Solve each equation by using the Square Root

ANSWER: {–8.45, –3.55}

4-5 Completing the Square Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.

2.

SOLUTION:

1.

Factor the perfect square trinomial.

SOLUTION:


Use the Square Root Property.


The solution set is {0.39, 7.61}.

The solution set is {–8.45, –3.55}.

ANSWER: {0.39, 7.61}

ANSWER: {–8.45, –3.55}

3.

SOLUTION:

2.

SOLUTION:







eSolutions Manual - Powered by Cognero

ANSWER:

ANSWER: {–12.87, –5.13}

Page 1

ANSWER: {0.39, 7.61} the Square 4-5 Completing

ANSWER: {–12.87, –5.13}

3.

4.

SOLUTION:

SOLUTION:






ANSWER: {–12.87, –5.13}

4.

SOLUTION:




ANSWER: {–2.77, –0.56}\

5. LASER LIGHT SHOW The area A in square feet 2

of a projected laser light show is given by A = 0.16d , where d is the distance from the laser to the screen in feet. At what distance will the projected laser light show have an area of 100 square feet?

SOLUTION: Substitute 100 for A, and find the value of d.


d cannot be negative.

Manual - Powered by Cognero eSolutions The solution set is {–2.77, –0.56}.

So, the distance is 25 feet. ANSWER:

Page 2

ANSWER: ANSWER: {–2.77, –0.56}\ 4-5 Completing the Square

2

25; (x – 5)

5. LASER LIGHT SHOW The area A in square feet 2

7.

of a projected laser light show is given by A = 0.16d , where d is the distance from the laser to the screen in feet. At what distance will the projected laser light show have an area of 100 square feet?

SOLUTION: To find the value of c, divide the coefficient of x by 2 and square the result. So: c = 6.25 Substitute 6.25 for c in the trinomial.

SOLUTION: Substitute 100 for A, and find the value of d.


ANSWER: 2

6.25; (x – 2.5)

d cannot be negative.

Solve each equation by completing the square.

So, the distance is 25 feet. ANSWER: 25 ft

8.

SOLUTION:

Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect square.

6.

SOLUTION: To find the value of c, divide the coefficient of x by 2 and square it. So: c = 25 Substitute 25 for c in the trinomial.


The solution set is {–4, 2}.

ANSWER: {–4, 2} ANSWER: 25; (x – 5)

2

9.

7.


SOLUTION: To find the value of c, divide the coefficient of x by 2

SOLUTION: Page 3

ANSWER: {–4, 2} 4-5 Completing the Square

ANSWER:

9.

10.

SOLUTION:

SOLUTION:



The solution set is

.

ANSWER:

The solution set is { –0.69, 2.19}.

10.

ANSWER: {–0.69, 2.19}

SOLUTION:

11.

SOLUTION:



Manual - Powered by Cognero eSolutions The solution set is { –0.69, 2.19}.

Page 4

ANSWER: {–0.69, 2.19}the Square 4-5 Completing

ANSWER: {–4.37, 1.37}

11.

12.

SOLUTION:

SOLUTION:



The solution set is

ANSWER:


13.

ANSWER: {–4.37, 1.37}

SOLUTION:

12.

SOLUTION:



ANSWER: {–6.45, –1.55}

Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.

The solution set is

eSolutions Manual - Powered by Cognero ANSWER:

14.

Page 5

ANSWER: {–6.45, –1.55} 4-5 Completing the Square

ANSWER: {–1.47, 7.47}

Solve each equation by using the Square Root Property. Round to the nearest hundredth if necessary.

16.

SOLUTION:

14.

SOLUTION:



The solution set is {–8.24, 0.24}.

ANSWER: {–8.24, 0.24}


ANSWER: {–5.16, 1.16}

17.

SOLUTION: 15.

SOLUTION:



ANSWER: {–7.65, –2.35}

ANSWER: {–1.47, 7.47}

18.

16.

SOLUTION:

SOLUTION: eSolutions Manual - Powered by Cognero


Page 6


ANSWER: {–1, 3}

18.

20.

SOLUTION:

SOLUTION:





ANSWER: {–8.24, –3.76}

ANSWER: {0.5, 4.5}

21.

19.

SOLUTION:

SOLUTION:



The solution set is {–1, 3}.


ANSWER: {–1, 3}

ANSWER: {4.67, 10.33}

20.

SOLUTION:

22.

SOLUTION:

Manual - Powered by Cognero eSolutions Use the Square Root Property.

Page 7


ANSWER: {4.67, 10.33}the Square 4-5 Completing

ANSWER: {–0.95, 3.95}

22.

24.

SOLUTION:

SOLUTION:



The solution set is {–17, –15}.

ANSWER: {–17, –15}


ANSWER: {–5.5, –1.5}

23.

25.

SOLUTION:

SOLUTION:



The solution set is {–0.95, 3.95}

The solution set is {4, 5} .

ANSWER: {–0.95, 3.95}

ANSWER: {4, 5}

24. Find the value of c that makes each trinomial a perfect square. Then write the trinomial as a perfect square.


Page 8

26.

.

ANSWER:

ANSWER: {4, 5} 4-5 Completing the Square

2

64; (x + 8)


28.

SOLUTION: To find the value of c, divide the coefficient of x by 2, and square it.

26.

SOLUTION: To find the value of c, divide the coefficient of x by 2, and square the result.

Substitute

in the trinomial.

Substitute 16 for c in the trinomial.

ANSWER:

ANSWER:

2

16; (x + 4)

27.

29.



Substitute 64 for c in the trinomial.

Substitute c = 20.25 in the trinomial.

ANSWER:

ANSWER:

2

64; (x + 8)

20.25; (x + 4.5)

28.




2

30.

SOLUTION:

Page 9

ANSWER:

ANSWER: {– 4.61, 2.61}

2

20.25; (x + 4.5) 4-5 Completing the Square


32.

30.

SOLUTION:

SOLUTION:



; The solution set is

The solution set is {– 4, – 2} .

.

ANSWER: {– 4, – 2}

ANSWER:

33.

31.

SOLUTION:

SOLUTION:



The solution set is {1, 3}.


ANSWER: {1, 3}

ANSWER: {– 4.61, 2.61}

32.


SOLUTION:

34.

SOLUTION:

Page 10

ANSWER:

ANSWER: {1, 3} 4-5 Completing the Square

34.

35.

SOLUTION:

SOLUTION:



The solution set is

.

The solution set is

.

ANSWER:

ANSWER:

35.

SOLUTION:

36.

SOLUTION:


The solution set is eSolutions Manual - Powered by Cognero

ANSWER:

. Page 11

ANSWER:

The solution set is {–1, 3}. ANSWER: {–1, 3}

4-5 Completing the Square

36.

38.

SOLUTION:

SOLUTION:

The solution set is

.

The solution set is

ANSWER: .

ANSWER:

39.

SOLUTION:

37.

SOLUTION:

The solution set is

.

[

ANSWER:

The solution set is {–1, 3}. ANSWER: {–1, 3}

40.

SOLUTION:

38.


Page 12

The solution set is

.

ANSWER:

ANSWER: ]{5 – 2i, 5 + 2i}


40.

42.

SOLUTION:

SOLUTION:

The solution set is

.

ANSWER: {3 – 3i, 3 + 3i}

The solution set is

.

ANSWER: {–0.39, 1.72}

41.

SOLUTION: 43.

SOLUTION:

The solution set is

.

ANSWER: ]{5 – 2i, 5 + 2i}


42.

SOLUTION:

The solution set is

.

ANSWER: eSolutions Manual - Powered by Cognero

Page 13

The solution set is

.

ANSWER:


43.

44.

SOLUTION:

SOLUTION:

]



ANSWER: {–0.71, 3.11}

45.

The solution set is

.

SOLUTION:

ANSWER:

The solution set is

. 44.

ANSWER:

SOLUTION:

]

46.

SOLUTION:


ANSWER: {–0.71, 3.11}


45.

Page 14

ANSWER:

ANSWER: {–0.89, 5.39}


46.

SOLUTION:

48. CCSS MODELING An architect’s blueprints call for a dining room measuring 13 feet by 13 feet. The customer would like the dining room to be a square, but with an area of 250 square feet. How much will this add to the dimensions of the room?

SOLUTION: The area of a square is given by the side length.

, where s is

Therefore:


ANSWER: {–1.39, 1.59}

Solve for x.

Therefore, about 2.81 ft should be added to the dimensions of the room.

47.

ANSWER: about 2.81 ft

SOLUTION: Solve each equation. Round to the nearest hundredth if necessary.

49.

SOLUTION: Write the equation in standard form and solve using the quadratic formula.

The quadratic formula is given by:


ANSWER: {–0.89, 5.39}

48. CCSS MODELING An architect’s blueprints call for aManual dining- Powered room measuring eSolutions by Cognero 13 feet by 13 feet. The customer would like the dining room to be a square, but with an area of 250 square feet. How much will this add to the dimensions of the room?


Page 15

dimensions of the room. ANSWER: {2.38, 4.62}

ANSWER: about 2.81 ft the Square 4-5 Completing

Solve each equation. Round to the nearest hundredth if necessary.

50.


49.






ANSWER: {–2.77, –0.56}

ANSWER: {2.38, 4.62}

51.

50.







ANSWER:

The solution set is {0.26, –1.26}.

ANSWER:

Page 16

ANSWER: {–1.26, 0.26}


52.

51.






The solution set is {0.26, –1.26}.

ANSWER: {–1.26, 0.26}

52.



ANSWER: {–1.44, 0.24}

53. FIREWORKS A firework’s distance d meters from 2

the ground is given by d = –1.5t + 25t, where t is the number of seconds after the firework has been lit.

a. How many seconds have passed since the firework was lit when the firework explodes if it explodes at the maximum height of its path?

b. What is the height of the firework when it explodes?

SOLUTION: a. The maximum occurs at the vertex. Find the x-coordinate of the vertex.

The x-coordinate of the vertex is given by

.

Here, a = –1.5 and b = 25. Therefore:


Manual - Powered by Cognero eSolutions ANSWER:

Page 17

a.

seconds


b. about 104.2 ft


53. FIREWORKS A firework’s distance d meters from 2

the ground is given by d = –1.5t + 25t, where t is the number of seconds after the firework has been lit.

54.

a. How many seconds have passed since the firework was lit when the firework explodes if it explodes at the maximum height of its path?

SOLUTION: To find the value of c, divide the coefficient of x by 2 and square it.

b. What is the height of the firework when it explodes?

SOLUTION: a. The maximum occurs at the vertex. Find the x-coordinate of the vertex.

Substitute the value of c in the trinomial.

The x-coordinate of the vertex is given by

.

Here, a = –1.5 and b = 25. Therefore:

ANSWER: 2

0.1225; (x + 0.35)

55.

That is


seconds should have passed when the

firework explodes at the maximum height.

b. Substitute

for t in the equation and solve for d.


The height of the firework is about 104.2 ft.

ANSWER: ANSWER: a.

seconds

2

2.56; (x – 1.6)

b. about 104.2 ft

FindManual the value of cbythat makes eSolutions - Powered Cognero

each trinomial a perfect square. Then write the trinomial as a perfect square.

56.

Page 18

SOLUTION: To find the value of c, divide the coefficient of x by 2

• Extend point Q.

ANSWER:

until it intersects the arc. Label this

2

• Construct rectangle ARQD.

2.56; (x – 1.6) 4-5 Completing the Square

b. ALGEBRAIC Let AD = x and CQ = 1. Use 56.

completing the square to solve

for x.


c. TABULAR Make a table of x and values for CQ = 2, 3, and 4.

d. VERBAL What do you notice about the xvalues? Write an equation you could use to determine x for CQ = n, where n is a nonzero real number.


SOLUTION: a.

ANSWER:

57. MULTIPLE REPRESENTATIONS In this problem, you will use quadratics to investigate golden rectangles and the golden ratio. a. GEOMETRIC

Therefore, DQ = x + 1.

• Draw square ABCD.

Substitute.

• Locate the midpoint of

b. AD = x. So, QR = x CQ = 1

. Label the midpoint P.

• Draw

• Construct an arc with a radius of from B clockwise past the bottom of the square.

• Extend point Q.

until it intersects the arc. Label this

• Construct rectangle ARQD.

b. ALGEBRAIC Let AD = x and CQ = 1. Use completing the square to solve

x cannot be negative. So:

for x.


c. TABULAR Make a table of x and values for CQ = 2, 3, and 4.

c.

Page 19

d. Sample answer: the x-values are multiples of

x cannot be negative. 4-5 Completing the Square So:

58. ERROR ANALYSIS Alonso and Aida are solving 2 x + 8x – 20 = 0 by completing the square. Is either of them correct? Explain your reasoning.

c.

d. Sample answer: the x-values are multiples of ;

ANSWER: a.

b.

SOLUTION: Alonso; Aida did not add 16 to each side; she added it only to the left side.

c.

ANSWER: Alonso; Aida did not add 16 to each side; she added it only to the left side.

d. Sample answer: the x-values are multiples of

2

58. ERROR ANALYSIS Alonso and Aida are solving 2 x + 8x – 20 = 0 by completing the square. Is either of them correct? Explain your reasoning.

59. CHALLENGE Solve x + bx + c = 0 by completing the square. Your answer will be an expression for x in terms of b and c.

SOLUTION:


Page 20

ANSWER: Alonso; Aida did not add 16 to each side; she added it only to the the left side. 4-5 Completing Square

e . 16(x + 2)2 = 0

2

2

f. (x + 4) = (x + 6)

2

59. CHALLENGE Solve x + bx + c = 0 by completing the square. Your answer will be an expression for x in terms of b and c.

SOLUTION:


SOLUTION: a. 2; rational; 16 is a perfect square so x + 2 and x are rational. b. 2; rational; 16 is a perfect square so x – 2 and x are rational. c. 2; complex; If the opposite of square is positive, the square is negative. The square root of a negative number is complex. d. 2; real; The square must equal 20. Since that is positive but not a perfect square, the solutions will be real but not rational. e . 1; rational; The expression must be equal to 0 and only -2 makes the expression equal to 0. f. 1; rational; The expressions (x + 4) and (x + 6) must either be equal or opposites. No value makes them equal, -5 makes them opposites. The only solution is -5.

ANSWER: a. 2; rational; 16 is a perfect square so x + 2 and x are rational. b. 2; rational; 16 is a perfect square so x – 2 and x are rational. c. 2; complex; If the opposite of square is positive, the square is negative. The square root of a negative number is complex. d. 2; real; The square must equal 20. Since that is positive but not a perfect square, the solutions will be real but not rational. e . 1; rational; The expression must be equal to 0 and only -2 makes the expression equal to 0. f. 1; rational; The expressions (x + 4) and (x + 6) must either be equal or opposites. No value makes them equal, -5 makes them opposites. The only solution is -5.

e . 16(x + 2)2 = 0

ANSWER:

60. CCSS ARGUMENTS Without solving, determine how many unique solutions there are for each equation. Are they rational, real, or complex? Justify your reasoning. a. (x + 2)2 = 16

2

b. (x - 2) = 16

c. -(x - 2)2 = 16 2

d. 36 - (x - 2) = 16

2

2

f. (x + 4) = (x + 6)

SOLUTION: a. 2; rational; 16 is a perfect square so x + 2 and x are rational.


61. OPEN ENDED Write a perfect square trinomial equation in which the linear coefficient is negative and the constant term is a fraction. Then solve the equation.

SOLUTION:

Page 21

f. 1; rational; The expressions (x + 4) and (x + 6) must either be equal or opposites. No value makes them equal, -5 makes them opposites. The only solution is -5.the Square 4-5 Completing

Completing the square allows you to rewrite one side of a quadratic equation in the form of a perfect square. Once in this form, the equation can be solved by using the Square Root Property.

61. OPEN ENDED Write a perfect square trinomial equation in which the linear coefficient is negative and the constant term is a fraction. Then solve the equation.

2

2

63. SAT/ACT If x + y = 2xy, then y must equal

A –1

B 0

SOLUTION: Sample answer:

C 1 D –x

E x

The solution set is

.

SOLUTION:

ANSWER: Sample answer:

62. WRITING IN MATH Explain what it means to complete the square. Include a description of the steps you would take.

SOLUTION: Completing the square allows you to rewrite one side of a quadratic equation in the form of a perfect square. Once in this form, the equation can be solved by using the Square Root Property.

The correct choice is E.

ANSWER: E

64. GEOMETRY Find the area of the shaded region.

ANSWER: Completing the square allows you to rewrite one side of a quadratic equation in the form of a perfect square. Once in this form, the equation can be solved by using the Square Root Property.

2

F 14 m

2

2

63. SAT/ACT If x + y = 2xy, then y must equal

A –1

B 0

C 1

G 18 m2

2

H 42 m

J 60 m2

SOLUTION: The area of the outer rectangle is given by:

D –x

eSolutions E xManual - Powered by Cognero

Page 22

SOLUTION:

The area of the inner rectangle is given by:

ANSWER: E 4-5 Completing the Square

ANSWER: H

64. GEOMETRY Find the area of the shaded region.

65. SHORT RESPONSE What value of c should be used to solve the following equation by completing the square?

SOLUTION:

2

F 14 m

G 18 m2

2

H 42 m

ANSWER: 125

J 60 m2

SOLUTION: The area of the outer rectangle is given by:

2

66. If 5 – 3i is a solution for x + ax + b = 0, where a and b are real numbers, what is the value of b?

A 10

The area of the inner rectangle is given by:

B 14

C 34

D 40

The area of the shaded region is

.

SOLUTION: Substitute 5 – 3i for x in the equation.

The correct choice is H.

Equate the real and the imaginary parts. ANSWER: H

65. SHORT RESPONSE What value of c should be used to solve the following equation by completing the square?

And: Substitute a = –10.

The correct choice is C.

SOLUTION:

ANSWER: C


Page 23

ANSWER: 39 + 80i

ANSWER: 125 4-5 Completing the Square

2

66. If 5 – 3i is a solution for x + ax + b = 0, where a and b are real numbers, what is the value of b?

68.

SOLUTION:

A 10

B 14

C 34

ANSWER: 24 – 34i

D 40

SOLUTION: Substitute 5 – 3i for x in the equation.

69.

SOLUTION:

Equate the real and the imaginary parts.

And: Substitute a = –10.

ANSWER:

The correct choice is C. ANSWER: C

Simplify.

Write a quadratic equation in standard form with the given root(s).

67.

70.

SOLUTION:

SOLUTION:

ANSWER: 39 + 80i

ANSWER: 2

eSolutions Manual - Powered by Cognero 68.

20x – 31x + 12 = 0

Page 24

ANSWER:

ANSWER: 2

5x – 28x – 12 = 0


Write a quadratic equation in standard form with the given root(s).

72.

70.

SOLUTION:

SOLUTION:

ANSWER:

ANSWER:

2

20x – 31x + 12 = 0

71.

SOLUTION:

2

28x + 31x + 6 = 0

73. TRAVEL Yoko is going with the Spanish Club to Costa Rica. She buys 10 traveler’s checks in denominations of $20, $50, and $100, totaling $370. She has twice as many $20 checks as $50 checks. How many of each denomination of traveler’s checks does she have?

SOLUTION: Let x, y, and z represent the number of $20, $50, and $100 checks respectively.

So: x + y + z = 10

And: 20x + 50y + 100z = 370

ANSWER: 2

Also: x = 2y

5x – 28x – 12 = 0

Solve the equations.

x = 6, y = 3, and z = 1.

72.

Therefore, Yoko has 1-$100, 3-$50, and 6-$20 checks.

SOLUTION:

ANSWER: 1 $100, 3 $50, and 6 $20 checks


74. SHOPPING Main St. Media sells all DVDs for one price and all books for another price. Alex bought 4 DVDs and 6 books for $170, while Matt boughtPage 3 25 DVDs and 8 books for $180. What is the cost of a DVD and the cost of a book?

ANSWER: 1 $100, 3 $50,the andSquare 6 $20 checks 4-5 Completing

ANSWER: DVD: $20; book: $15

74. SHOPPING Main St. Media sells all DVDs for one price and all books for another price. Alex bought 4 DVDs and 6 books for $170, while Matt bought 3 DVDs and 8 books for $180. What is the cost of a DVD and the cost of a book?

Graph each inequality.

75.

SOLUTION: The boundary of the graph is the graph of y = 4x – 3. Since the inequality symbol is , the line will be solid. Test the point (0, 0).

SOLUTION: Let x and y are the cost of a DVD and a book.

Multiply the first and the second equation by 3 and – 4 respectively then add.

Shade the region that includes (0, 0).

The graph of the inequality

is:

Substitute 15 for y in the first equation and solve for x.

ANSWER:

The cost of a DVD is $20.

The cost of a book is $15.

ANSWER: DVD: $20; book: $15

Graph each inequality.

76.

75.

SOLUTION: The boundary of the graph is the graph of y = 4x – 3. Since the inequality symbol is , the line will be solid. Test the point (0, 0).

SOLUTION: The boundary of the graph is the graph of 2x – 3y = 6. Since the inequality symbol is <, the line will be dashed. Test the point (0, 0).


Page 26

Shade the region that contains (0, 0).


76.

77.

SOLUTION: The boundary of the graph is the graph of 2x – 3y = 6. Since the inequality symbol is <, the line will be dashed. Test the point (0, 0).

SOLUTION: The boundary of the graph is the graph of . Since the inequality symbol is line will be solid.

, the

Test the point (0, 0).

Shade the region that contains (0, 0). The graph of the inequality is:

Shade the region that does not contain (0, 0). The graph of the inequality is:

ANSWER:

ANSWER:

77.

SOLUTION: The boundary of the graph is the graph of . Since the inequality symbol is line will be solid.

Write the piecewise function shown in each graph. , the

Test the point (0, 0).

78.


Shade the region that does not contain (0, 0).

Page 27 SOLUTION: The left portion is the graph of the constant function f (x) = –4. There is a dot at –2, so the constant


Write the piecewise function shown in each graph.

79.

78.

SOLUTION: The left portion is the graph of the constant function f (x) = –8. There is a dot at –2, so the constant function is defined for x ≤ –2. The center portion is the graph of f (x) = 2x – 6. There is circle at –2 and a dot at 5, so the function is defined for –2 < x ≤ 5. The right portion is the graph of f (x) = 2x – 18. There is a circle at –5, so the function is defined for > 5. The piece-wise function is:

SOLUTION: The left portion is the graph of the constant function f (x) = –4. There is a dot at –2, so the constant function is defined for x ≤ –2. The center portion is the graph of f (x) = –2x + 3. There are circles at –2 and 1, so the function is defined for –2 < x < 1. The right portion is the graph of f (x) = x – 5. There is a dot at 2, so the function is defined for x ≥ 2. The piece-wise function is:

ANSWER:

ANSWER:

80.

79.

SOLUTION: The left portion is the graph of the constant function f eSolutions - Powered Cognero (x) =Manual is abydot at –2, so the constant –8. There function is defined for x ≤ –2.

SOLUTION: The left portion is the graph of f (x) = x + 12. There is dot at –6, so the function is defined for x ≤ –6. The center portion is the graph of the constant function f (x) = 8. There are circles at –6 and 2,Page so 28 the constant function is defined for –6 < x < 2. The right portion is the graph of f (x) = –2.5x + 15.


ANSWER: –4

82. a = –2, b = –7, c = 3

SOLUTION: Substitute a = –2, b = –7, and c = 3.

80.

SOLUTION: The left portion is the graph of f (x) = x + 12. There is dot at –6, so the function is defined for x ≤ –6. The center portion is the graph of the constant function f (x) = 8. There are circles at –6 and 2, so the constant function is defined for –6 < x < 2. The right portion is the graph of f (x) = –2.5x + 15. There is a dot at 2, so the function is defined for x ≥ 2. The piece-wise function is:

ANSWER: 73

83. a = –5, b = –8, c = –10

SOLUTION: Substitute a = –5, b = –8, and c = –10.

ANSWER:

ANSWER: –136

2

Evaluate b – 4ac for the given values of a, b, and c.

81. a = 5, b = 6, c = 2

SOLUTION: Substitute a = 5, b = 6, and c = 2.

ANSWER: –4

82. a = –2, b = –7, c = 3


SOLUTION: Substitute a = –2, b = –7, and c = 3.

Page 29

Solve each equation by using the Square Root

Recommend Documents