡ 4x + 1 ¡p ¡p 3 x2 + 5 3 = 3 x2 ¡ 4x + 1 3 x2 + 5 = x2 ¡ 4x + 1 ¡x2 ¡ x2 5 = ¡4x + 1 ¡1 ¡1 4 = ¡4x ¡4 ¡ 4 ¡1 = x x = ¡1
Odd index; we don 0t need to check our results Raise both sides to the third power Subtract x2 on both sides of the equation Subtract 1 from both sides of the equation Divide both sides by ¡ 4
Our Solution
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CCBC ~ Math 083
Section 3.7
3.7 Practice - Solving Radical Equations Solve. p 1) 2x + 3 ¡ 3 = 0 p 3) 6x ¡ 5 ¡ x = 0 p 5) 3 + x = 6x + 13 p 7) 3 ¡ 3x ¡ 1 = 2x p p 9) 4x + 5 ¡ x + 4 = 0 p p 11) 2x ¡ 4 ¡ x + 3 = 0 p 13) 4 x ¡ 3 =2 p 15) 5 6x ¡ 2 =-2
p 5x + 1 ¡ 4 = 0 p p 4) x + 2 ¡ 3x = 0 p 6) x ¡ 1 = 7 ¡ x p p 8) 2x + 2 = 5x ¡ 1 p p 10) 3x + 4 ¡ x + 2 = 0 p 12) 3 3x + 1 =-2 p 14) 4 7x ¡ 5 =-2 p p 16) 3 2x ¡ 1 = 3 7x + 9 2)
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CCBC ~ Math 083
Section 3.7
3.7 Answers - Solving Radical Equations 1) 3
7)
2) 3
8) 1
3) 1, 5
9)
4) 1
10) -1
5) 2
11) 7
6) 3
13) 19
1 4
14) no solution
1 ¡3
15) -5 16) ¡2
12) -3
128
CCBC ~ Math 083
Section 3.8
Section 3.8 - Complex Numbers Objective: Add, subtract, multiply, divide, and simplify expressions using complex numbers. World View Note: When mathematics was rst used, the primary purpose was for counting. Thus zero, negative numbers, fractions or irrational numbers were not originally used. However, the ancient Egyptians quickly developed the need for a part and so they made up a new type of number, the ratio or fraction. The Ancient Greeks did not believe in irrational numbers (people were killed for believing otherwise). The Mayans of Central America later made up the number zero when they found use for it as a placeholder. Ancient Chinese Mathematicians made up negative numbers when they found use for them. In mathematics, when the current number system does not provide the tools to solve the problems the culture is working with, mathematicians create new ways for dealing with the problem. Throughout history this has been the case with the need for a number that is nothing (0), smaller than zero (negatives), between integers (fractions), and between fractions (irrational numbers). This is also the case for the square roots of negative numbers. To work with the square root of negative numbers and to solve equations such as x2 = ¡1; mathematicians have created a new number system called the complex numbers. De nition of the Imaginary Unit i: i =
p
¡1 where i2 = ¡1
With this de nition, the square root of a negative number can now be expressed as a multiple of i. We will use the product rule and simplify the negative as a factor of negative one. This process is shown in the following examples. Example 1. Write in terms of i . p ¡16 p = ¡1 16 = 4i
Consider the negative as a factor of ¡1 Take each root; square root of ¡1 is i Our Answer
Example 2. Write in terms of i . p ¡24 p = ¡1 4 6 p = 2i 6
Find perfect square factors; including ¡1 Square root of ¡1 is i; square root of 4 is 2 Our Answer
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CCBC ~ Math 083
Section 3.8
Then, mathematicians created a new number system called the set of complex numbers. A complex number is one that contains both a real and imaginary part. De nition of Complex Numbers: a + bi where a and b are real numbers We call a the real part and b the imaginary part. Examples of complex numbers p include 2 + 5i; ¡3 + i 5 ; ¡6i because ¡6i = 0 ¡ 6i; and 3 because 3 = 3 + 0i . When performing operations (addition, subtraction, multiplication, division) on complex numbers, we use the same methods as for polynomials. When adding and subtracting complex numbers, we combine like terms by adding or subtracting the real parts, adding or subtracting the imaginary parts, and expressing the answer in the form of a complex number a + bi: Example 3. Add. Combine real parts 2 + 4; imaginary parts 5i ¡ 7i Our Answer
(2 + 5i) + (4 ¡ 7i) = 6 ¡ 2i
It is important to notice what operation we are doing. Students often see the parentheses and think that means FOIL. We only use FOIL to multiply. This problem is an addition problem, so we simply add the real and imaginary parts. For subtraction of complex numbers, the idea is the same, but we need to remember to rst distribute the negative onto all the terms in the parentheses. Example 4. Subtract. (4 ¡ 8i) ¡ (3 ¡ 5i) = 4 ¡ 8i ¡ 3 + 5i = 1 ¡ 3i
Distribute the negative Combine real parts 4 ¡ 3; imaginary parts ¡8i + 5i Our Answer
Addition and subtraction can be combined into one problem. Example 5. Perform the indicated operations. (5i) ¡ (3 + 8i) + (¡4 + 7i) = 5i ¡ 3 ¡ 8i ¡ 4 + 7i = ¡7 + 4i
Distribute the negative Combine real parts ¡ 3 ¡ 4; imaginary parts 5i ¡ 8i + 7i Our Answer
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CCBC ~ Math 083
Section 3.8
Multiplying with complex numbers is the same as multiplying polynomials, but we replace i2 with ¡ 1. Example 6. Multiply. (3i)(7i) = 21i2 = 21(¡1) = ¡21
Multiply Replace i2 with ¡ 1 Multiply Our Answer
When multiplying complex radicals, it is important that we rst rewrite as multiples of i. Example 7. Multiply. p p ¡6 ¡3 p p = (i 6 )(i 3 ) p = i2 18 p = ¡ 18 p =¡ 92 p = ¡3 2
p Simplify each root using i= ¡1 Multiply Replace i2with ¡ 1 Simplify the radical Take square root of 9 Our Answer
Example 8. Multiply. 5i(3i ¡ 7) = 15i2 ¡ 35i = 15(¡1) ¡ 35i = ¡15 ¡ 35i
Distribute Replace i2 with ¡ 1 Multiply Our Answer
Example 9. Multiply. (2 ¡ 4i)(3 + 5i) = 6 + 10i ¡ 12i ¡ 20i2 = 6 + 10i ¡ 12i ¡ 20(¡1) = 6 + 10i ¡ 12i + 20 = 26 ¡ 2i
FOIL Replace i2 with ¡ 1 Multiply Combine real parts 6 + 20; imaginary parts 10i ¡ 12i Our Answer
Remember when squaring a binomial, we write as a product of two same binomials and then FOIL.
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CCBC ~ Math 083
Section 3.8
Example 10. Multiply. (4 ¡ 5i)2 = (4 ¡ 5i)(4 ¡ 5i) = 16 ¡ 20i ¡ 20i + 25i2 = 16 ¡ 20i ¡ 20i + 25(¡1) = 16 ¡ 20i ¡ 20i ¡ 25 = ¡9 ¡ 40i
Write as a product of two same binomials FOIL Replace i2 with ¡ 1 Multiply Combine real parts 16 ¡ 25; imaginary parts ¡20i ¡ 20i Our Answer
Example 11. Multiply. (2 + 3i)(2 ¡ 3i) = 4 ¡ 6i + 6i ¡ 9i2 = 4 ¡ 6i + 6i ¡ 9(¡1) = 4 ¡ 6i + 6i + 9 = 13
FOIL Replace i2with ¡ 1 Multiply Combine real parts 4 + 9; imaginary parts ¡ 6i + 6i Our Answer
Notice how the product of the two complex numbers above resulted in a real number. The complex numbers a + bi and a ¡ bi are called complex conjugates of each other. Notice that (a + bi )(a ¡ bi ) = a2+b2. When we multiply complex conjugates, the result is always a real number. Dividing with complex numbers also has one thing we need to be careful of. If i is p ¡1 , and it is in the denominator of a fraction, then we have a radical in the denominator! This means we will want to rationalize our denominator so there are no i's. This is done by multiplying numerator and denominator by the conjugate of the denominator. Example 12. Divide. 2 ¡ 6i 4 + 8i
Multiply by conjugate of denominator; 4 ¡ 8i
=
(2 ¡ 6i) (4 ¡ 8i) (4 + 8i) (4 ¡ 8i)
FOIL in numerator; denominator is dierence of squares
=
8 ¡ 16i ¡ 24i + 48i2 16 ¡ 64i2
Replace i2 with ¡ 1
=
8 ¡ 16i ¡ 24i + 48(¡1) 16 ¡ 64(¡1)
Multiply
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CCBC ~ Math 083
Section 3.8
=
8 ¡ 16i ¡ 24i ¡ 48 16 + 64
Combine real and imaginary parts
=
¡40 ¡ 40i 80
Reduce; divide each term by 80
=¡
40 40i ¡ 80 80
1 1 =¡ ¡ i 2 2
Reduce and write in the form of a complex number Our Answer
Example 13. Divide. 7 + 3i ¡5i
Multiply by conjugate of denominator; 5i
=
(7 + 3i) 5i ¡5i 5i
Distribute 5i in numerator
=
35i + 15i2 ¡25i2
Replace i2 = ¡1
=
35i + 15(¡1) ¡25(¡1)
Multiply
=
35i ¡ 15 25
Reduce; divide each term by 25
=¡
15 35 + i 25 25
3 7 =¡ + i 5 5
Reduce and write in the form of a complex number Our Answer
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CCBC ~ Math 083
Section 3.8
3.8 Practice - Complex Numbers Write in terms of i . p 1) ¡81
2)
Multiply. p p 3) ¡4 ¡9 p p 5) ¡12 ¡2
p ¡45
p p ¡10 ¡2 p p 6) ¡3 27 4)
Perform the indicated operation, writing the answer in the form of a complex number a + bi . 7) 3 ¡ (¡8 + 4i)
8) (¡8i) ¡ (7i) ¡ (5 ¡ 3i)
11) (¡6i) ¡ (3 + 7i)
12) (5 ¡ 4i) + (8 ¡ 4i)
10) (¡4 ¡ i) + (1 ¡ 5i)
9) (7i) ¡ (3 ¡ 2i)
13) (3 ¡ 3i) + (¡7 ¡ 8i)
14) (i) ¡ (2 + 3i) ¡ 6
15) (3i)(¡8i)
16) (16i)(¡2i)
17) (6i)(¡9i)
18) (¡7i)2
19) (¡5i)(¡10i)
20) (¡7 ¡ 4i)(¡8 + 6i) 22) (8 ¡ 6i)(¡4 + 2i)
21) (6 + 5i)2
24) (¡2 + i)(3 ¡ 5i)
23) (¡4 + 5i)(2 ¡ 7i)
9
26) i
25) (1 + 5i)(2 + i) 5
27) 6i
28)
¡3 + 2i ¡3i
30)
¡4 + 2i 3i
31) ¡i
32)
4i ¡10 + i
33) 7 ¡ 6i
34) 1 ¡ 5i
35) 10 ¡ 7i
36)
37) 3 + 2i
38) 1 + i
29)
¡3 ¡ 6i 4i
10 ¡ i 8
9i
7
5 ¡ 3i
4 4 + 6i
1 + 7i
6¡i
3 + 8i
39) 4 ¡ 3i
40) 2 ¡ 5i
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CCBC ~ Math 083
Section 3.8
3.8 Answers - Complex Numbers 1) 9i p 2) 3i 5
19) ¡50
35)
70 149
3) ¡6 p 4) ¡2 5 p 5) ¡2 6
21) 11 + 60i
36)
4 13
¡ 13 i
22) ¡ 20 + 40i
37)
9 13
¡ 13 i
7) 11 ¡ 4i
25) ¡3 + 11i
6) 9i
8) ¡5 ¡ 12i 9) ¡3 + 9i
10) ¡3 ¡ 6i
11) ¡3 ¡ 13i 12) 13 ¡ 8i
20) 80 ¡ 10i
23) 27 + 38i
24) ¡1 + 13i 26) ¡9i 5
2
28) ¡ 3 ¡ i 3
3
29) ¡ 2 + i 4 4
30) 3 + 3 i
15) 24
32) 101 ¡ 101 i
16) 32
31) 1+10i 4
40
56
48
17) 54
33) 85 + 85 i
18) ¡49
34) ¡ 26 + 26 i
45
19
38) 4 + 3i 39)
27 25
14
+ 25 i
24
13) ¡4 ¡ 11i 14) ¡8 ¡ 2i
6
31
40) 29 + 29 i
27)¡ 6 i
2
49
+ 149 i
9
135
CCBC ~ Math 083
Chapter 4
Chapter 4: Quadratic Equations and Graphs Section 4.1 - Solving Equations with Exponents............................................138 Section 4.2 - Completing the Square..............................................................143 Section 4.3 - Quadratic Formula....................................................................150 Section 4.4 - Parabolas...................................................................................157 Section 4.5 - Quadratic Applications ............................................................165
137
CCBC ~ Math 083
Section 4.1
Section 4.1 - Solving Equations with Exponents Objective: Solve equations with exponents using the odd root property and the even root property. Another type of equation we can solve is one with exponents. As you might expect, we can clear exponents by using roots. This is done with very few unexpected results when the exponent is odd. We solve these problems using the odd root property. p Odd Root Property: If an = b; then a = n b when n is odd: Example 1. Solve the equation. x5 = 32 p p 5 x5 = 5 32 x=2
Use odd root property Simplify radicals Our Solution
Example 2. Solve the equation. x3 = ¡8 p p 3 x3 = 3 ¡8 x = ¡2
Use odd root property Simplify radicals Our Solution
However, when the exponent is even (such as the case of a square root), we will have two results. One result will be positive and one result will be negative. This is because both a2 = b and (¡a)2 = b. So, when solving a2 = b, we will have two p p ¡ solutions: one positive (a = + b ), and one negative a = ¡ b . We write these p two solutions using the shorthand notation a = b . p Even Root Property: If an = b; then a = n b when n is even:
When n = 2, we will often refer to this property as the square root property. Example 3. Solve the equation. x4 = 16 p p 4 x4 = 4 16 x = 2
Use even root property () Simplify radicals Our Solutions
World View Note: In 1545, French Mathematicain Gerolamo Cardano published his book The Great Art, or the Rules of Algebra which included the solution of an equation with a fourth power. However, many people considered the idea of taking a quantity to the fourth power to be absurd because of their belief that there are only three dimensions.
138
CCBC ~ Math 083
Section 4.1
Example 4. Solve the equation. (2x + 4)2 = 36 p (2x + 4)2 = 36 2x + 4 = 6 2x + 4 = 6 or 2x + 4 = ¡6 ¡4 ¡4 ¡4 ¡ 4 2x = 2 or 2x = ¡10 2 2 2 2 x = 1 or x = ¡5 p
Use even root property () Simplify radicals To avoid sign errors; we need two equations One equation for +; one equation for ¡ Subtract 4 from both sides Divide both sides by 2 Our Solutions
In the previous example we needed two equations to simplify, because when we took the root, our solutions were two rational numbers, 6 and ¡6. If the roots do not simplify to rational numbers, we may keep the in the equation. Example 5. Solve the equation. (6x ¡ 9)2 = 45 p (6x ¡ 9)2 = 45 p 6x ¡ 9 = 3 5 +9 +9 p 6x = 9 3 5 6 6 p 93 5 x= 6 p 3 5 x= 2
p
Use even root property () Simplify radicals Use one equation; because radical did not simplify to rational Add 9 to both sides Divide both sides by 6 Simplify; divide each term by 3 Our Solutions
When solving with exponents, it is important to rst isolate the part with the exponent before taking any roots. Example 6. Solve the equation. (x + 4)3 ¡ 6 = 119 +6 +6 (x + 4)3 = 125 p p 3 (x + 4)3 = 3 125 x+4=5 ¡4 ¡ 4 x=1
Isolate part with exponent Add 6 to both sides Use odd root property Simplify radicals Solve Subtract 4 from both sides Our Solution
139
CCBC ~ Math 083
Section 4.1
Example 7. Solve the equation. (6x + 1)2 + 6 = 10 ¡6 ¡ 6 (6x + 1)2 = 4 p p (6x + 1)2 = 4 6x + 1 = 2 6x + 1 = 2 or 6x + 1 = ¡2 ¡1 ¡1 ¡1 ¡ 1 6x = 1 or 6x = ¡3 6 6 6 6 1 1 x = or x = ¡ 6 2
Isolate part with exponent Subtract 6 from both sides Use even root property () Simplify radicals To avoid sign errors; we need two equations Solve each equation Subtract 1 from both sides Divide both sides by 6 Our Solutions
140
CCBC ~ Math 083
Section 4.1
4.1 Practice - Solving Equations with Exponents Solve. 1) x2 = 75
2) x3 = ¡8
3) x2 + 5 = 13
4) 4x3 ¡ 2 = 106 6) (x ¡ 4)2 = 49
5) 3x2 + 1 = 73 7) (x + 2)5 = ¡243
8) (5x + 1)4 = 16
9) (2x + 5)3 ¡ 6 = 21
10) (2x + 1)2 + 3 = 21
141
CCBC ~ Math 083
p 1) 5 3
2) ¡2 p 3) 2 2 4) 3
Section 4.1
4.1 Answers - Solving Equations with Exponents p 9) ¡1 5) 2 6 6) ¡3; 11 7) ¡5 1
10)
3
8) 5 ; ¡ 5
142
p ¡1 3 2 2
CCBC ~ Math 083
Section 4.2
Section 4.2 - Completing the Square Objective: Solve quadratic equations by completing the square. When solving quadratic equations in the past, we have used factoring to solve for our variable. This is exactly what is done in the next example. Example 1. Solve the equation. x2 + 5x + 6 = 0 (x + 3)(x + 2) = 0 x + 3 = 0 or x + 2 = 0 ¡2 ¡ 2 ¡3 ¡ 3 x = ¡3 or x = ¡2
Factor using ac method Set each factor equal to zero Solve each equation Our Solutions
However, the problem with factoring is that not all equations can be solved by factoring. Consider the following equation: x2 ¡ 2x ¡ 7 = 0. The expression on the left side, x2 ¡ 2x p ¡ 7; cannot p be factored; however, there are two solutions to this equation: 1 + 2 2 and 1 ¡ 2 2 . To nd these two solutions, we will use a method known as completing the square. When completing the square, we will change the quadratic into a perfect square that can easily be solved with the square root property. The next example reviews the square root property. Example 2. Solve the equation. (x + 5)2 = 18 p (x + 5)2 = 18 p x + 5 = 3 2 ¡5 ¡5 p x = ¡5 3 2
p
Use square root property Simplify radicals Subtract 5 from both sides Our Solutions
To complete the square, or make our problem into the form of the previous example, we will be searching for the third term of a trinomial. If a quadratic is of the form x2 + b x + c, and a perfect square, the third term, c, can be easily ¡1 found by the formula 2 b 2. This is shown in the following examples, where we nd the number that completes the square, and then factor that perfect square trinomial.
143
CCBC ~ Math 083
Section 4.2
Example 3. Find the value of c that makes this expression a perfect square trinomial; then, factor that perfect square trinomial. 2 1 2 b and our b = 8 x + 8x + c c= 2 The third term 2 that completes the square is 16: 1 8 = (4)2 = 16 2 x2 + 8x + 16 Our expression is a perfect square; factor = (x + 4)(x + 4) = (x + 4)2 Our Answer Example 4. Find the value of c that makes this expression a perfect square trinomial; then, factor that perfect square trinomial. 2 1 2 x ¡ 7x + c c= b and our b = ¡7 2 49 The third term that completes the square is : 4 2 1 7 2 49 ¡7 = ¡ = 2 2 4 49 x2 ¡ 7x + Our expression is a perfect square; factor 4 7 7 = x¡ x¡ 2 2 2 7 = x¡ Our Answer 2 Example 5. Find the value of c that makes this expression a perfect square trinomial; then, factor that perfect square trinomial. 2 1 5 5 2 c= b and our b = x + x+c 2 3 3 25 The third term that completes the square is : 36 2 1 5 2 5 25 = = 6 36 2 3 5 25 x2 + x + Our expression is a perfect square; factor 3 36 5 2 = x+ Our Answer 6 The process in the previous examples, combined with the even root property, is used to solve quadratic equations by completing the square. The following six steps describe the process used to solve a quadratic equation by completing the square, along with a practice example to demonstrate each step.
144
CCBC ~ Math 083
Section 4.2
Steps for Solving by Completing the Square 1. Separate the constant term from the variable terms
3 2 18 x + 3x 3 2
2. If a = / 1, then divide each term by a 3. Find the value that completes the square:
Example 3x2 + 18x ¡ 6 = 0 +6 + 6 2 3x + 18x =6
¡1 2
b
2
6
=3 =2
x + 6x ¡ 1 2 6 = 32 = 9 2 x2 + 6x
=2 4. Add the resulting value to both sides of the equation +9 +9 2 x + 6x + 9 = 11 5. Factor the perfect square trinomial (x + 3)2 = 11 p p (x + 3)2 = 11 p x + 3 = 11 6. Solve by using the even root property ¡3 ¡3 p x = ¡3 11 The advantage of this method is that it can be used to solve any quadratic equation. The following examples show how completing the square can give us rational solutions, irrational solutions, and even complex solutions. Example 6. Solve the equation. 2x2 + 20x + 48 = 0 ¡48 ¡ 48 2 2x + 20x =¡48 2 2 2 x + 10x 2
=¡24
x2 + 10x =¡24 +25 +25 2 x + 10x + 25 = 1 (x + 5)2 = 1 p p (x + 5)2 = 1 x + 5 = 1 ¡5 ¡5 x = ¡5 1 x = ¡4 or ¡6
Separate constant term from variable terms Subtract 48 on both sides of the equation Divide each term by 2
1 Find the value that completes the square: b 2 2 1 Our b = 10; 10 = (5)2 = 25 2 Add 25 to both sides of the equation Factor the perfect square trinomial Solve using the even root property Simplify radicals Subtract 5 from both sides Evaluate both x = ¡5 + 1 and x = ¡5 ¡ 1 Our Solutions
145
2
CCBC ~ Math 083
Section 4.2
Example 7. Solve the equation. x2 ¡ 3x ¡ 2 = 0 +2 + 2 2 x ¡ 3x =2
x2 ¡ 3x +
9 9 =2+ 4 4
9 17 = 4 4 3 2 17 x¡ = 2 4
x2 ¡ 3x +
s
3 x¡ 2
2
=
r
17 4
p 3 17 x¡ = 2 2 3 2
+
x=
3
+
3 2 p
2
Separate the constant term from variable terms Add 2 to both sides No need to divide since a = 1 2 1 Find the value that completes the square: b 2 2 1 3 2 9 = Our b = ¡3; ¡3 = ¡ 2 2 4 9 Add to both sides of the equation 4 Need common denominator (4) on right side of equation 2 4 9 8 9 17 + = + = 1 4 4 4 4 4 Factor the perfect square trinomial Solve using the even root property
Simplify radicals Add
3 to both sides of the equation 2
Notice that we already have a common denominator 17
Our Solutions
As Example 7 has shown, when solving by completing the square, we will often need to use fractions and be comfortable nding common denominators and adding fractions together. Sometimes when solving by completing the square, the solutions are complex numbers, as is the case in Example 8.
146
CCBC ~ Math 083
Section 4.2
Example 8. Solve the equation. x2 ¡ 6x + 30 = 0 ¡30 ¡30 2
x ¡ 6x
=¡30
x2 ¡ 6x
=¡30 +9 +9 x2 ¡ 6x + 9 = ¡21 (x ¡ 3)2 = ¡21 p p (x ¡ 3)2 = ¡21 p x ¡ 3 = i 21 +3 +3 p x = 3 i 21
Separate constant term from variable terms Subtract 30 on both sides of the equation
1 Find the value that completes the square: b 2 2 1 ¡6 = (¡3)2 = 9 Our b = ¡6; 2 Add 9 to both sides of the equation
2
Factor the perfect square trinomial Solve with even root property Simplify radicals Add 3 to both sides
Our Solutions
Once we get comfortable completing the square using the six steps, any quadratic equation can be easily solved.
147
CCBC ~ Math 083
Section 4.2
4.2 Practice - Completing the Square Find the value of c that makes each expression a perfect square trinomial; then, factor that perfect square trinomial. 1) x2 ¡ 30x + c
2) a2 ¡ 24a + c
3) m2 ¡ 36m + c
4) x2 ¡ 34x+c
5) x2 ¡ 15x + c
6) r2 ¡ 9 r + c
1
7) y 2 ¡ y + c
8) p2 ¡ 17p + c
9) x2 ¡ 16x + 55 = 0
10) n2 ¡ 8n ¡ 12 = 0
11) v 2 ¡ 8v + 45 = 0
12) b2 + 2b + 43 = 0
13) x2 + 5x = 7
14) 3k 2 + 2k ¡ 4 = 0
Solve each equation by completing the square.
15) ¡4z 2 + z + 1 = 0
16) 8a2 + 16a ¡ 1 = 0
17) x2 + 10x ¡ 57 = 4
18) p2 ¡ 16p ¡ 52 = 0
19) n2 ¡ 16n + 67 = 4
20) m2 ¡ 8m ¡ 3 = 6
21) 2x2 + 4x + 38 = ¡6
22) 6r2 + 12r ¡ 24 = ¡6
23) 8b2 + 16b ¡ 37 = 5
24) 6n2 ¡ 12n ¡ 14 = 4
25) x2 = ¡10x ¡ 29
26) v 2 = 14v + 36
27) n2 = ¡21 + 10n
28) a2 ¡ 56 = ¡10a
29) 3k 2 + 9 = 6k
30) 5n2 = ¡10n + 15
31) p2 ¡ 8p = ¡55
32) x2 + 8x + 15 = 8
33) 7n2 ¡ n + 7 = 7n + 6n2
34) n2 + 4n = 12
35) 8n2 + 16n = 64
36) b2 + 7b ¡ 33 = 0
37) 5x2 + 8x ¡ 40 = 8
38) m2 = ¡15 + 9m
39) 4b2 ¡ 15b + 56 = 3b2
40) 10v 2 ¡ 15v = 27 + 4v 2 ¡ 6v
148
CCBC ~ Math 083
1) 225; (x ¡ 15)2 2) 144; (a ¡ 12)2
Section 4.2
4.2 Answers - Completing the Square p p 21) ¡1 + i 21 ; ¡1 ¡ i 21 22) 1; ¡3
3) 324; (m ¡ 18)2
3
4) 289; (x ¡ 17)2 5)
225 15 ; (x ¡ 2 )2 4
6)
1 ; (r 324 1
24) 3; ¡1
25) ¡5 + 2i; ¡5 ¡ 2i p p 26) 7 + 85 ; 7 ¡ 85
1
¡ 18 )2
27) 7; 3
1
7) 4 ; (y ¡ 2 )2 8)
7
23) 2 ; ¡ 2
28) 4; ¡14 p p 29) 1 + i 2 ; 1 ¡ i 2
289 17 ; (p ¡ 2 )2 4
9) 11; 5 p p 10) 4 + 2 7 ; 4 ¡ 2 7 p p 11) 4 + i 29 ; 4 ¡ i 29 p p 12) ¡1 + i 42 ; ¡1 ¡ i 42
30) 1; ¡3 p p 31) 4 + i 39 ; 4 ¡ i 39
13)
34) 2; ¡6
14) 15) 16)
32) ¡1:¡7 33) 7; 1
p p ¡5 + 53 ¡5 ¡ 53 ; 2 2 p p ¡1 + 13 ¡1 ¡ 13 ; 3 3 p p 1 + 17 1 ¡ 17 ; 8 8 p p ¡4 + 3 2 ¡4 ¡ 3 2 ; 4 4
35) 2; ¡4 36) 37)
p p 17) ¡5 + 86 ; ¡5 ¡ 86 p p 18) 8 + 2 29 ; 8 ¡ 2 29
38)
p p ¡7 + 181 ¡7 ¡ 181 ; 2 2
12 ; ¡4 5 p p 9 + 21 9 ¡ 21 ; 2 2
39) 8; 7 3
19) 9; 7
40) 3; ¡ 2
20) 9; ¡1
149
CCBC ~ Math 083
Section 4.3
Section 4.3 - Quadratic Formula Objective: Solve quadratic equations by using the quadratic formula. The general form of a quadratic is a x2 + bx + c = 0, where a, b, and c are real numbers and a= / 0. We will now develop a new formula to solve this equation for x by completing the square of the general quadratic. Example 1. Solve for x by completing the square. ax2 + bc + c = 0 ¡c ¡ c 2 ax + bx =¡c a a a b ¡c x2 + x = a a
b b2 b2 c x2 + x + 2 = 2 ¡ a 4a 4a a
b b2 b2 ¡ 4ac x2 + x + 2 = a 4a 4a2 b 2 b2 ¡ 4ac x+ = 2a 4a2 s
b x+ 2a
2
r
=
b2 ¡ 4ac 4a2
p b b2 ¡ 4ac x+ = 2a 2a p ¡b b2 ¡ 4ac x= 2a
Separate constant term from variable terms Subtract c from both sides Divide each term by a Find the value that completes the square: 2 1 b 2 b b2 = = 2 2 a 2a 4a Add that value to both sides of the equation Get common denominator (4a2) on the right side of the equation: b2 c 4a b2 4ac b2 ¡ 4ac ¡ = ¡ = 4a2 a 4a 4a2 4a2 4a2 Factor the perfect square trinomial Solve using the even root property
Simplify radicals
Subtract
b from both sides 2a
Our new formula
150
CCBC ~ Math 083
Section 4.3
This result is a very important one to us. Just as we solved a quadratic equation by completing the square, we can use this formula to solve any quadratic equation. Once we identify what the values of a; p b; and c are in the quadratic equation, ¡b
b2 ¡ 4ac
we can substitute those values into x = and we will get our two solu2a tions. This formula is known as the quadratic formula.
Quadratic Formula: If ax2 + bx + c = 0; then x =
¡b
p
b2 ¡ 4ac : 2a
World View Note: Indian mathematician Brahmagupta gave the rst explicit formula for solving quadratics in 628 AD. However, at that time mathematics was not done with variables and symbols, so the formula he gave was written in words.
We can use the quadratic formula to solve any quadratic equation. This is demonstrated in the following examples. Example 2. Solve the equation using the quadratic formula. x2 + 3x + 2 = 0
x=
¡(3)
p
(3)2 ¡ 4(1)(2) 2(1)
x=
¡3
p 2
9¡8
p ¡3 1 x= 2 ¡3 1 2 ¡3 + 1 ¡3 ¡ 1 x= or x = 2 2 ¡2 ¡4 x= or x = 2 2 x=
x = ¡1 or ¡2
a = 1; b = 3; c = 2; use quadratic formula Evaluate the exponent and multiply
Evaluate the subtraction under radical sign Evaluate the root Evaluate +and ¡ to get the two answers Simplify the fractions; if possible Our Solutions
As we are solving a quadratic equation by using the quadratic formula, it is important to remember that the equation must rst be set equal to zero.
151
CCBC ~ Math 083
Section 4.3
Example 3. Solve the equation using the quadratic formula. 25x2 = 30x + 11 ¡30x ¡ 11 ¡30x ¡ 11 25x2 ¡ 30x ¡ 11 = 0
x=
¡(¡30)
p
(¡30)2 ¡ 4(25)(¡11) 2(25)
p 30 900 + 1100 x= 50 p 30 2000 x= 50 p 30 20 5 x= 50 p 32 5 x= 5
First set the equation equal to zero Subtract 30x and 11 from both sides of the equation a = 25; b = ¡30; and c = ¡11; use quadratic formula Evaluate the exponent and multiply
Evaluate the addition under radical sign Simplify the root Reduce fraction by dividing each term by 10 Our Solutions
Example 4. Solve the equation using the quadratic formula. 3x2 + 4x + 8 = 2x2 + 6x ¡ 5 ¡2x2 ¡ 6x + 5 ¡ 2x2 ¡ 6x + 5 x2 ¡ 2x + 13 = 0 x=
¡(¡2)
p
First set the equation equal to zero Subtract 2x2 and 6x and add 5 a = 1; b = ¡2; c = 13; use quadratic formula
(¡2)2 ¡ 4(1)(13) 2(1)
Evaluate the exponent and multiply
p 2 4 ¡ 52 x= 2
Evaluate the subtraction under radical sign
x=
2
p 2
¡48
Simplify the root
p 2 4i 3 x= 2
Reduce fraction by dividing each term by 2
p x = 1 2i 3
Our Solutions
152
CCBC ~ Math 083
Section 4.3
When we use the quadratic formula, we don't necessarily get two unique answers. We can end up with only one solution if the square root simpli es to zero. Example 5. Solve the equation using the quadratic formula. 4x2 ¡ 12x + 9 = 0 x=
¡(¡12)
p
(¡12)2 ¡ 4(4)(9) 2(4)
p 12 144 ¡ 144 x= 8 p 12 0 x= 8 x=
12 0 8
x=
12 8
x=
3 2
a = 4; b = ¡12; c = 9; use quadratic formula Evaluate the exponent and multiply
Evaluate the subtraction under radical sign Simplify the root Evaluate +and ¡ to get the two answers;
They are identical values; so; only one instance needs to be considered Reduce fraction Our Solution
If a term is missing from the quadratic, we can still solve with the quadratic formula; we simply use zero for that term. The order is important. So, if the term with x is missing, we have b = 0; if the constant term is missing, we have c = 0. Example 6. Solve the equation using the quadratic formula. 3x2 + 7 = 0
x=
¡(0)
p
(0)2 ¡ 4(3)(7) 2(3) p ¡84 x= 6 p 2i 21 x= 6 p i 21 x= 3
a = 3; b = 0 (missing term); c = 7; use quadratic formula Evaluate the exponent and multiply
Simplify the root Reduce the fraction; divide 2i and 6 by 2 Our Solutions
153
CCBC ~ Math 083
Section 4.3
We have covered three dierent methods that can be used to solve a quadratic equation: factoring, completing the square, and using the quadratic formula. It is important to be familiar with all three as each has its own advantage to solving quadratic equations. The following table walks you through a suggested process and an example of each method to decide which would be best for you to use when solving a quadratic equation. 1. If it can easily factor, solve by factoring:
2. If a = 1 and b is even, complete the square:
3. Otherwise, solve by the quadratic formula:
x2 ¡ 5x + 6 = 0 (x ¡ 2)(x ¡ 3) = 0 x = 2 or x = 3 x2 + 2x = 4 ¡ 1 2 2 = 12 = 1 2 x2 + 2x + 1 = 5 (x + 1)2 = 5 p x+1= 5 p x = ¡1 5 x2 ¡ 3x + 4 = 0 x= x=
p
(¡3)2 ¡ 4(1)(4) 2(1) p 3i 7 2 3
The above table is merely a suggestion for deciding how to solve a quadratic equation. Remember that either method, completing the square or using the quadratic formula, will always work to solve any quadratic equation. Factoring only works if the expression can be factored.
154
CCBC ~ Math 083
Section 4.3
4.3 Practice - Quadratic Formula Solve each equation using the quadratic formula. 1) 4a2 + 6 = 0
2) 3k 2 + 2 = 0
3) 2x2 ¡ 8x ¡ 2 = 0
4) 6n2 ¡ 1 = 0
5) 9m2 ¡ 16 = 0
6) 5p2 + 2p + 6 = 0
7) 3r2 ¡ 2r ¡ 1 = 0
8) 2x2 ¡ 2x ¡ 15 = 0
9) 4n2 ¡ 36 = 0
10) 3b2 + 6 = 0
13) 2a2 + 3a + 14 = 6
14) 6n2 ¡ 3n + 3 = ¡4
15) 3k 2 + 3k ¡ 4 = 7
16) 4x2 ¡ 14 = ¡2
19) 2p2 + 6p ¡ 16 = 4
20) m2 + 4m ¡ 48 = ¡3
12) 2x2 + 4x + 12 = 8
11) v 2 ¡ 4v ¡ 5 = ¡8
18) 4n2 + 5n = 7
17) 7x2 + 3x ¡ 16 = ¡2
22) 3b2 ¡ 3 = 8b
21) 3n2 + 3n = ¡3
24) 3r2 + 4 = ¡6r
23) 2x2 = ¡7x + 49 25) 5x2 = 7x + 7
26) 6a2 = ¡5a + 13
27) 8n2 = ¡3n ¡ 8
28) 6v 2 = 4 + 6v 30) x2 = 8
29) 2x2 + 5x = ¡3 31) 4a2 ¡ 64 = 0
32) 2k 2 + 6k ¡ 16 = 2k
35) ¡5n2 ¡ 3n ¡ 52 = 2 ¡ 7n2
36) 7m2 ¡ 6m + 6 = ¡m
39) 2n2 ¡ 9 = 4
40) 6b2 = b2 + 7 ¡ b
33) 4p2 + 5p ¡ 36 = 3p2
34) 12x2 + x + 7 = 5x2 + 5x
37) 7r2 ¡ 12 = ¡3r
38) 3x2 ¡ 3 = x2
155
CCBC ~ Math 083
Section 4.3
4.3 Answers - Quadratic Formula 1) 2)
p p i 6 i 6 ; ¡ 2 2 p p i 6 i 6 ;¡ 3 3
21)
1
22) 3; ¡ 3
p p 3) 2 + 5 ; 2 ¡ 5 4) 5) 6)
7
23) 2 ; ¡7 24)
4 4 ; ¡3 3 p p 6 6 ; ¡ 2 2 p p ¡1 + i 29 ¡1 ¡ i 29 ; 5 5
25) 26)
7) 1; ¡ 3
27)
1+
28)
1
8)
p p 31 1 ¡ 31 ; 2 2
3
31) 4; ¡4
11) 3; 1
32) 2; ¡4
12) ¡1 + i; ¡1 ¡ i
14) 15) 16) 17) 18)
p p ¡3 + i 3 ¡3 ¡ i 3 ; 3 3 p p 7 + 3 21 7 ¡ 3 21 ; 10 10 p p ¡5 + 337 ¡5 ¡ 337 ; 12 12 p p ¡3 + i 247 ¡3 ¡ i 247 ; 16 16 p p 3 + 33 3 ¡ 33 ; 6 6
29) ¡1; ¡ 2 p p 30) 2 2 ; ¡2 2
9) 3; ¡3 p p 10) i 2 ; ¡i 2
13)
p p ¡1 + i 3 ¡1 ¡ i 3 ; 2 2
p p ¡3 + i 55 ¡3 ¡ i 55 ; 4 4 p p 3 + i 159 3 ¡ i 159 ; 12 12 p p ¡3 + 141 ¡3 ¡ 141 ; 6 6
33) 4; ¡9 34)
p p 2 + 3i 5 2 ¡ 3i 5 ; 7 7 9
35) 6; ¡ 2 36)
p p 3; ¡ 3
37)
p p ¡3 + 401 ¡3 ¡ 401 ; 14 14 p p ¡5 + 137 ¡5 ¡ 137 ; 8 8
38) 39)
19) 2; ¡5
40)
20) 5; ¡9
156
p p 5 + i 143 5 ¡ i 143 ; 14 14 p p ¡3 + 345 ¡3 ¡ 345 ; 14 14 p p 6 6 ;¡ 2 2 p p 26 26 ;¡ 2 2 p p ¡1 + 141 ¡1 ¡ 141 ; 10 10
CCBC ~ Math 083
Section 4.4
Section 4.4 - Parabolas Objective: Graph parabolas using the vertex, x -intercepts, and y-intercept. Just as we drew pictures of the solutions for lines or linear equations, we can draw a picture of the solutions to quadratics as well. One way we can start to do that is to make a table of values. Example 1. Graph the parabola. y = x2 ¡ 4x + 3 x y 0 1 2 3 4 2 y = (0) + 4(0) + 3 = 0 ¡ 0 + 3 = 3 y = (1)2 ¡ 4(1) + 3 = 1 ¡ 4 + 3 = 0 y = (2)2 ¡ 4(2) + 3 = 4 ¡ 8 + 3 = ¡1 y = (3)2 ¡ 4(3) + 3 = 9 ¡ 12 + 3 = 0 y = (4)2 ¡ 4(4) + 3 = 16 ¡ 16 + 3 = 3 x y 0 3 1 0 2 ¡1 3 0 4 3
Make a table of values
We will test 5 values to get an idea of the shape
Plug 0 in for x and evaluate Plug 1 in for x and evaluate Plug 2 in for x and evaluate Plug 3 in for x and evaluate Plug 4 in for x and evaluate
Our completed table:
Plot the points (0; 3); (1; 0); (2; ¡1); (3; 0); and (4; 3). Connect the dots with a smooth curve. Our Graph
When we have x2 in our equations, the graph will no longer be a straight line. Quadratics have a graph that looks like a U shape that is called a parabola. World View Note: The rst major female mathematician was Hypatia of Egypt who was born around 370 AD. She studied conic sections. The parabola is one type of conic section.
157
CCBC ~ Math 083
Section 4.4
The above method to graph a parabola works for any equation; however, it can be very tedious to nd all the correct points to get the correct bend and shape. For this reason, we identify several key points on a graph and in the equation to help us graph parabolas more e ciently. These key points are described below. Point A: y-intercept: Where the graph crosses the vertical y-axis.
B
Points B and C: x -intercepts: Where the graph crosses the horizontal x -axis
C
Point D: Vertex: The point where the graph curves and changes directions.
A D
We will use the following method to nd each of the points on our parabola. To graph the parabola y = ax2 + bx + c; nd the following points: 1. y-intercept: Found by making x = 0, this simpli es down to y = c 2. x -intercepts: Found by making y = 0, this means solving 0 = ax2 + bx + c ¡b
3. Vertex: Let x = 2a to nd x (the x ¡ coordinate of the vertex). Then plug this value into the equation to nd its corresponding y value; which is the y ¡ coordinate of the vertex. 4. If a is a positive number, then the vertex will be the minimum point of the parabola and the graph will open upward (U-shaped); If a is a negative number, then the vertex will be the maximum point of the parabola and the graph will open downward (upside down U-shaped). After nding these points we can connect the dots with a smooth curve to nd our graph! Example 2. Graph the parabola. y = x2 + 4x + 3 y=3 0 = x2 + 4x + 3 0 = (x + 3)(x + 1) x + 3 = 0 or x + 1 = 0 ¡3 ¡ 3 ¡1 ¡ 1 x = ¡3 or x = ¡1
Find the key points y = c is the y ¡ intercept; point (0; 3) To nd the x ¡ intercepts; we solve the equation Factor completely Set each factor equal to zero Solve each equation Our x ¡ intercepts; points (¡3; 0) and (¡1; 0)
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¡4 ¡4 = = ¡2 2(1) 2 y = (¡2)2 + 4(¡2) + 3 x=
y=4¡8+3 y = ¡1 (¡2; ¡1)
¡b 2a Plug this value into the equation to nd the y ¡ coordinate Evaluate y ¡ value of vertex Vertex as a point
To nd the vertex; rst use x =
Graph points (0,3), (-3,0), and (-1,0), as well as the vertex at (-2,-1). Connect the dots with a smooth curve in a U shape to get our parabola. Our Graph
If the a in y = ax2 + bx + c is a negative value, the parabola will end up being an upside-down U. The process to graph it is identical, we just need to be very careful of how our signs operate. Remember, if a is negative, then ax2 will also be negative because we only square the x, not the a. Example 3. Graph the parabola. y = ¡3x2 + 12x ¡ 9 y = ¡9
Find the key points y ¡ intercept is y = c; point (0; ¡9)
0 = ¡3x2 + 12x ¡ 9 0 = ¡3(x2 ¡ 4x + 3) 0 = ¡3(x ¡ 3)(x ¡ 1) x ¡ 3 = 0 or x ¡ 1 = 0 +3 + 3 +1 + 1 x = 3 or x=1
To nd x ¡ intercepts; solve this equation Factor out GCF rst; then factor rest Set each factor with a variable equal to zero Solve each equation
¡12 ¡12 = =2 2(¡3) ¡6 y = ¡3(2)2 + 12(2) ¡ 9
¡b 2a Plug this value into the equation to nd the y ¡ coordinate Evaluate
x=
y = ¡3(4) + 24 ¡ 9 y = ¡12 + 24 ¡ 9 y=3 (2; 3)
Our x ¡ intercepts; points (3; 0) and (1; 0) To nd the vertex; rst use x =
y ¡ value of vertex Vertex as a point
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Section 4.4
Graph the points (0,-9), (3,0), and (1,0), as well as the vertex at (2,3). Connect the dots with a smooth curve in an upside-down U shape to get our parabola. Our Graph
It is important to remember the graph of all quadratics is a parabola with the same U shape (they could be upside-down). If you plot your points and we cannot connect them in the correct U shape, then at least one of your points must be wrong. Go back and check your work to be sure they are correct! Just as all quadratics (equation with y = x2) have the same U-shape and all linear equations (equations such as y = x) have the same line shape when graphed, different functions have dierent shapes to them. Below are some common functions (some we have yet to cover!) with their graph shape drawn. We will discuss the concept of functions in more detail in another lesson. Absolute Value y = jxj
Square Root p y= x
Quadratic
Cubic
y = x2
y = x3
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Section 4.4
Exponential
Logarithmic
y = ax
y = logax
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Section 4.4
4.4 Practice - Parabolas Find the vertex and intercepts. Use this information to graph each parabola. 1) y = x2 ¡ 2x ¡ 8
2) y = x2 ¡ 2x ¡ 3
5) y = ¡2x2 + 12x ¡ 18
6) y = ¡2x2 + 12x ¡ 10
3) y = 2x2 ¡ 12x + 10
4) y = 2x2 ¡ 12x + 16
7) y = ¡3x2 + 24x ¡ 45
8) y = ¡3x2 + 12x ¡ 9
9) y = ¡x2 + 4x + 5
10) y = ¡x2 + 4x ¡ 3
11) y = ¡x2 + 6x ¡ 5
12) y = ¡2x2 + 16x ¡ 30
13) y = ¡2x2 + 16x ¡ 24
14) y = 2x2 + 4x ¡ 6
15) y = 3x2 + 12x + 9
16) y = 5x2 + 30x + 45
17) y = 5x2 ¡ 40x + 75
18) y = 5x2 + 20x + 15
19) y = ¡5x2 ¡ 60x ¡ 175
20) y = ¡5x2 + 20x ¡ 15
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Section 4.4
4.4 Answers - Parabolas
1)
12)
(-2,0) (0,-8)
(4,0)
(4,2) (5,0)
(3,0) 7)
(3,0)
(1,-9)
(4,3) (5,0)
(0,-30)
2)
(0,-45) (-1,0) (0, -3)
(3, 0)
13)
(2,3) (1,0)
(1, -4)
(3,0)
3)
(0,10) (1,0)
(-9,0) (5,0) (3,-8)
4)
(0,5) (-1,0)
(4,0)
(5,0)
(0,-6)
(-1,-8)
(1,0)
(-3,0) (-1,0)
(3,0) (0,-3)
(-2,-3)
11)
16)
(3,4)
(1,0)
(1,0)
(2,1)
(3,0)
(1,0) (3,8)
(-3,0)
15)
5)
6)
(6,0)
(0,-24)
(2,9)
10)
(3, -2)
(0,-18)
(2,0)
14)
9)
(0,16) (2,0)
(4,8)
8)
(0,45)
(5,0)
(0,-5)
(-3,0)
(5,0)
(-10,0)
163
(0,9)
CCBC ~ Math 083
Section 4.4
17)
19)
(-6,5) (0,75)
(-7,0) (3,0) (5,0)
(-5,0) (0,-175)
(4,-5) 18)
20)
(2,5)
(0,15) (1,0)
(-3,0)
(3,0)
(-1,0) (-2,-5)
(-15,0)
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Section 4.5
Section 4.5 - Quadratic Applications Objective: Solve quadratic application problems. The vertex of the parabola formed by the graph of a quadratic equation is either a maximum point or a minimum point, depending on the sign of a. If a is a positive number, then the vertex is the minimum; if a is a negative number, then the vertex is a maximum. An example of a maximum would be the highest height of a ball that has been thrown into the air. An example of a minimum would be to minimize average cost to a company for a product that has been produced. Example 1. Answer each of the following questions. Terry is on the balcony of her apartment, which is 150 feet above the ground. She tosses a ball vertically upward. The ball's height above the ground as it travels is a quadratic de ned by the equation s = ¡16t2 + 64t + 150, where t is the amount of time (in seconds) the ball has been in ight and s is the height of the ball (in feet) at any particular time. a. How many seconds will it take for the ball to reach its maximum height above the ground? s = ¡16t2 + 64t + 150 t=
¡(64) ¡64 = =2 2(¡16) ¡32 2 seconds
The time t is unknown; a = ¡16; b = 64; c = 150 ¡b to nd the amount of time; t; that has passed when 2a the ball reaches its maximum height
Use
So; the ball reaches its maximum height after 2 seconds
b. What is the ball's maximum height above the ground? s = ¡16t2 + 64t + 150 s = ¡16(2)2 + 64(2) + 150 s = ¡16(4) + 64(2) + 150
The time that has passed; t; is 2 seconds Substitute the value 2 in for t everywhere in the equation Simplify
s = ¡64 + 128 + 150 = 214 214 feet
So; the maximum height of the ball is 214 feet
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Section 4.5
c. How long does it take for the ball to hit the ground? Round to the nearest tenth of a second, if necessary.
t=
¡(64)
s = ¡16t2 + 64t + 150
The time that has passed; t; is unknown; when the ball hits the ground; its height s is zero
0 = ¡16t2 + 64t + 150
Set the quadratic equation equal to zero and solve the equation; a = ¡16; b = 64; c = 150
p
(64)2 ¡ 4(¡16)(150) 2(¡16)
p ¡64 4096 + 9600 t= ¡32
Use the quadratic formula to determine the time p ¡b b2 ¡ 4ac t= 2a Simplify
p ¡64 13696 t= ¡32 t = ¡1.7 or t = 5.7 5.7 seconds
Time cannot be negative; so; t = ¡1.7 is extraneous So; the time lapsed when the ball hits the ground is 5.7 seconds
Example 2. Solve. Arthur sells used cell phones. He has determined that his average cost to package and ship cell phones to customers is given by the equation C = 2x2 ¡ 60x + 1700, where x is the number of cell phones packaged and shipped every two weeks, and C is the average cost. How many cell phones must Arthur package and ship during the two-week period in order to minimize the average cost? C = 2x2 ¡ 60x + 1700
x=¡
b (¡60) 60 =¡ = = 15 2a 2(2) 4
C = 2(15)2 ¡ 60(15) + 1700
The number of cell phones; x; is unknown; a = 2; b = ¡60; c = 1700 ¡b to nd the number of cell phones that will minimize 2a the cost
Use
Substitute the value 15 for x everywhere in the equation
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C = 2(15)2 ¡ 60(15) + 1700 C = 2(225) ¡ 60(15) + 1700 C = 450 ¡ 900 + 1700 C = 1250
Section 4.5
Simplify using the Order of Operations
So; 1250 cell phones need to be packaged and shipped to minimize the average cost
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Section 4.5
4.5 Practice - Quadratic Applications Use the following information to answer questions 1 and 2. George is standing on the top of a 275 foot building. He throws a ball straight up into the air. The ball's initial velocity is given as 48 ft./sec. The height, in feet, of the ball after t seconds is given by h = ¡16t2 + 48t + 275. 1) How long will it take for the ball to reach its maximum height above the ground? 2) What is the maximum height that the ball reaches? Use the following information to answer questions 3 5. Shelly is standing on a platform 100 feet above the ground. She tosses a baseball straight up into the air. The formula h = ¡16t2 + 64t + 100 models the ball's height h above the ground t seconds after it was thrown. 3) How long will it take for the ball to reach its maximum height above the ground? 4) What is the maximum height that the ball reaches? 5) How many seconds does it take for the ball to nally hit the ground (rounded to the nearest tenth of a second)? Use the following information to answer questions 6 8. Les is standing on the ground. He launches a model rocket straight up into the air. The formula s = ¡16t2 + 64t models the rocket's height s above the ground t seconds after it was launched. 6) How long will it take for the rocket to reach its maximum height above the ground? 7) What is the maximum height that the rocket reaches? 8) How many seconds does it take for the ball to nally hit the ground (rounded to the nearest tenth of a second)? Use the following information to answer questions 9 10. The formula P = ¡0.001x2 +2.45x ¡ 525 models the pro t P for Mama Anna's Restaurant's lasagne meal each week. 9) How many lasagne meals should the restaurant sell each week in order to maximize its pro t? 10) What would be the maximum weekly pro t if they sell the necessary number of meals (rounded to the nearest cent)?
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Section 4.5
Use the following information to answer questions 11 12. The formula P = 0.001t2 ¡0.24t + 59.90 closely models common stock XYZ's closing price, P each of its days trading on the market for the calendar year of 2015. 11) After how many days was XYZ stock at its lowest value? 12) What was the stock's lowest price for 2015?
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Section 4.5
4.5 Answers - Quadratic Applications '
1) 1.5 seconds
6) 2 seconds
2) 311 feet
7) 64 feet
3) 2 seconds
8) 4 seconds
4) 164 feet
9) 1225 meals
5) 5.2 seconds
10) S975.63
170
11) 120 days 12) S45.50
CCBC ~ Math 083
Chapter 5
Chapter 5: Functions and Exponential and Logarithmic Equations Section 5.1 - Functions..................................................................................172 Section 5.2 - Operations on Functions...........................................................184 Section 5.3 - Exponential Functions and Equations.......................................190 Section 5.4 - Logarithmic Functions and Equations.......................................197 Section 5.5 - Compound Interest....................................................................205
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Section 5.1
Section 5.1 - Functions Objective: Identify functions and use correct notation to evaluate functions at numerical and variable values. An equation represents the relationship between variables and numbers. There are many dierent types of equations that we can work with in algebra. Examples of several relationships are below: p (x ¡ 3)2 (y + 2)2 ¡ = 1 and y = x2 ¡ 2x + 7 and y + x ¡ 7 = xy 9 4 There is a special classi cation of relationships known as functions. Functions have at most one output for any input. Generally x is the variable that we put into an equation to evaluate and nd y. For this reason x is considered an input variable and y is considered an output variable. This means the de nition of a function, in terms of equations in x and y could be stated as: there is at most one y value corresponding with any x value. Once we know a relationship is a function, we may be interested in what values can be put into the equations. The domain is the set of values that are put into an equation (generally the x values). When nding the domain, it is often easier to consider what cannot happen in a given function, then exclude appropriate values. The range is the set of values that are output values (generally the y values). Example 1. Which of the following set of ordered pairs represent functions? a) f(1; a); (2; b); (3; c); (4; a)g is a function, since there are no two pairs with the same rst component. The domain is the set {1,2,3,4} and the range is {a,b,c}. b) f(1; a); (2; b); (1; c); (4; a)g is not a function, since there are two pairs with the same rst component but dierent second components: (1; a) and (1; c).
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Section 5.1
A mapping shows how the elements are paired. It is like a ow chart for a function, showing the input and output values. Example 2. Which of the diagrams represent functions?
a) This diagram represents a function since each element in the domain corresponds to exactly one element in the range.
b) This diagram represents a function since each element in the domain corresponds to at most one element in the range.
c) This diagram does not represent a function since one element of the domain corresponds with more than one element of the range; (a,I), (a,III).
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Section 5.1
Vertical Line Test A great way to visualize this de nition is to look at the graphs of a few relationships. Because x values correspond with vertical lines, we will draw a vertical line through the graph. If the vertical line crosses the graph more than once, that means we have too many possible y values. If the graph crosses the graph at most once, then we say the relationship is a function. Example 3. Which of the following graphs are graphs of functions?
Drawing a vertical line through this graph will only cross the graph once. It is a function.
Drawing a vertical line through this graph will cross the graph twice. This is not a function.
Drawing a vertical line through this graph will cross the graph at most once. It is a function.
We can look at the above idea algebraically by taking a relationship and solving it for y. If we have only one solution, then it is a function. Example 4. Determine whether the relation represents a function. Is 3x2 ¡ y = 5 a function? ¡3x2 ¡ 3x2 ¡y = ¡3x2 + 5 ¡1 ¡1 ¡1 y = 3x2 ¡ 5
Solve the relation for y Subtract 3x2 from both sides Divide each term by ¡1 Only one solution for y:
Yes!
It is a function
Example 5. Determine whether the relation represents a function. Is
y 2 ¡ x = 5 a function? +x + x y2 = x + 5 p p y2 = x + 5 p y= x+5 No!
Solve the relation for y Add x to both sides Square root of both sides Simplify Two solutions for y (one +; one ¡) Not a function
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Section 5.1
Once we know we have a function, often we will change the notation used to emphasize the fact that is is a function. Instead of writing the output value as y, we will use function notation which is written f(x). We read this function notation as f of x . So in the example above, instead of writing y = 3x2 ¡ 5, we could have written f (x) = 3x2 ¡ 5. It is important to point out that f(x) does not mean f times x . It is a notation that names the function with the rst letter (function f ) and then in parentheses, we are give information about what variable is used as the input in the function (variable x ). The rst letter naming the function can be anything we want it to be. For example, you will often see g(x), read g of x . One way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x -axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values.
We can observe that the graph extends horizontally from -5 to the right without bound, so the domain is [¡5; 1) orfxjx ¡5g. The vertical extent of the graph is all range values 5 and below, so the range is (¡1, 5] or fy jy 6 5g. Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range.
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Section 5.1
Example 6. Find the domain and range given the following graph.
Answer
Domain: (-3,1] or ¡3 < x 6 1 Range: [-4,0] or ¡4 6 y 6 0
Notice that the open circle at (-3,0) indicates that the point (-3,0) does not belong to the graph. The closed circle at (1,-4) indicates that the point (1,-4) does belong to the graph.
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Section 5.1
Example 7. Find the domain of the function: f(x) =
3x ¡ 1
x2 + x ¡ 6
x2 + x ¡ 6 = 0 (x + 3)(x ¡ 2) = 0 x + 3 = 0 or x ¡ 2 = 0 +2 + 2 ¡3 ¡3 x = ¡3 or x=2 All real numbers except x = ¡3 and 2
Find values for x for which the denominator is equal to 0 Solve by factoring Set each factor equal to zero Solve each equation Exclude these values from the domain of f Our Answer
The notation in the previous example tells us that x can be any value except for ¡3 and 2. If x were one of those two values, the function would be unde ned. Example 8. Find the domain of the function: f (x) = 3x2 ¡ x All real numbers or R or (¡1; 1)
With this function; there are no excluded values Our Solution
In the above example there are no real numbers that make the function unde ned. This means any number can be used for x. Example 9. Find the domain of the function: p f (x) = 2x ¡ 3 2x ¡ 3 > 0 +3 + 3 2x > 3 2 2 3 x> 2
Square roots cannot be negative Set up an inequality Solve
Our Answer
3
The notation in the above example states that our variable can be 2 or any 3 number larger than 2 . But any number smaller would make the function unde ned because the radicand would be a value less than zero.
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Section 5.1
Another use of function notation is to easily evaluate functions. If we want to substitute a numerical value or an expression for a variable, we simply replace the variable with what we want to plug in. This process is shown in the following examples. Example 10. Evaluate the function. Let f (x) = 3x2 ¡ 4x; nd f (¡2): f (¡2) =3(¡2)2 ¡ 4(¡2) =3(4) ¡ 4(¡2) =12 + 8 f (¡2) = 20
Substitute ¡2 in for x in the function Evaluate; using order of operations Multiply Add Our Answer
Example 11. Evaluate the function. Let h(x) = 32x¡6; nd h(4): h(4) = 32(4)¡6 =38¡6 =32 h(4) = 9
Substitute 4 in for x in the function Simplify exponent; mutiplying rst Subtract in exponent Evaluate exponent Our Answer
Example 12. Evaluate the function. Let k(a) = 2ja + 4j; nd k(¡7): k(¡7) = 2j¡7 + 4j =2j¡3j =2(3) k(¡7) = 6
Substitute ¡7 in for a in the function Add inside absolute value Evaluate absolute value Multiply Our Answer
As the above examples show, the function can take many dierent forms, but the pattern to evaluate the function is always the same: replace the variable with what is in parentheses and simplify. We can also substitute expressions into functions using the same process. Often the expressions use the same variable, so it is important to remember each occurrence of the variable is replaced by whatever is in parentheses. Example 13. Evaluate the function. Let g (x) = x4 + 1; nd g(3x): g(3x) = (3x)4 + 1 g(3x) = 81x4 + 1
Replace x in the function with (3x) Simplify exponent Our Answer
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Section 5.1
Example 14. Evaluate the function. Let p(t) = t2 ¡ t; nd p(t + 1): p(t + 1) = (t + 1)2 ¡ (t + 1) = t2 + 2t + 1 ¡ (t + 1) = t2 + 2t + 1 ¡ t ¡ 1 p(t + 1) = t2 + t
Replace each t in the function with (t + 1) Square binomial Distribute negative sign Combine like terms Our Answer
It is important to become comfortable with function notation and learn how to use it as we transition into more advanced algebra topics.
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Section 5.1
5.1 Practice - Functions Determine which of the following represent functions. 1) a)
b)
c)
d)
e) y = 3x ¡ 7 p g) y + x = 2
f) y 2 ¡ x2 = 1 h) x2 + y 2 = 1
Specify the domain of each of the following functions. p 2) f (x) = ¡5x + 1 3) f (x) = 5 ¡ 4x 1
4) s(t) = t2
5) f (x) = x2 ¡ 3x ¡ 4 p 7) f (x) = x ¡ 16
1
6) s(t) = t2 + 1 ¡2
8) f (x) = x2 ¡ 3x ¡ 4
9) h(x) =
x
10 y(x) = x2 ¡ 25
180
p 3x ¡ 12 x2 ¡ 25
CCBC ~ Math 083
Section 5.1
Evaluate each function. 11) g(x) = 4x ¡ 4; Find g(0)
12) g(n) = ¡3 5¡n; Find g(2)
15) f (n) = ¡2j¡n ¡ 2j+1; Find f (¡6)
16) f (n) = n ¡ 3; Find f (10)
19) f (t) = jt + 3j; Find f (10)
20) w(x) = x2 + 4x; Find w(¡5)
13) f (x) = j3x + 1j +1; Find f (0)
14) f (x) = x2 + 4; Find f (¡9)
17) f (t) = 3 ¡ 2; Find f (¡2) t
18) f (a) ¡ 3a¡1 ¡ 3; Find f (2)
21) w(n) = 4n + 3; Find w(2) 23) w(n) = 2n+2; Find w(¡2)
22) w(x) = ¡4x + 3; Find w(6)
25) p(n) = ¡3jnj; Find p(7)
26) k(a) = a + 3; Find k(¡1)
24) p(x) = ¡jxj+1; Find p(5)
27) p(t) = ¡t3 + t; Find p(4)
28) k(x) = ¡2 42x¡2; Find k(2)
29) k(n) = jn ¡ 1j; Find k(3)
30) p(t) = ¡2 42t+1 + 1; Find p(¡2)
31) h(x) = x + 2; Find h(¡4x) 3
32) h(n) = 4n + 2; Find h(n + 2)
33) h(x) = 3x + 2; Find h(¡1 + x)
a
34) h(a) = ¡3 2a+3; Find h( 4 )
35) h(t) = 2j¡3t ¡ 1j+2; Find h(n ) 2
x
36) h(x) = x2 + 1; Find h( 4 )
37) g(x) = x + 1; Find g(3x)
38) h(t) = t2 + t; Find h(t2)
39) g(x) = 5x; Find g(¡3 ¡ x)
n
40) h(n) = 5n¡1 + 1; Find h( 2 )
For each of the functions in problems 41 - 43, nd the following: a) Domain b) Range c)f (¡1) d) f (0) e) the x value(s) such that f (x) = 0
41)
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Section 5.1
42)
43)
Determine whether each relation is a function. Identify the domain and range for each relation. 44) f(¡2; 4); (3; 6); (4; 6); (9; 0)g
45) f(0; ¡5); (2; ¡3); (4; ¡1); (6; 1); (8; 3)g 46) f(5; 6); (7; 0); (9; ¡1)(7; 1)g
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Section 5.1
5.1 Answers - Functions 1) a. yes b. yes c. no d. no e. yes f. no g. yes h. no 2) all real numbers 5
3) x 6 4 4) t = /0 5) all real numbers 6) all real numbers 7) x > 16 8) x = / ¡1; 4
9) x > 4; x = /5 10) x = / 5 11) ¡4 12)
3 ¡ 25
22) ¡21
d)f (0) = 4
23) 1
e)x = ¡2
24) ¡4
42) a) x > ¡2.5
25) ¡21
b) y > ¡4
26) 2
c)f (¡1) = ¡3
27) ¡60
d)f (0) = 0
28) ¡32
e)x = ¡2; 0; 2
29) 2 30)
43) a) x 6 1
31 32
31) ¡64x3 + 2 32) 4n + 10
33) ¡1 + 3x 34) ¡3 2
a+12 4
b) y > 0 c)f (¡1) = 2 d)f (0) = 1.5 e)x = 1 44) Function;
13) 2
35) 2j¡3n2 ¡ 1j+2
14) 85
36) 1 + 16 x2
15) ¡7
37) 3x + 1
Range f0; 4; 6g
38) t4 + t2
45) Function;
39) 5¡3¡x
Domain f0; 2; 4; 6; 8g
16) 7
17
17) ¡ 9 18) ¡6 19) 13 20) 5 21) 11
1
40) 5
¡2+n 2
+1
41) a) ¡3 6 x < 2
b) ¡5 6 y 6 4 c)f (¡1) = 3
183
Domain f¡2; 3; 4; 9g
Rangef¡5; ¡3; ¡1; 1; 3g 46) Not a Function; Domain f5; 7; 9g; Rangef¡1; 0; 1; 6g
CCBC ~ Math 083
Section 5.2
Section 5.2 - Operations on Functions Objective: Combine functions using sum, dierence, product, and quotient of functions. Several functions can work together in one larger function. There are four common operations that can be performed on functions. The four basic operations on functions are addition, subtraction, multiplication, and division. The notation for these functions is as follows. Addition
(f + g)(x) = f (x) + g(x)
Subtraction
(f ¡ g)(x) = f (x) ¡ g(x)
Multiplication Division
(f g)(x) = f (x) g(x) f (x) f for g(x) = /0 (x) = g(x) g
When we want to evaluate the sum/dierence/product/quotient of two functions, we can simply evaluate the two functions independently and then perform the given operation with both results. Example 1. Perform the indicated operation. Let f (x) = x2 ¡ x ¡ 2 and g(x) = x + 1: Find (f + g)(¡3):
Evaluate (f + g) at x = ¡3 Evaluate f at x = ¡3: f (¡3) = (¡3)2 ¡ (¡3) ¡ 2 f (¡3) = 9 + 3 ¡ 2 f (¡3) = 10 Evaluate g at x = ¡3: g(¡3) = (¡3) + 1 g(¡3) = ¡2 Add the two results together Add Our Answer
= f (¡3) + g(¡3) = (10) + (¡2) =8
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Section 5.2
The process is the same regardless of the operation being performed. Example 2. Perform the indicated operation. Let h(x) = 2x ¡ 4 and k(x) = ¡3x + 1: Find (h k)(5)
Evaluate (h k) at x = 5 Evaluate h at x = 5: h(5) = 2(5) ¡ 4 h(5) = 10 ¡ 4 h(5) = 6 Evaluate k at x = 5: k(5) = ¡3(5) + 1 k(5) = ¡15 + 1 k(5) = ¡14 Multiply the two results together Multiply Our Answer
= h(5) k(5) = (6)(¡14) = ¡84
Often as we add, subtract, multiply, or divide functions, we do so in a way that keeps the variable. If there is no number to plug into the functions, we will simply use each equation, in parentheses, and simplify the expression. Example 3. Perform the indicated operation. Let f (x) = 2x ¡ 4 and g(x) = x2 ¡ x + 5: Find (f ¡ g)(x):
= f (x) ¡ g(x) = (2x ¡ 4) ¡ (x2 ¡ x + 5) = 2x ¡ 4 ¡ x2 + x ¡ 5 = ¡x2 + 3x ¡ 9
Find the dierence of two functions
Replace f (x) with (2x ¡ 4) and g(x) with (x2 ¡ x + 5) Distribute the negative Combine like terms Our Answer
The parentheses are very important when we are replacing f(x) and g(x) with a variable. In the previous example, we needed the parentheses to know to distribute the negative.
185
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Section 5.2
Example 4. Perform the indicated operation. Let f (x) = x2 ¡ 4x ¡ 5 and g(x) = x ¡ 5: f (x): Find g
Find the quotient of two functions
=
f (x) g(x)
=
(x2 ¡ 4x ¡ 5) (x ¡ 5)
Simplify the fraction; we must rst factor
=
(x ¡ 5)(x + 1) (x ¡ 5)
Divide out common factor of x ¡ 5
=x+1
Replace f (x) with (x2 ¡ 4x ¡ 5) and g(x) with (x ¡ 5)
Our Answer
Example 5. Perform the indicated operation. Let f (x) = 2x ¡ 1 and g(x) = x + 4: Find (f + g)(x2): = f (x2) + g(x2) = [2(x2) ¡ 1] + [(x2) + 4] = 2x2 ¡ 1 + x2 + 4 = 3x2 + 3
Find the sum of two functions Replace x in f (x) and g(x) with x2 Distributing the + does not change the problem Combine like terms Our Answer
Example 6. Perform the indicated operation. Let f (x) = 2x ¡ 1 and g(x) = x + 4: Find (f g)(3x):
Find the product of two functions Replace x in f(x) and g(x) with 3x Multiply out 2(3x) FOIL Combine like terms Our Answer
= f (3x) g(3x) = [2(3x) ¡ 1] [(3x) + 4] = (6x ¡ 1)(3x + 4) = 18x2 + 24x ¡ 3x ¡ 4 = 18x2 + 21x ¡ 4
World View Note: The term function came from Gottfried Wihelm Leibniz, a German mathematician from the late 17th century.
186
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Section 5.2
5.2 Practice - Operations on Functions Perform the indicated operations. 2) f (x) = ¡3x2 + 3x g(x) =2x + 5 Find f (¡4) g(¡4)
1) g(x) = a3 + 5a2 f (x) = 2a + 4 Find g(3) + f (3) 3) g(x) = 3a + 3 f (x) = 2a ¡ 2 Find (g + f )(9)
4) g(x) = 4x + 3 h(x) =x3 ¡ 2x2 Find (g ¡ h)(¡1)
5) g(x) = x + 3 f (x) =¡x + 4 Find (g ¡ f )(3)
6) g(x) = ¡4x + 1 h(x) =¡2x ¡ 1 Find g(5) + h(5)
7) g(x) = x2 + 2 f(x) =2x + 5 Find (g ¡ f )(0)
8) g(x) = 3x + 1 f (x) =x3 + 3x2 Find g(2) f (2)
9) g(t) = t ¡ 3 h(t) =¡3t3 + 6t Find g(1) +h(1)
10) f (n) = n ¡ 5 g(n) =4n + 2 Find (f + g)(¡8)
11) h(t) = t + 5 g(t) =3t ¡ 5 Find (h g)(5)
12) g(a) = 3a ¡ 2 h(a) =4a ¡ 2 Find (g + h) (¡10)
13) h(n) = 2n ¡ 1 g(n) =3n ¡ 5 Find h(0) g(0)
14) g(x) = x2 ¡ 2 h(x) =2x + 5 Find g(¡6) + h(¡6)
15) f (a) = ¡2a ¡ 4 g(a) =a2 + 3 f Find ( g )(7)
16) g(n) = n2 ¡ 3 h(n) =2n ¡ 3 Find (g ¡ h)(n)
18) g(x) = 2x ¡ 3 h(x) =x3 ¡ 2x2 + 2x Find (g ¡ h)(x)
17) g(x) = ¡x3 ¡ 2 h(x) =4x Find (g ¡ h)(x)
20) g(t) = t ¡ 4 h(t) =2t Find (g h)(t)
19) f (x) = ¡3x + 2 g(x) =x2 + 5x Find (f ¡ g)(x)
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Section 5.2
Perform the indicated operations. 21) g(x) = 4x + 5 h(x) =x2 + 5x Find g(x) h(x)
22) g(t) = ¡2t2 ¡ 5t h(t) =t + 5 Find g(t) h(t)
25) g(n) = n2 + 5 f (n) =3n + 5 Find g(n) f (n)
26) f (x) = 2x + 4 g(x) =4x ¡ 5 Find f (x) ¡ g(x)
23) f (x) = x2 ¡ 5x g(x) =x + 5 Find (f + g)(x)
24) f (x) = 4x ¡ 4 g(x) =3x2 ¡ 5 Find (f + g)(x)
27) g(a) = ¡2a + 5 f (a) =3a + 5 g Find ( f )(a)
28) g(t) = t3 + 3t2 h(t) =3t ¡ 5 Find g(t) ¡ h(t)
30) f (x) = 4x + 2 g(x) =x2 + 2x Find f (x) g(x)
29) h(n) = n3 + 4n g(n) =4n + 5 Find h(n) + g(n) 31) g(n) = n2 ¡ 4n h(n) =n ¡ 5 Find g(n2) h(n2)
32)g(n) = n + 5 h(n) =2n ¡ 5 Find (g h)(¡3n)
33) f (x) = 2x g(x) =¡3x ¡ 1 Find (f + g)(¡4 ¡ x)
34) g(a) = ¡2a h(a) =3a Find g(4n) h(4n)
35) f (t) = t2 + 4t g(t) =4t + 2 Find f (t2) + g(t2)
36) h(n) = 3n ¡ 2 g(n) =¡3n2 ¡ 4n n n Find h( 3 ) g( 3 )
37) g(a) = a3 + 2a h(a) =3a + 4 g Find ( h )(¡x)
38) g(x) = ¡4x + 2 h(x) =x2 ¡ 5 Find g(x2) + h(x2)
39) f(n) = ¡3n2 + 1 g(n) =2n + 1 n Find (f ¡ g)( 3 )
40) f (n) = 3n + 4 g(n) =n3 ¡ 5n n n Find f ( 2 ) ¡ g( 2 )
188
CCBC ~ Math 083
Section 5.2
5.2 Answers - Operations on Functions 1) 82
22) ¡2t3 ¡ 15t2 ¡ 25t
2) 20
23) x2 ¡ 4x + 5
24) 3x2 + 4x ¡ 9
3) 46 4) 2
25)
5) 5
26) ¡2x + 9
6) ¡30
27)
7) ¡3
n2 + 5 3n + 5
¡2a + 5 3a + 5
28) t3 + 3t2 ¡ 3t + 5
8) 140
29) n3 + 8n + 5
9) 1
30)
10) ¡43
4x + 2 x2 + 2x
11) 100
31) n6 ¡ 9n4 + 20n2
12) ¡74
33) x + 3
13)
32) 18n2 ¡ 15n ¡ 25
1 5
2
34) ¡ 3
14) 27 15)
35) t4 + 8t2 + 2
9 ¡ 26
16) n2 ¡ 2n 17) ¡x ¡ 4x ¡ 2 3
36)
3n ¡ 6 ¡n2 ¡ 4n
37)
¡x3 ¡ 2x ¡3x + 4
38) x4 ¡ 4x2 ¡ 3
18) ¡x3 + 2x2 ¡ 3 19) ¡x2 ¡ 8x + 2
39)
¡n2 ¡ 2n 3
20) 2t2 ¡ 8t
40)
32 + 23n ¡ n3 8
21) 4x3 + 25x2 + 25x
189
CCBC ~ Math 083
Section 5.3
Section 5.3 - Exponential Functions and Equations Objective: Graph exponential functions and solve exponential equations by nding a common base. As our study of algebra gets more advanced, we begin to study more involved functions. One pair of inverse functions we will look at are exponential functions and logarithmic functions. Here we will look at exponential functions and then we will consider logarithmic functions in another lesson. Exponential functions have the form f (x) =ax where a > 0 and a = / 1. Notice that exponential functions have the variable in the exponent. It is important not to confuse exponential functions with polynomial functions where the variable is in the base such as f (x) = x2. Graphs of Exponential Functions Example 1. Evaluate f (x) = 2x and g(x) = f (x) = 2x x ¡2 ¡1 0 1 2
y = f (x) 1 1 2¡2 = 22 = 4 1
¡ 1 x 2
at x = ¡2; ¡1; 0; 1; 2 and graph. ¡ 1 x g(x) = 2 x y = g(x) ¡1 ¡2 2 ¡2 = 22 = 4 ¡1 ¡1 2 ¡1 = 21 = 2 ¡ 1 0 0 =1 ¡ 21 1 1 1 = ¡ 21 2 21 =4 2 2
1
2¡1 = 21 = 2 20 = 1 21 = 2 22 = 4
The domain of f(x) = ax consists of all real numbers, written in interval notation as (¡1; 1). The range of f (x) = ax consists of all positive real numbers, written in interval notation as (0; 1).
If a > 1, then the graph goes up to the right and is an increasing function, called an exponential growth function. If 0 < a < 1; then the graph goes down to the right and is a decreasing function, called an exponential decay function.
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Section 5.3
The graphs of all exponential functions of the form f (x) = ax pass through the point (0,1). The graph of an exponential function f (x) = ax approaches, but does not touch, the x-axis. We say the x -axis, or y=0, is a horizontal asymptote of the graph of the function.
World View Note One common application of exponential functions is population growth. According to the 2009 CIA World Factbook, the country with the highest population growth rate is a tie between the United Arab Emirates (north of Saudi Arabia) and Burundi (central Africa) at 3.69%. There are 32 countries with negative growth rates, the lowest being the Northern Mariana Islands (north of Australia) at ¡7.08%. Solving exponential equations cannot be done using the skill set we have seen in the past. For example, if 3x = 9, we cannot take the x ¡ root of 9 because we do not know what the index is and this doesn't get us any closer to nding x. However, we may notice that 9 is 32. We can then conclude that if 3x = 32 then x = 2. This is the process we will use to solve exponential equations. If we can re-write a problem so that the bases match, then the exponents must also match. Example 2. Solve the equation. 42x = 4x+3 2x = x + 3 2x = x + 3 ¡x ¡ x x=3
Same bases; set exponents equal Solve Subtract x from both sides Our Solution
Example 3. Solve the equation. 52x+1 = 125 52x+1 = 53 2x + 1 = 3 ¡1 ¡ 1 2x = 2 2 2 x=1
Rewrite 125 as 53 Same bases; set exponents equal Solve Subtract 1 from both sides Divide both sides by 2 Our Solution
Sometimes we may have to do work on both sides of the equation to get a common base. As we do so, we will use various exponent properties to help. First we will use the exponent property that states (ax)y = axy.
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CCBC ~ Math 083
Section 5.3
Example 4. Solve the equation. 83x = 32 (23)3x = 25 29x = 25 9x = 5 9 9 5 x= 9
Rewrite 8 as 23 and 32 as 25 Multiply exponents 3 and 3x Same bases; set exponents equal Solve Divide both sides by 9 Our Solution
As we multiply exponents, we may need to distribute if there are several terms involved. Example 5. Solve the equation. 33x+5 = 814x+1 33x+5 = (34)4x+1 33x+5 = 316x+4 3x + 5 = 16x + 4 ¡16x ¡16x ¡13x + 5 = 4 ¡5 ¡5 ¡13x = ¡1 ¡13 ¡13 1 x= 13
Rewrite 81 as 34 Multiply exponents 4(4x + 1) by distributing Same bases; set exponents equal Solve Subtract 16x from both sides Subtract 5 from both sides Divide both sides by ¡ 13 Our Solution
Another useful property of exponents is that negative exponents will give us a rec1 iprocal, a¡n = an . Example 6. Solve the equation. 2x 1 = 37x¡1 9 (3¡2)2x = 37x¡1 3¡4x = 37x¡1 ¡4x = 7x ¡ 1 ¡7x ¡ 7x ¡11x = ¡1 ¡11 ¡11 1 x= 11
1 as 3¡2 (Use the negative exponent property to ip) 9 Multiply exponents ¡2 and 2x Same bases; set exponents equal Subtract 7x from both sides Rewrite
Divide by both sides by ¡11 Our Solution
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CCBC ~ Math 083
Section 5.3
If we have several factors with the same base on one side of the equation, we can add the exponents using the property that states axa y = ax+ y. Example 7. Solve the equation. 54x 52x¡1 = 53x+11 56x¡1 = 53x+11 6x ¡ 1 = 3x + 11 ¡3x ¡3x 3x ¡ 1 = 11 +1 +1 3x = 12 3 3 x=4
Add exponents on left; combing like terms Same bases; set exponents equal Solve Subtract 3x from both sides Add 1 to both sides Divide both sides by 3 Our Solution
It may take a bit of practice to get used to knowing which base to use. We will use our properties of exponents to help us simplify. Again, below are the properties we used: (ax) y = axy
and
1 = a¡n an
and
axay = ax+y
In all of the problems we have solved here, we were able to nd a common base. However, this is not always possible. For example, 2 = 10x cannot be written using common bases. To solve problems like this, we will need to use the inverse of an exponential function. The inverse is called a logarithmic function, which we will discuss in another section.
193
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Section 5.3
5.3 Practice - Exponential Functions and Equations Solve each equation. 1) 31¡2n = 31¡3n
2) 42x = 16
3) 42a = 1
4) 16¡3p = 64¡3p
1
1
5) ( 25 )¡k = 125¡2k¡2
1
6) 625¡n¡2 = 125
1
7) 62m+1 = 36
8) 62r ¡3 = 6r ¡3
9) 6¡3x = 36
10) 52n = 5¡n
11) 64b = 25
12) 216¡3v = 363v
1
13) ( 4 )x = 16
14) 27¡2n¡1 = 9
15) 43a = 43
16) 4¡3v = 64
17) 363x = 2162x+1
18) 64x+2 = 16
19) 92n+3 = 243
1
20) 162k = 64
21) 33x¡2 = 33x+1 23) 3
¡2x
25) 5
m+2
27)
=3
22) 243 p = 27¡3p
3
=5
24) 42n = 42¡3n
¡m
26) 6252x = 25
1 ( 36 )b¡1 = 216
28) 2162n = 36
29) 62¡2x = 62 31) 42
¡3n¡1
=
1
30) ( 4 )3v¡2 = 641¡v
1 4
32)
33) 43k¡3 42¡2k = 16¡k
216 6¡2a
= 63a 1
34) 322p¡2 8 p = ( 2 )2p
1
35) 9¡2x ( 243 )3x = 243¡x
36) 32m 33m = 1
1
37) 64n¡2 16n+2 = ( 4 )3n¡1
38) 32¡x 33m = 1
39) 5¡3n¡3 52n = 1
1
40) 43r 4¡3r = 64
Graph the following exponential functions and write the domain and range in interval notation. 41) f (x) = 3x ¡ 1 x 42) f (x) = 3
43) f (x) = 2x + 3
194
CCBC ~ Math 083
Section 5.3
5.3 Answers - Exponential Functions and Equations 1) 0
15) 1
29) 0
2) ¡1
16) ¡1
30) No solution
4) 0
18) ¡ 3
17) No solution
3) 0
4
31) 1
1
32) 3
3
33)
1 3
34)
2 3
5) ¡ 4
3
19) ¡ 4
5
20) ¡ 4
6) ¡ 4
21) No solution
3
7) ¡ 2
22) 0
8) 0 9)
24)
10) 0 11)
35) 0
3
23) ¡ 2
2 ¡3
5 6
12) 0 13) ¡2 5
14) ¡ 6
36) 0
2 5
3 8
25) ¡1
37)
26)
38) ¡1
1 4
1
27) ¡ 2 28)
39) ¡3
1 3
40) No solution
41) a)Domain: All real numbers; Range: y > 0
b) Domain: All real numbers; Range: y > 0
195
CCBC ~ Math 083
Section 5.3
c) Domain: All real numbers; Range: y > 3
196
CCBC ~ Math 083
Section 5.4
Section 5.4 - Logarithmic Functions and Equations Objectives: Convert between logarithms and exponents and use that relationship to solve basic logarithmic equations. Graph logarithmic functions. The inverse of an exponential function is a function known as a logarithm. Logarithms are studied in detail in advanced algebra. Here, we will take an introductory look at how logarithms work. When working withpradicals, we found that there were two ways to write radicals. The expression n am could be written as m a n . Each form has its advantages, thus we need to be comfortable using both the radical form and the rational exponent form. Similarly, an exponent can be written in two forms, each with its own advantages. We are very familiar with the rst form, ax = b, where a is the base, b can be thought of as our answer, and x is the exponent. The second way to write this is with a logarithm, logab = x. The word log tells us that we are in this form. The parts of the equation all still mean the same thing: a is the base, b can be thought of as our answer, and x is the exponent. Notice a logarithm is an exponent; so, logarithmic form will let us isolate an exponent. De nition. Logarithmic Function with Base a For x > 0, a > 0; and a = / 1,
y = loga x if and only if x = ay
The function given by: f (x) = loga x read as log base a of x is called the logarithmic function with base a. Using this idea, the equation 52 = 25 could also be written in equivalent logarithmic form as log525 = 2. Both mean the same thing, both are still the same exponent problem, but just as roots can be written in radical form or rational exponent form, both our forms have their own advantages. The most important thing to be comfortable doing with logarithms and exponents is to be able to switch back and forth between the two forms. This process is shown in the next few examples. Example 1. Write the exponential equation in logarithmic form. 2 = 10x log102 = x
Identify base 10; answer 2; and exponent x Our Answer
197
CCBC ~ Math 083
Section 5.4
Example 2. Write each exponential equation in logarithmic form. m3 = 5 logm5 = 3
Identify base m; answer 5; and exponent 3 Our Answer
72 = b log7b = 2
Identify base 7; answer b; and exponent 2 Our Answer
4 16 2 = 81 3 16 log 2 = 4 3 81
2 16 Identify base ; answer ; and exponent 4 3 81 Our Answer
Example 3. Write each logarithmic equation in exponential form. log416 = 2 42 = 16
Identify base 4; answer 16; and exponent 2 Our Answer
log3x = 7 37 = x
Identify base 3; answer x; and exponent 7 Our Answer
1 2 1 2 9 =3
log93 =
Identify base 9; answer 3; and exponent
1 2
Our Answer
Once we are comfortable switching between logarithmic and exponential form, we are able to evaluate logarithmic expressions and solve exponential equations. We will rst evaluate logarithmic expressions. An easy way to evaluate a logarithm is to set the logarithm equal to x and change it into an exponential equation. Example 4. Evaluate the expression. Evaluate log264 log264 = x 2x = 64 2x = 26 x=6
Set logarithm equal to x Change to exponential form Write using same bases; 64 = 26 Same bases; set exponents equal Our Answer
198
CCBC ~ Math 083
Section 5.4
Example 5. Evaluate the expression. Evaluate log1255 log1255 = x 125x = 5 (53)x = 5 53x = 5 3x = 1 3 3 1 x= 3
Set logarithm equal to x Change to exponential form Write using same bases; 125 = 53 Multiply exponents Same bases; set exponents equal (5 = 51) Solve Divide both sides by 3 Our Answer
Example 6. Evaluate the expression. 1 Evaluate log3 27 1 log3 = x 27 1 3x = 27 3x = 3¡3 x = ¡3
Set logarithm equal to x Change to exponential form 1 Write using same bases; = 3¡3 27 Same bases; set exponents equal Our Answer
Solving logarithmic equations is done in a very similar way, by simply changing the equation into exponential form and solving the resulting equation. Example 7. Solve the equation. log5x = 2 52 = x 25 = x
Change to exponential form Evaluate exponent Our Solution
Example 8. Solve the equation. log2(3x + 5) = 4 24 = 3x + 5 16 = 3x + 5 ¡5 ¡5 11 = 3x 3 3 11 =x 3
Change to exponential form Evaluate exponent Solve Subtract 5 from both sides Divide both sides by 3 Our Solution
199
CCBC ~ Math 083
Section 5.4
Example 9. Solve the equation. logx8 = 3 x3 = 8 x=2
Change to exponential form Cube root of both sides Our Solution
There is one base of the logarithm that is used more often than any other base, mainly base 10. Similar to square roots and not writting the common index of 2 in the radical, we don't write the common base of 10 in the logarithm. So if we are working on a problem with no base written, we will always assume that the base is 10. The other important log is the natural", or base-e, log, denoted as ln(x ) and usually pronounced as ell-enn-of-x ". (Note: That's ell-enn", not one-enn or eye-enn"!) Just as the number e arises naturally in math and the sciences, so also does the natural log, which is why you need to be familiar with it. The number e is a constant similar in idea to pi () in that it goes on forever without repeat or pattern, but just as pi () naturally occurs in several geometry applications, e appears in many exponential applications which we will see in the next section. Example 10. Solve the equation. log x = ¡2 10¡2 = x 1 =x 100
Rewrite as exponent; 10 is base Evaluate; remember negative exponent is fraction Our Solution
Example 11. Solve the equation. ln x = 3 Rewrite as exponent; e is base e3 = x Our solution
This lesson has introduced the idea of logarithms, changing between logarithms and exponents, evaluating logarithms, and solving basic logarithmic equations. In an advanced algebra course, logarithms will be studied in much greater detail.
200
CCBC ~ Math 083
Section 5.4
Graphs of Logarithmic Functions To sketch the graph of y = loga x, you can use the fact that the graphs of inverse functions are re ections of each other in the line y = x.
The domain of f (x) = logax consists of all positive real numbers, (0; 1). The range of f (x) = logax consists of all real numbers, (¡1; 1). Example 12. Graph f (x) = log3 x. From our de nition earlier in the section: For x > 0, a > 0; and a = / 1,
y = loga x if and only if x = ay
We may rewite f (x) = log3 x as an exponential 3f (x) = x ! 3y = x y = log3 x
! 3y = x so we evaluate the equation at y = ¡2; ¡1; 0; 1; 2
x y=f(x) 1 ¡2 3 = 9 -2 1
3¡1 = 3 30 = 1 31 = 3 32 = 9
-1 0 1 2
Domain: x > 0 or (0; 1) Range: y all real numbers or (¡1; 1)
201
CCBC ~ Math 083
Section 5.4
202
CCBC ~ Math 083
Section 5.4
5.4 Practice - Logarithmic Functions and Equations Rewrite each equation in exponential form. 2) ln a = ¡16
1) log9 81 = 2 1
3) log7 49 =¡2
4) log16 256 = 2
5) log13 169=2
6) ln 1 = 0
Rewrite each equations in logarithmic form. 7) 80 = 1
1
8) 17¡2 = 289
9) 152 = 225
1
10) 144 2 = 12
1
11) 64 6 = 2
12) 192 = 361
Evaluate each expression. 13) log125 5 15) log343
14) log5 125 16) log7 1
1 7
1 64
17) log4 16
18) log4
19) log6 36
20) log36 6
21) log2 64
22) log3 243
Solve each equation. 23) log5 x = 1
24) log8 k = 3
25) log2 x = ¡2
26) log n = 3
27) log11 k = 2
28) log4 p = 4
29) log9 (n + 9) = 4
30) log11 (x ¡ 4) = ¡1
31) log5 (¡3m) = 3
32) log2 ¡8r = 1
33) log11 (x + 5) = ¡1
34) log7 ¡3n = 4
35) log4 (6b + 4) = 0
36) log11 (10v + 1) = ¡1
37) log5 (¡10x + 4) = 4
38) log9 (7 ¡ 6x) = ¡2
39) log2 (10 ¡ 5a) = 3
40) log8 (3k ¡ 1) = 1
203
CCBC ~ Math 083
Section 5.4
5.4 Answers - Logarithmic Functions and Equations 1) 92 = 81
29) 6552
1
15) ¡ 3
2) e¡16 = a
16) 0
1
3) 7¡2 = 49
17) 2
4) 162 = 256
18) ¡3
5) 13 = 169 2
30)
45 11
31) ¡
125 3 1
32) ¡ 4
19) 2
54
33) ¡ 11
6) e0 = 1
20)
7) log8 1 = 0
21) 6
8) log17 289 = ¡2
34) ¡
22) 5
9) log15 225=2
23) 5
35) ¡ 2
1
10) log144 12= 11) log64 2=
1 2
1 6
12) log19 361=2
1 2
1 1
36) ¡ 11
24) 512 25)
621
37) ¡ 10
1 4
26) 1000
38)
283 243 2 5
1 3
27) 121
39)
14) 3
28) 256
40) 3
13)
2401 3
204
CCBC ~ Math 083
Section 5.5
Section 5.5 - Compound Interest Objective: Calculate nal account balances using the formulas for compound and continuous interest. One application of exponential functions involves compound interest. When money is invested in an account (or borrowed as a loan), a certain amount is added to the balance. This money added to the balance is called the interest. Once that interest is added to the balance, it will earn more interest during the next compounding period. This idea of earning interest on interest is called compound interest. For example, if you invest S100 at 10% interest compounded annually, after one year you will earn S10 in interest, giving you a new balance of S110. The next year you will earn 10% of S110 or S11, giving you a new balance of S121. The third year you will earn 10% of S121 or S12.10, giving you a new balance of S133.10. This pattern will continue each year until you close the account. There are several ways interest can be paid. The rst way, as described above, is compounded annually. In this model, the interest is paid once per year. But interest can be compounded more often. Some common compounding periods include semi-annually (twice per year), quarterly (four times per year, such as quarterly taxes), monthly (12 times per year, such as a savings account), weekly (52 times per year), or even daily (365 times per year, such as some student loans). When interest is compounded in any of these ways, we can calculate the balance after any amount of time using the following formula: r nt Compound Interest Formula: A = P 1 + n A = Final Amount P = Principal (starting balance) r = Annual nterest rate (as a decimal) n = number of compounding periods per year t = time (in years) Example 1. If you take a car loan for S25000 with an annual interest rate of 6.5% compounded quarterly, no payments required for the rst ve years, what will your balance be at the end of those ve years?
0.065 45 A = 25000 1 + 4 A = 25000(1.01625)45
Identify information given: P = 25000; r = 0.065; n = 4; t = 5 Plug each value into formula; evaluate parentheses Multiply exponents
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CCBC ~ Math 083
A = 25000(1.01625)20 A = 25000(1.38041977:::) A = 34510.49 S34; 510.49
Section 5.5
Evaluate exponent Multiply Our Answer
We can also nd a missing part of the equation by using our techniques for solving equations. Example 2. What principal amount will grow to S3000 if invested at 6.5% compounded weekly for 4 years?
0.065 524 3000 = P 1 + 52 3000 = P (1.00125)524 3000 = P (1.00125)208 3000 = P (1.296719528:::) 1.296719528::: 1.296719528::: 2313.53 = P S2313.53
Identify information given: A = 3000; r = 0.065; n = 52; t = 4 Evaluate parentheses Multiply exponent Evaluate exponent Divide each side by 1.296719528::: Solution for P Our Answer
It is interesting to compare the same investments made at several dierent types of compounding periods. The next few examples do just that. Example 3. If S4000 is invested in an account paying 3% interest compounded monthly, what is the balance after 7 years?
0.03 127 A = 4000 1 + 12 A = 4000(1.0025)127 A = 4000(1.0025)84 A = 4000(1.2333548) A = 4933.42 S4933.42
Identify information given: P = 4000; r = 0.03; n = 12; t = 7 Plug each value into formula; evaluate parentheses Multiply exponents Evaluate exponent Multiply Our Answer
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CCBC ~ Math 083
Section 5.5
To investigate what happens to the balance if the compounding happens more often, we will consider the same problem, this time with interest compounded daily. Example 4. If S4000 is invested in an account paying 3% interest compounded daily, what is the balance after 7 years?
0.03 3657 A = 4000 1 + 365 A = 4000(1.00008219:::)3657 A = 4000(1.00008219:::)2555 A = 4000(1.23366741::::) A = 4934.67 S4934.67
Identify information given P = 4000; r = 0.03; n = 365; t = 7 Plug each value into formula; evaluate parentheses Multiply exponent Evaluate exponent Multiply Our Answer
While this dierence is not very large, it is still a bit higher. The table below shows the result for the same problem with dierent compounding periods. Compounding Annually Semi-Annually Quarterly Monthly Weekly Daily
Balance S4919.50 S4927.02 S4930.85 S4933.42 S4934.41 S4934.67
As the table illustrates, the more often interest is compounded, the higher the final balance will be because we are calculating interest on interest. So once interest is added into the account, it will start earning interest for the next compounding period and thus giving a higher nal balance. The next question one might consider is what is the maximum number of compounding periods possible? We actually have a way to calculate interest compounded an in nite number of times a year. This is when the interest is compounded continuously. When we see the word continuously we will know that we cannot use the rst formula. Instead we will use the following formula: Interest Compounded Continuously: A = Pert A = Final Amount P = Principal (starting balance) e = a constant approximately equal to 2.71828183:::: r = Annual Interest rate (written as a decimal) t = time (years)
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CCBC ~ Math 083
Section 5.5
The number e is a constant similar in idea to pi () in that it goes on forever without repeat or pattern, but just as pi () naturally occurs in several geometry applications, e appears in many exponential applications, continuous interest being one of them. If you have a scienti c calculator you probably have an ex button (often using the 2nd or shift key, then hit ln) that will be useful in calculating interest compounded continuously. Example 5. If S4000 is invested in an account paying 3% interest compounded continuously, what is the balance after 7 years?
A = 4000e0.037 A = 4000e0.21 A = 4000(1.23367806:::) A = 4934.71 S4934.71
Identify information given: P = 4000; r = 0.03; t = 7 Plug each value into formula; multiply exponent Evaluate e0.21 Multiply Our Answer
Albert Einstein once said that the most powerful force in the universe is compound interest. Consider the following example, illustrating how powerful compound interest can be. Example 6. If you invest S6.16 in an account paying 12% interest compounded continuously for 100 years, and that is all you have to leave your children as an inheritance, what will the nal balance be that they will receive?
A = 6.16e0.12100 A = 6.16e12 A = 6.16(162; 544.79) A = 1; 002; 569.52 S1; 002; 569.52
Identify information given P = 6.16; r = 0.12; t = 100 Plug each value into formula; multiply exponent Evaluate e12 Multiply Our Answer
In 100 years that one time investment of S6.16 investment grew to over one million dollars! That's the power of compound interest!
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CCBC ~ Math 083
Section 5.5
5.5 Practice - Compound Interest 1) Find each of the following: a. S500 invested at 4% compounded annually for 10 years. b. S600 invested at 6% compounded annually for 6 years. c. S750 invested at 3% compounded annually for 8 years. d. S1500 invested at 4% compounded semiannually for 7 years. e. S900 invested at 6% compounded semiannually for 5 years. f. S950 invested at 4% compounded semiannually for 12 years. g. S2000 invested at 5% compounded quarterly for 6 years. h. S2250 invested at 4% compounded quarterly for 9 years. i. S3500 invested at 6% compounded quarterly for 12 years. j. All of the above compounded continuously. 2) What principal will amount to S2000 if invested at 4% interest compounded semiannually for 5 years? 3) What principal will amount to S3500 if invested at 4% interest compounded quarterly for 5 years? 4) What principal will amount to S3000 if invested at 3% interest compounded semiannually for 10 years? 5) What principal will amount to S2500 if invested at 5% interest compounded semiannually for 7.5 years? 6) What principal will amount to S1750 if invested at 3% interest compounded quarterly for 5 years? 7) A thousand dollars is left in a bank savings account drawing 7% interest, compounded quarterly for 10 years. What is the balance at the end of that time? 8) A thousand dollars is left in a credit union drawing 7% compounded monthly. What is the balance at the end of 10 years? 9) S1750 is invested in an account earning 13.5% interest compounded monthly for a 2 year period. What is the balance at the end of 9 years? 10) You lend out S5500 at 10% compounded monthly. If the debt is repaid in 18 months, what is the total owed at the time of repayment? 11) A S10; 000 Treasury Bill earned 16% compounded monthly. If the bill matured in 2 years, what was it worth at maturity? 12) You borrow S25000 at 12.25% interest compounded monthly. If you are unable to make any payments the rst year, how much do you owe, excluding penalties? 13) A savings institution advertises 7% annual interest, compounded daily, How much more interest would you earn over the bank savings account or credit union in problems 7 and 8?
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CCBC ~ Math 083
Section 5.5
14) An 8.5% account earns continuous interest. If S2500 is deposited for 5 years, what is the total accumulated? 15) You lend S100 at 10% continuous interest. If you are repaid 2 months later, what is owed?
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CCBC ~ Math 083
Section 5.5
5.5 Answers - Compound Interest 1) a. 740.12; 745.91 b. 851.11; 859.99 c. 950.08; 953.44 d. 1979.22; 1984.69
f. 1528.02; 1535.27 g. 2694.70; 2699.72
j. See the second answer given in each of the problems a - i.
h. 3219.23; 3224.99
e. 1209.52; 1214.87
i. 7152.17; 7190.52
2) 1640.70
7) 2001.60
3) 2868.41
8) 2009.66
4) 2227.41
9) 2288.98
5) 1726.16
10) 6386.12
6) 1507.08
11) 13742.19
12) 28240.43 13) 12.02; 3.96
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14) 3823.98 15) 101.68
