# MHR Principles of Mathematics 10 Solutions 1 - Weebly

MHR • Principles of Mathematics 10 Solutions 9. Chapter 6 Section 1 Question 6 Page 270 . a) () () 2 2 222 222 2 10 20 10 20 10 5 5 20 10 5 5 20 55 yx...

Chapter 6

Question 1

Page 262

a)

b)

c)

d)

MHR • Principles of Mathematics 10 Solutions

1

Question 2

Page 262

a)

b)

c)

d)

6

2 MHR • Principles of Mathematics 10 Solutions

Question 3

Page 262

a) The square roots of 100 are 10 and –10.

b) The square roots of 36 are 6 and –6.

c) The square roots of 75 are 8.7 and –8.7.

d) The square root of –9 is not a real number.

Question 4

Page 262 b) ± 32 + 42 = ± 9 + 16

a) ± 25 − 16 = ± 9

= ±3

= ± 25 = ±5 d) ± 82 − 3 ( 5 ) = ± 64 − 15

c) ± 62 − 11 = ± 36 − 11

= ± 25 = ±5

= ± 49 = ±7

Question 5

(

a) 2 x 2 + 10 x + 12 = 2 x 2 + 5 x + 6

)

Page 262 b) x 2 + 16 x + 28 = ( x + 14 )( x + 2 )

= 2 ( x + 3)( x + 2 )

c) x 2 − 16 x + 63 = ( x − 9 )( x − 7 )

(

d) 4 x 2 − 36 = 4 x 2 − 9

)

= 4 ( x − 3)( x + 3)

e) 81x 2 − 49 = ( 9 x − 7 )( 9 x + 7 )

(

f) 3 x 2 − 6 x − 24 = 3 x 2 − 2 x − 8

)

= 3 ( x − 4 )( x + 2 )

g) 16 x 2 − 8 x + 1 = ( 4 x − 1)

2

h) 12 x 2 + 19 x + 4 = 12 x 2 + 16 x + 3 x + 4

(

)

= 12 x 2 + 16 x + ( 3 x + 4 ) = 4 x ( 3x + 4 ) + ( 3x + 4 ) = ( 3 x + 4 )( 4 x + 1)

MHR • Principles of Mathematics 10 Solutions

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Question 6

Page 262

(

a) x 2 − 4 x − 32 = ( x − 8 )( x + 4 )

b) 2 x 2 + 12 x + 18 = 2 x 2 + 6 x + 9

= 2 ( x + 3)

2

d) 144 x 2 − 312 x + 169 = (12 x − 13)

c) 64 x 2 − 1 = ( 8 x − 1)( 8 x + 1) e) 49 x 2 + 70 x + 25 = ( 7 x + 5 )

)

2

2

f) 2 x 2 − 9 x − 5 = 2 x 2 − 10 x + x − 5

(

)

= 2 x 2 − 10 x + ( x − 5 ) = 2 x ( x − 5) + ( x − 5) = ( x − 5 )( 2 x + 1)

(

h) 6 x 2 + 21x + 9 = 3 2 x 2 + 7 x + 3

g) 9 x 2 − 18 x + 8 = 9 x 2 − 12 x − 6 x + 8 = ( 9 x − 12 x ) + ( −6 x + 8 ) 2

)

= 3 ⎡⎣ 2 x 2 + 6 x + x + 3⎤⎦

= 3x ( 3x − 4 ) − 2 ( 3x − 4 )

(

)

= 3 ⎡⎣ 2 x 2 + 6 x + ( x + 3) ⎤⎦

= ( 3 x − 4 )( 3 x − 2 )

= 3 ⎡⎣ 2 x ( x + 3) + ( x + 3) ⎤⎦ = 3 ( x + 3)( 2 x + 1)

Question 7

Page 262

a) The first number is x. The second number is

1 x. 2

b) The area of the original garden is A. The area of the new garden is 3A. c) The number of price reductions is n. The new price, in dollars, is 12 – n. d) The first number is x. The second number is x2 + (x + 1)2. e) The width is w. The length is 2w – 3.

Question 8

Page 262

a) x + y = 100

b) 2(l + w) = 50

c) P = 8w

d) xy = ( 2 x )

= 4 x2 e) x + y + 3 = 12

4 MHR • Principles of Mathematics 10 Solutions

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Chapter 6 Section 1:

Maxima and Minima

Chapter 6 Section 1

Question 1

Page 270

a) y = x 2 + 2 x + 5

= ( x + 1) + 4 2

MHR • Principles of Mathematics 10 Solutions

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b) y = x 2 + 4 x + 7

= ( x + 2) + 3 2

6 MHR • Principles of Mathematics 10 Solutions

c) y = x 2 + 6 x + 3

= ( x + 3) − 6 2

MHR • Principles of Mathematics 10 Solutions

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Chapter 6 Section 1

Question 2

a) c = 9; x 2 + 6 x + 9 = ( x + 3)

b) c = 49; x 2 + 14 x + 49 = ( x + 7 )

2

c) c = 36; x 2 − 12 x + 36 = ( x − 6 ) e) c = 1; x 2 + 2 x + 1 = ( x + 1)

Page 270

2

d) c = 25; x 2 − 10 x + 25 = ( x − 5)

Question 3

a) y = x 2 + 6 x − 1

Page 270 b) y = x 2 + 2 x + 7

= ( x2 + 6x ) − 1

= ( x2 + 2x ) + 7

= ( x 2 + 6 x + 32 ) − 32 − 1

= ( x 2 + 2 x + 12 ) − 12 + 7

= ( x 2 + 6 x + 32 − 32 ) − 1

= ( x 2 + 2 x + 12 − 12 ) + 7

= ( x + 3) − 10

= ( x + 1) + 6

c) y = x 2 + 10 x + 20

d) y = x 2 + 2 x − 1

2

2

= ( x 2 + 10 x ) + 20

= ( x2 + 2x ) − 1

= ( x 2 + 10 x + 52 ) − 52 + 20

= ( x 2 + 2 x + 12 ) − 12 − 1

= ( x 2 + 10 x + 52 − 52 ) + 20

= ( x 2 + 2 x + 12 − 12 ) − 1

= ( x + 5) − 5

= ( x + 1) − 2

2

2

e) y = x 2 − 6 x − 4

f) y = x 2 − 8 x − 2

= ( x2 − 6x ) − 4

= ( x2 − 8x ) − 2

)

= x 2 − 6 x + ( −3) − ( −3) − 4 2

2

2

)

− 6 x + ( −3) − ( −3) − 4 2

2

= ( x − 3) − 13

( = (x

g) y = x 2 − 12 x + 8

)

= x 2 − 12 x + ( −6 ) − ( −6 ) + 8 2

2

2

2

2

)

− 12 x + ( −6 ) − ( −6 ) + 8 2

2

= ( x − 6 ) − 28 2

8 MHR • Principles of Mathematics 10 Solutions

2

)

− 8 x + ( −4 ) − ( −4 ) − 2 2

2

= ( x 2 − 12 x ) + 8

)

= x 2 − 8 x + ( −4 ) − ( −4 ) − 2

= ( x − 4 ) − 18

2

( = (x

2

f) c = 1600; x 2 − 80 x + 1600 = ( x − 40 )

2

Chapter 6 Section 1

( = (x

2

2

2

Chapter 6 Section 1

Question 4

a) y = x 2 + 6 x + 2

Page 270 b) y = x 2 + 12 x + 30

= ( x2 + 6x ) + 2

= ( x 2 + 12 x ) + 30

= ( x 2 + 6 x + 32 ) − 32 + 2

= ( x 2 + 12 x + 62 ) − 62 + 30

= ( x 2 + 6 x + 32 − 32 ) + 2

= ( x 2 + 12 x + 62 − 62 ) + 30 = ( x + 6) − 6

= ( x + 3) − 7 2

2

The vertex is (–3, –7).

The vertex is (–6, –6).

c) y = x 2 − 8 x + 13

d) y = x 2 − 6 x + 17

= ( x 2 − 8 x ) + 13

( = (x

= ( x 2 − 6 x ) + 17

)

( = (x

= x 2 − 8 x + ( −4 ) − ( −4 ) + 13 2

2

2

)

− 8 x + ( −4 ) − ( −4 ) + 13 2

2

= ( x − 4) − 3

2

2

2

)

− 6 x + ( −3) − ( −3) + 17 2

2

= ( x − 3) + 8

2

2

The vertex is (4, –3). Chapter 6 Section 1

)

= x 2 − 6 x + ( −3) − ( −3) + 17

The vertex is (3, 8). Question 5

Page 270

a) y = ( x − 5) + 1 2

The vertex is (5, 1). Graph D b) y = ( x − 1) + 4 2

The vertex is (1, 4). Graph A c) y = ( x + 1) + 4 2

The vertex is (–1, 4). Graph B d) y = ( x + 5) − 1 2

The vertex is (–5, –1). Graph C

MHR • Principles of Mathematics 10 Solutions

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Chapter 6 Section 1

Question 6

Page 270

a) y = x 2 + 10 x + 20

= ( x 2 + 10 x ) + 20

= ( x 2 + 10 x + 52 − 52 ) + 20 = ( x 2 + 10 x + 52 ) − 52 + 20 = ( x + 5) − 5 2

The vertex is (–5, –5).

b) y = x 2 − 16 x + 60 = ( x 2 − 16 x ) + 60

( = (x

)

= x 2 − 16 x + ( −8 ) − ( −8 ) + 60 2

2

2

)

− 16 x + ( −8 ) − ( −8 ) + 60 2

2

= ( x − 8) − 4 2

The vertex is (8, –4).

10 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 1

Question 7

a) y = − x 2 + 80 x − 100

b) y = − x 2 − 6 x + 4

= − ( x 2 − 80 x ) − 100

(

Page 270

= − ( x2 + 6x ) + 4

)

= − x 2 − 80 x + ( −40 ) − ( −40 ) − 100 2

2

= − ( x 2 − 80 x + 1600 ) − ( −1)( −40 ) − 100 2

= − ( x 2 + 6 x + 32 − 32 ) + 4

= − ( x 2 + 6 x + 9 ) − ( −1) ( 32 ) + 4 = − ( x + 3) + 13 2

= − ( x − 40 ) + 1500 2

d) y = 2 x 2 − 16 x + 15

c) y = 3x 2 + 90 x + 50

= 2 ( x 2 − 8 x ) + 15

= 3 ( x 2 + 30 x ) + 50

= 3 ( x 2 + 30 x + 152 − 152 ) + 50

= 3 ( x + 30 x + 225) − 3 (15 ) + 50 2

2

= 3 ( x + 15) − 625 2

(

)

= 2 x 2 − 8 x + ( −4 ) − ( −4 ) + 15 2

2

= 2 ( x 2 − 8 x + 16 ) − 2 ( −4 ) + 15 2

= 2 ( x − 4 ) − 17 2

e) y = −7 x 2 + 14 x − 3

= −7 ( x 2 − 2 x ) − 3

(

)

= −7 x 2 − 2 x + ( −1) − ( −1) − 3 2

2

= −7 ( x 2 − 2 x + 1) − ( −7 )( −1) − 3 2

= −7 ( x − 1) + 4 2

MHR • Principles of Mathematics 10 Solutions

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Chapter 6 Section 1

Question 8

b) y = − x 2 + 14 x − 50

a) y = − x 2 − 10 x − 9

= − ( x 2 − 14 x ) − 50

= − ( x 2 + 10 x ) − 9

(

= − ( x 2 + 10 x + 52 − 52 ) − 9

)

= − x 2 − 14 x + ( −7 ) − ( −7 ) − 50

= − ( x + 10 x + 25) − ( −1) ( 5 ) − 9 2

Page 271

2

2

= − ( x 2 − 14 x + 49 ) − ( −1)( −7 ) − 50

2

2

= − ( x + 5) + 16 2

= − ( x − 7) − 1 2

The maximum point is at (–5, 16).

The maximum point is at (7, –1).

c) y = 2 x 2 + 120 x + 75

d) y = 3x 2 − 24 x + 10

(

= 2 ( x 2 + 60 x ) + 75

)

= 3 x 2 − 8 x + 10

= 2 ( x 2 + 60 x + 302 − 302 ) + 75

(

)

= 3 x 2 − 8 x + ( −4 ) − ( −4 ) + 10 2

2

= 2 ( x + 60 x + 900 ) − 2 ( 30 ) + 75

= 3 x 2 − 8 x + 16 − 3 ( −4 ) + 10

= 2 ( x + 30 ) − 1725

= 3 ( x − 4 ) − 38

2

2

2

(

)

2

2

The minimum point is at (–30, –1725). The minimum point is at (4, –38). e) y = −5 x 2 − 200 x − 120

= −5 ( x 2 + 40 x ) − 120

= −5 ( x 2 + 40 x + 202 − 202 ) − 120

= −5 ( x 2 + 40 x + 400 ) − ( −5) ( 202 ) − 120 = −5 ( x + 20 ) + 1880 2

The maximum point is at (–20, 1880).

12 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 1

Question 9

Page 271

a) The minimum point is at (–3, –10).

b) The minimum point is at (3.8, 3.5).

c) The maximum point is at (1.3, –2.3).

d) The minimum point is at (0.1, 0.5).

e) The maximum point is at (0.8, 23.3).

f) The minimum point is at (–0.1, 12.9).

MHR • Principles of Mathematics 10 Solutions

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Chapter 6 Section 1

Question 10

Page 271

a) y = − x 2 − 2 x − 6

= − ( x2 + 2 x ) − 6

= − ( x 2 + 2 x + 12 − 12 ) − 6

= − ( x 2 + 2 x + 25) − ( −1) (12 ) − 6 = − ( x + 1) − 5 2

The vertex is (–1, –5). Other points will vary. Possible other points are (–2, –6) and (0, –6). b) y = 4 x 2 + 24 x + 41

= 4 ( x 2 + 6 x ) + 41

= 4 ( x 2 + 6 x + 32 − 32 ) + 41

= 4 ( x 2 + 6 x + 9 ) − 4 ( 32 ) + 41 = 4 ( x + 3) + 5 2

The vertex is (–3, 5). Other points will vary. Possible other points are (–2, 9) and (–4, 9). c) y = 5 x 2 − 30 x + 41

= 5 ( x 2 − 6 x ) + 41

(

)

= 5 x 2 − 6 x + ( −3) − ( −3) + 41 2

2

= 5 ( x 2 − 6 x + 9 ) − 5 ( −3) + 41 2

= 5 ( x − 3) − 4 2

The vertex is (3, –4). Other points will vary. Possible other points are (2, 1) and (4, 1). d) y = −3x 2 + 12 x − 13

= −3 ( x 2 − 4 x ) − 13

(

)

= −3 x 2 − 4 x + ( −2 ) − ( −2 ) − 13 2

2

= −3 ( x 2 − 4 x + 4 ) − ( −3)( −2 ) − 13 2

= −3 ( x − 2 ) − 1 2

The vertex is (2, –1). Other points will vary. Possible other points are (1, –4) and (3, –4).

14 MHR • Principles of Mathematics 10 Solutions

e) y = 2 x 2 + 8 x + 3

= 2 ( x2 + 4 x ) + 3

= 2 ( x 2 + 4 x + 22 − 22 ) + 3

= 2 ( x 2 + 4 x + 4 ) − 2 ( 22 ) + 3 = 2 ( x + 2) − 5 2

The vertex is (–2, –5). Other points will vary. Possible other points are (0, 3) and (–3, –3).

Chapter 6 Section 1

Question 11

Page 271

a) y = −2 x 2 − 3x + 7

= −2 ( x 2 + 1.5 x ) + 7

= −2 ( x 2 + 1.5 x + 0.752 − 0.752 ) + 7

= −2 ( x 2 + 1.5 x + 0.752 ) − ( −2 ) ( 0.752 ) + 7 = −2 ( x + 0.75) + 8.125 2

The vertex is (–0.75, 8.125). Other points will vary. Possible other points are (0, 7) and (–1, 8). b) y = 3x 2 − 9 x + 11

= 3 ( x 2 − 3x ) + 11

= 3 ( x 2 − 3x + 1.52 − 1.52 ) + 11

= 3 ( x 2 − 3x + 1.52 ) − 3 (1.52 ) + 11 = 3 ( x − 1.5) + 4.25 2

The vertex is (1.5, 4.25). Other points will vary. Possible other points are (1, 5) and (2, 5). c) y = − x 2 + 8 x − 10

= − ( x 2 − 8 x ) − 10

(

)

= − x 2 − 8 x + ( −4 ) − ( −4 ) − 10 2

2

= − ( x 2 − 8 x + 16 ) − ( −1)( −4 ) − 10 2

= − ( x − 4) + 6 2

The vertex is (4, 6). Other points will vary. Possible other points are (3, 5) and (6, 2).

MHR • Principles of Mathematics 10 Solutions

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d) y = 4 x 2 − 16 x + 11

= 4 ( x 2 − 4 x ) + 11

(

)

= 4 x 2 − 4 x + ( −2 ) − ( −2 ) + 11 2

2

= 4 ( x 2 − 4 x + 4 ) − 4 ( −2 ) + 11 2

= 4 ( x − 2) − 5 2

The vertex is (2, –5). Other points will vary. Possible other points are (1, –1) and (3, –1). e) y = −5 x 2 − 30 x − 48

= −5 ( x 2 + 6 x ) − 48

= −5 ( x 2 + 6 x + 32 − 32 ) − 48

= −5 ( x 2 + 6 x + 32 ) − ( −5) ( 32 ) − 48 = −5 ( x + 3) − 3 2

The vertex is (–3, –3). Other points will vary. Possible other points are (–4, –8) and (–2, –8).

Chapter 6 Section 1

Question 12

Page 271

y = − x2 + 4 x + 1 = − ( x2 − 4 x ) + 1

(

)

= − x 2 − 4 x + ( −2 ) − ( −2 ) + 1 2

2

= − ( x 2 − 4 x + 4 ) − ( −1)( −2 ) + 1 2

= − ( x − 2) + 5 2

The maximum height of 5 m occurs at a horizontal distance of 2 m.

Chapter 6 Section 1

Question 13

Page 271

For h = −4.9t 2 + 10t + 1 , the maximum height of 6.1 m occurs at time t = 1.0 s.

For h = −0.0163x 2 + 0.577 x + 1 , the maximum height of 6.1 m occurs at a horizontal distance of x = 17.7 m.

16 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 1

Question 14

Page 271

y = − x2 + 2 x + 3 = − ( x2 − 2 x ) + 3

(

)

= − x 2 − 2 x + ( −1) − ( −1) + 3 2

2

= − ( x 2 − 2 x + 1) − ( −1)( −1) + 3 2

= − ( x − 1) + 4 2

The diver's maximum height is 4 m. Chapter 6 Section 1

Question 15

Page 271

C = 2t 2 − 84t + 1025 = 2 ( t 2 − 42t ) + 1025

= 2 ( t 2 − 42t + 212 − 212 ) + 1025

= 2 ( t 2 − 42t + 212 ) − 2 ( 212 ) + 1025 = 2 ( t − 21) + 143 2

The minimum cost of running the machine is \$143 at a running time of 21 h. Chapter 6 Section 1

Question 16

Page 272

a) The price of a garden ornament is 4 – 0.5x. The number of garden ornaments sold is 120 + 20x. b) R = ( 4 − 0.5 x )(120 + 20 x ) c) R = ( 4 − 0.5 x )(120 + 20 x )

= −10 x 2 + 20 x + 480 = −10 ( x 2 − 2 x ) + 480

(

)

= −10 x 2 − 2 x + ( −1) − ( −1) + 480 2

2

= −10 ( x 2 − 2 x + 1) − ( −10 )( −1) + 480 2

= −10 ( x − 1) + 490 2

To maximize revenue at \$490, the artisan should charge \$3.50 per ornament. d)

Both forms of the equation produce the same graph.

MHR • Principles of Mathematics 10 Solutions

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Chapter 6 Section 1

Question 17

a) y = 1.5 x 2 + 6 x − 7

Page 272 b) y = −0.1x 2 − 2 x + 1

= 1.5 ( x 2 + 4 x ) − 7

= −0.1( x 2 + 20 x ) + 1

= 1.5 ( x 2 + 4 x + 4 ) − 1.5 ( 22 ) − 7

= −0.1( x 2 + 20 x + 102 ) − ( −0.1) (102 ) + 1

= 1.5 ( x + 2 ) − 13

= −0.1( x + 10 ) + 11

= 1.5 ( x 2 + 4 x + 22 − 22 ) − 7

= −0.1( x 2 + 20 x + 102 − 102 ) + 1

2

2

The minimum point is at (–2, –13).

The maximum point is at (–10, 11).

c) y = 0.3x 2 + 3x

d) y = −1.25 x 2 + 5 x

= 0.3 ( x 2 + 10 x )

(

= −1.25 x 2 − 4 x

= 0.3 ( x 2 + 10 x + 52 − 52 )

= 0.3 ( x + 10 x + 5 ) − 0.3 ( 5 2

2

2

( = −1.25 ( x

)

= −1.25 x 2 − 4 x + ( −2 ) − ( −2 )

)

= 0.3 ( x + 5) − 7.5 2

2

2

)

2

2

e) y = 0.5 x 2 − 6 x + 12

f) y = −0.02 x 2 − 0.6 x − 9

)

= −0.02 ( x 2 + 30 x ) − 9

= 0.5 x 2 − 12 x + 12

)

= 0.5 x 2 − 12 x + ( −6 ) − ( −6 ) + 12 2

2

2

)

− 12 x + ( −6 ) − 0.5 ( −6 ) + 12 2

2

= −0.02 ( x 2 + 30 x + 152 − 152 ) − 9

= −0.02 ( x 2 + 30 x + 152 ) − ( −0.02 ) (152 ) − 9 = −0.02 ( x + 15) − 4.5 2

= 0.5 ( x − 6 ) − 6 2

The minimum point is at (6, –6).

Chapter 6 Section 1

The maximum point is at (–15, –4.5).

Question 18

Page 272

y = 0.2 x 2 − 1.6 x + 4.2

(

)

= 0.2 x 2 − 8 x + 4.2

( = 0.2 ( x

)

= 0.2 x 2 − 8 x + ( −4 ) − ( −4 ) + 4.2 2

2

2

)

− 8 x + ( −4 ) − 0.2 ( −4 ) + 4.2 2

2

= −1.25 ( x − 2 ) + 5

The maximum point is at (2, 5).

( = 0.5 ( x

)

− 4 x + ( −2 ) − ( −1.25)( −2 )

The minimum point is at (–5, –7.5).

(

2

2

= 0.2 ( x − 4 ) + 1 2

The depth of the half-pipe is 4.2 – 1, or 3.2 m.

18 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 1

Question 19

a) c = 6, h = 4; x 2 + 8 x + 16 = ( x + 4 )

Page 272

2

b) b = 12, h = 6; x 2 + 12 x + 36 = ( x + 6 )

2

or b = –12, h = –6; x 2 − 12 x + 36 = ( x − 6 )

2

c) b = –10, c = 27; ( x − 5 ) + 2 = x 2 − 10 x + 25 + 2 2

= x 2 − 10 x + 27 Chapter 6 Section 1

Question 20

Page 272

d = 0.15v 2 − 9v + 195 = 0.15 ( v 2 − 60v ) + 195

= 0.15 ( v 2 − 60v + 302 − 302 ) + 195

= 0.15 ( v 2 − 60v + 302 ) − 0.15 ( 302 ) + 195 = 0.15 ( v − 30 ) + 60 2

Minimum drag occurs at a speed of 30 km/h. Chapter 6 Section 1

Question 21

Page 272

y = − x 2 + 10 x − 21 = − ( x 2 − 10 x ) − 21

(

)

= − x 2 − 10 x + ( −5) − ( −5) − 21 2

2

= − ( x 2 − 10 x + 25) − ( −1)( −5) − 21 2

= − ( x − 5) + 4 2

The maximum height of the spring is 4 m. It will hit the ceiling before reaching the box. Chapter 6 Section 1

Question 22

Page 272

Maximum productivity occurs at 13.7 h of training.

MHR • Principles of Mathematics 10 Solutions

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Chapter 6 Section 1

Question 23

Page 273

Let the length be represented by x. The width will be 10 – x.

A = x (10 − x ) = − x 2 + 10 x = − ( x 2 − 10 x )

( = −(x

= − x 2 − 10 x + ( −5) − ( −5) 2

2

)

2

)

− 10 x + ( −5) − ( −1)( −5) 2

2

= − ( x − 5) + 25 2

The maximum area of 25 cm2 occurs at a length of 5 cm and a width of 5 cm. Bend the pipe cleaner into a square. Chapter 6 Section 1

Question 24

Page 273

Let x represent the width of the field. The length is 200 – 2x. A = x ( 200 − 2 x ) = −2 x 2 + 200 x = −2 ( x 2 − 100 x )

( = −2 ( x

= −2 x 2 − 100 x + ( −50 ) − ( −50 ) 2

2

)

2

)

− 100 x + ( −50 ) − ( −2 )( −50 ) 2

2

= −2 ( x − 50 ) + 5000 2

The maximum area of 5000 m2 occurs with a width of 50 m and a length of 100 m. Chapter 6 Section 1

Question 25

Page 273

20 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 1

Question 26

Page 273

Let the equation of the parabola be y = ax 2 + bx + c . Since the y-intercept is 5, c = 5. Substitute the coordinates of the points (2, –3) and (–1, 12) into y = ax 2 + bx + 5 to form a linear system of equations that can be solved for a and b. y = ax 2 + bx + 5 y = ax 2 + bx + 5 −3 = a ( 2 ) + b ( 2 ) + 5

12 = a ( −1) + b ( −1) + 5

2

4a + 2b = −8 2 a + b = −4

2

7 = a−b a−b = 7

c

2 a + b = −4 a−b = 7

d

c d c+d

3a = 3 a =1 Substitute a = 1 into equation d. a−b = 7 1− b = 7 b = −6 y = x2 − 6x + 5

( = (x

)

= x 2 − 6 x + ( −3) − ( −3) + 5 2

2

2

)

− 6 x + ( −3) − ( −3) + 5 2

2

= ( x − 3) − 4 2

The vertex is at (3, –4). Chapter 6 Section 1

Question 27

Page 273

y = ax 2 + bx + c b ⎞ ⎛ = a ⎜ x2 + x ⎟ + c a ⎠ ⎝ 2 2 ⎛ b ⎛ b ⎞ ⎛ b ⎞ ⎞ = a ⎜ x2 + x + ⎜ ⎟ − ⎜ ⎟ ⎟ + c ⎜ a ⎝ 2a ⎠ ⎝ 2a ⎠ ⎟⎠ ⎝ 2 2 ⎛ 2 b ⎛ b ⎞ ⎞ ⎛ b ⎞ = a⎜ x + x + ⎜ ⎟ ⎟ − a⎜ ⎟ + c ⎜ a ⎝ 2a ⎠ ⎟⎠ ⎝ 2a ⎠ ⎝ 2

b ⎞ b2 ⎛ = a⎜ x + +c ⎟ − 2a ⎠ 4a ⎝ The x-coordinate of the vertex is −

b . 2a

MHR • Principles of Mathematics 10 Solutions

21

Chapter 6 Section 1

Question 28

Page 273

Refer to the yellow triangle shown. Since the triangle inscribed in the circle is an equilateral triangle, you can show that this is a 30º-60º-90º. The hypotenuse is R, the height is R S , and the base is . Apply the Pythagorean 2 2 theorem to find S. 2

2

⎛S ⎞ ⎛R⎞ 2 ⎜ ⎟ +⎜ ⎟ = R ⎝2⎠ ⎝ 2⎠ S 2 3R 2 = 4 4 2 S = 3R 2 S = 3R The side length of the triangle is 3R . Chapter 6 Section 1

Question 29

Page 273

There are 4 × 3 = 12 ways to draw the names. Two of these include Jane and Farhad. The 2 1 = . Answer D. probability that they will go together is 12 6 Chapter 6 Section 1

Question 30

Page 273

10001000 = 103000

(

= 10100

)

30

22 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 2

Chapter 6 Section 2

Question 1

Page 279

a) x = −5 or x = −2

b) x = 3 or x = −4

c) x = 1 or x = 7

d) x = 0 or x = −9

e) x = −

g) x =

3 or x = 5 2

1 4 or x = − 2 3

5 3 or x = 3 4

Chapter 6 Section 2 a)

f) x =

Question 2

Page 279

x 2 + 8 x + 12 = 0

( x + 2 )( x + 6 ) = 0 x + 2 = 0 or x + 6 = 0 x = −2 or x = −6 Check by substituting both solutions in the original equation. For x = –2: For x = –6: R.S. = 0 L.S. = x 2 + 8 x + 12 L.S. = x 2 + 8 x + 12 = ( −2 ) + 8 ( −2 ) + 12

= ( −6 ) + 8 ( −6 ) + 12

=0

=0

2

R.S. = 0

2

L.S. = R.S.

L.S. = R.S.

The roots are –2 and –6. b)

h 2 + 9h + 18 = 0

( h + 3)( h + 6 ) = 0 h + 3 = 0 or h + 6 = 0 h = −3 or h = −6 Check by substituting both solutions in the original equation. For h = –3: For h = –6: 2 R.S. = 0 L.S. = h + 9h + 18 L.S. = h 2 + 9h + 18 = ( −3) + 9 ( −3) + 18

= ( −6 ) + 9 ( −6 ) + 18

=0

=0

2

L.S. = R.S.

R.S. = 0

2

L.S. = R.S.

The roots are –3 and –6.

MHR • Principles of Mathematics 10 Solutions

23

c) m 2 + 3m = 0 m ( m + 3) = 0 m + 3 = 0 or m = 0 m = −3

Check by substituting both solutions in the original equation. For m = –3: For m = 0: 2 R.S. = 0 L.S. = m + 3m L.S. = m 2 + 3m = ( −3) + 3 ( −3)

= ( 0) + 3( 0)

=0

=0

2

R.S. = 0

2

L.S. = R.S.

L.S. = R.S.

The roots are –3 and –0. d)

w2 − 18w + 56 = 0

( w − 4 )( w − 14 ) = 0 w−4=0 w=4

or w − 14 = 0 or w = 14

Check by substituting both solutions in the original equation. For w = 4: For w = 14: R.S. = 0 L.S. = w2 − 18w + 56 L.S. = w2 − 18w + 56 = ( 4 ) − 18 ( 4 ) + 56

R.S. = 0

= (14 ) − 18 (14 ) + 56 2

2

=0

=0

L.S. = R.S.

L.S. = R.S.

The roots are 4 and 14. e) x 2 − 2 x = 0 x ( x − 2) = 0 x−2 =0 x=2

or x = 0

Check by substituting both solutions in the original equation. For x = 2: For x = 0: 2 R.S. = 0 L.S. = x − 2 x L.S. = x 2 − 2 x = (2) − 2 (2)

= (0) − 2 (0)

=0

=0

2

R.S. = 0

2

L.S. = R.S.

The roots are 2 and 0.

24 MHR • Principles of Mathematics 10 Solutions

L.S. = R.S.

f)

c 2 − 17c + 30 = 0

( c − 2 )( c − 15) = 0 c−2=0 c=2

or c − 15 = 0 or c = 15

Check by substituting both solutions in the original equation. For c = 2: For c = 15: R.S. = 0 L.S. = c 2 − 17c + 30 L.S. = c 2 − 17c + 30

R.S. = 0

= ( 2 ) − 17 ( 2 ) + 30

= (15 ) − 17 (15 ) + 30

=0

=0

2

2

L.S. = R.S.

L.S. = R.S.

The roots are 2 and 15. g)

n 2 + 9n − 22 = 0

( n + 11)( n − 2 ) = 0 n + 11 = 0 or n − 2 = 0 n = −11 or n=2 Check by substituting both solutions in the original equation. For n = –11: For n = 2: 2 R.S. = 0 L.S. = n + 9n − 22 L.S. = n 2 + 9n − 22 = ( −11) + 9 ( −11) − 22

= ( 2 ) + 9 ( 2 ) − 22

=0

=0

2

R.S. = 0

2

L.S. = R.S.

L.S. = R.S.

The roots are –11 and 2. h) y 2 − 11y = 0

y ( y − 11) = 0 y − 11 = 0 or y = 0 y = 11 Check by substituting both solutions in the original equation. For y = 11: For y = 0: R.S. = 0 L.S. = y 2 − 11 y L.S. = y 2 − 11 y = (11) − 11(11)

= ( 0 ) − 11( 0 )

=0

=0

2

L.S. = R.S.

R.S. = 0

2

L.S. = R.S.

The roots are 11 and 0.

MHR • Principles of Mathematics 10 Solutions

25

Chapter 6 Section 2

Question 3

3 x 2 + 28 x + 9 = 0

a)

Page 280 4k 2 + 19k + 15 = 0

b)

4k 2 + 4k + 15k + 15 = 0

3 x 2 + 27 x + x + 9 = 0

( 3x

2

)

( 4k

+ 27 x + ( x + 9 ) = 0

2

)

+ 4k + (15k + 15 ) = 0

3x ( x + 9 ) + ( x + 9 ) = 0

4k ( k + 1) + 15 ( k + 1) = 0

( x + 9 )( 3x + 1) = 0

( k + 1)( 4k + 15) = 0

x+9=0

or 3x + 1 = 0

x = −9 or

x=−

k +1 = 0 1 3

8 y 2 − 22 y + 15 = 0

c)

8 y 2 − 12 y − 10 y + 15 = 0

(8 y

2

or 4k + 15 = 0 15 k = −1 or k =− 4

− 12 y ) + ( −10 y + 15) = 0

16b 2 − 1 = 0

d)

( 4b + 1) + ( 4b − 1) = 0

4 y ( 2 y − 3 ) − 5 ( 2 y − 3) = 0

( 2 y − 3)( 4 y − 5) = 0 2 y − 3 = 0 or 4 y − 5 = 0 3 5 y= or y= 2 4

4b + 1 = 0

or 4b − 1 = 0 1 1 b = − or b= 4 4

e) 10m 2 + 30m = 0

f) 4 x 2 − 12 x + 9 = 0

10m ( m + 3) = 0

10m = 0 m=0

( 2 x − 3)

or m + 3 = 0 or

m = −3

2x − 3 = 0 3 x= 2

26 MHR • Principles of Mathematics 10 Solutions

2

=0

Chapter 6 Section 2 a)

Question 4

x 2 + 5 x = −4

Page 280 b)

x2 + 5x + 4 = 0

c 2 + 8c + 15 = 0

( x + 1)( x + 4 ) = 0

( c + 3)( c + 5) = 0 c + 3 = 0 or c + 5 = 0 c = −3 or c = −5

x + 1 = 0 or x + 4 = 0 x = −1 or x = −4 c)

k 2 = 13k − 12

d)

b 2 + 1 = −2b b 2 + 2b + 1 = 0

k 2 − 13x + 12 = 0

( k − 12 )( k − 1) = 0

( b + 1)

k − 12 = 0 or k − 1 = 0 k = 12 or k =1 e)

8c + 15 = −c 2

2

=0

b +1 = 0 b = −1

m 2 = 300 − 20m

f)

y2 = 7 y

m 2 + 20m − 300 = 0

y2 − 7 y = 0

( m − 10 )( m + 30 ) = 0

y ( y − 7) = 0

m − 10 = 0 or m + 30 = 0 m = 10 or m = −30

y −7 = 0 y=7

or y = 0

MHR • Principles of Mathematics 10 Solutions

27

Chapter 6 Section 2

Question 5 2m 2 = −7 m − 6

a)

9 x2 = x + 8

b)

2m 2 + 7 m + 6 = 0

9 x2 − x − 8 = 0

2m 2 + 4m + 3m + 6 = 0

9x2 − 9x + 8x − 8 = 0

( 2m

2

)

(9x

+ 4m + ( 3m + 6 ) = 0

2

)

− 9 x + (8 x − 8) = 0

2m ( m + 2 ) + 3 ( m + 2 ) = 0

9 x ( x − 1) + 8 ( x − 1) = 0

( m + 2 )( 2m + 3) = 0

( x − 1)( 9 x + 8) = 0

m+2=0

or 2m + 3 = 0 3 m = −2 or m=− 2

c)

Page 280

4 y 2 − 12 y = −9

x −1 = 0

or 9 x + 8 = 0 8 x = 1 or x=− 9 −5 = 2 p − 16 p 2

d)

4 y 2 − 12 y + 9 = 0

( 2 y − 3)

2

16 p 2 − 2 p − 5 = 0 16 p 2 + 8 p − 10 p − 5 = 0

=0

(16 p

2

+ 8 p ) + ( −10 p − 5) = 0

8 p ( 2 p + 1) − 5 ( 2 p + 1) = 0

( 2 p + 1)(8 p − 5) = 0 2y − 3 = 0 3 y= 2

2 p +1 = 0

or 8 p − 5 = 0 1 5 p = − or p= 2 8

12m 2 = 10 − 37m

e)

(12m

3w2 + 22 w = −7

f)

12m 2 + 37m − 10 = 0

3w2 + 22 w + 7 = 0

12m 2 + 40m − 3m − 10 = 0

3w2 + 21w + w + 7 = 0

2

( 3w

+ 40m ) + ( −3m − 10 ) = 0

2

)

+ 21w + ( w + 7 ) = 0

4m ( 3m + 10 ) − 1( 3m + 10 ) = 0

3w ( w + 7 ) + ( w + 7 ) = 0

( 3m + 10 )( 4m − 1) = 0

( w + 7 )( 3w + 1) = 0

3m + 10 = 0

or 4m − 1 = 0 10 1 m=− or m= 3 4

w+7 =0 w = −7

28 MHR • Principles of Mathematics 10 Solutions

or 3w + 1 = 0 or

w=−

1 3

Chapter 6 Section 2

Question 6

Page 280

a) − x 2 − 10 x − 16 = 0

Divide both sides by − 1.

x + 10 x + 16 = 0 2

( x + 8)( x + 2 ) = 0 x+8=0

or x + 2 = 0

x = −8 or

x = −2

b) 3t 2 + 24t + 45 = 0

Divide both sides by 3.

t + 8t + 15 = 0 2

( t + 3)( t + 5) = 0 t +3=0

or t + 5 = 0

t = −3 or

t = −5

c) 6d 2 + 15d = −9 6d 2 + 15d + 9 = 0

Divide both sides by 3.

2d 2 + 5d + 3 = 0 2d 2 + 2d + 3d + 3 = 0

( 2d

2

+ 2d ) + ( 3d + 3) = 0

2d ( d + 1) + 3 ( d + 1) = 0

( d + 1)( 2d + 3) = 0 d +1 = 0

or 2d + 3 = 0 3 d = −1 or d =− 2

d) −10 g 2 + 32 g = 6 −10 g 2 + 32 g − 6 = 0

Divide both sides by − 2.

5g 2 − 16 g + 3 = 0 5 g 2 − 15g − g + 3 = 0

( 5g

2

− 15g ) + ( − g + 3) = 0

5 g ( g − 3) − 1 ( g − 3 ) = 0

( g − 3)( 5g − 1) = 0 g −3=0

or 5 g − 1 = 0 1 g = 3 or g= 5

MHR • Principles of Mathematics 10 Solutions

29

Chapter 6 Section 2

Question 7

Page 280

− x2 + 2x + 3 = 0 x2 − 2x − 3 = 0

( x + 1)( x − 3) = 0 x + 1 = 0 or x − 3 = 0 x = −1 or x=3 The ball has travelled 3 m horizontally when it hits the ground. Chapter 6 Section 2

Question 8

Page 280

( x + 10 )( 2 x − 3) = 54 2 x − 3x + 20 x − 30 = 54 2

2 x 2 + 17 x − 84 = 0 2 x 2 + 24 x − 7 x − 84 = 0

(2x

2

+ 24 x ) + ( −7 x − 84 ) = 0

2 x ( x + 12 ) − 7 ( x + 12 ) = 0

( x + 12 )( 2 x − 7 ) = 0 x + 12 = 0

or 2x − 7 = 0 7 x = −12 or x= 2

The negative answer is inadmissible. If x is 3.5 cm, the area is 54 cm2. Chapter 6 Section 2

Question 9

a) ( x − 5 )( x − 4 ) = 0 Chapter 6 Section 2

Page 280 b) ( x + 2 )( x − 3) = 0

Question 10

Page 280

a) ( x − 6 )( x + 7 ) = 0 x 2 + x − 42 = 0

b) If you multiply both sides of the equation in part a) by 3, the roots will remain the same. The new equation is equivalent to the equation in part a). Chapter 6 Section 2

Question 11

Page 280

( 3x − 2 )( 5 x + 4 ) = 0 15 x 2 + 12 x − 10 x − 8 = 0 15 x 2 + 2 x − 8 = 0

30 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 2

Question 12

Page 280

a) Answers will vary. For example:

x2 − x − 2 = 0 ( x + 1)( x − 2) = 0

b) Answers will vary. For example:

6x2 + x − 1 = 0 (2 x + 1)(3x − 1) = 0

Chapter 6 Section 2

Question 13

Page 280

Answers will vary. For example: x 2 − x + 3 = 0 There are no integers whose product is 3 and sum is –1. Chapter 6 Section 2

Question 14

Page 280

x 2 + ( x − 1) = 292 2

x 2 + x 2 − 2 x + 1 = 841 2 x 2 − 2 x − 840 = 0

Divide both sides by 2.

x − x − 420 = 0 2

( x + 20 )( x − 21) = 0 x + 20 = 0

or x − 21 = 0

x = −20 or x = 21 The negative answer is inadmissible. One side measures 21 cm, and the other measures 20 cm. Chapter 6 Section 2

Question 15

Page 281

n = 0 will also satisfy the equation. If Chris wants to divide out a common factor, it should not contain any variables. Chris should subtract 15n from both sides of the equation, then divide both sides of the equation by 3, and then solve the equation by factoring.

MHR • Principles of Mathematics 10 Solutions

31

Chapter 6 Section 2 a) For n = 1: S = n ( n + 1)

Question 16

Page 281

For n = 2: S = n ( n + 1)

= 1(1 + 1)

= 2 ( 2 + 1)

=2

=6

The formula works for n = 1 and n = 2. b) S = n ( n + 1)

= 5 ( 5 + 1) = 30 The sum of the first five even numbers is 30. n ( n + 1) = 306

c)

n 2 + n − 306 = 0

( n − 17 )( n + 18) = 0 n − 17 = 0

or n + 18 = 0

n = 17 or n = −18 The value of n is 17. Chapter 6 Section 2

Question 17

Page 281

( 3.90 − 0.10n )(120 + 20n ) = 700 468 + 78n − 12n − 2n 2 = 700 −2n 2 + 66n − 232 = 0

Divide both sides by − 2.

n − 33n + 116 = 0 2

( n − 4 )( n − 29 ) = 0 n−4 =0

or n − 29 = 0

n=4 n = 29 or Either 4 or 29 price reductions result in revenue of \$700. Chapter 6 Section 2

Question 18

Page 281

Solutions for the Achievement Checks are shown in the Teacher Resource.

32 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 2

Question 19

Page 281

Let the width be represented by x. The length is x + 7. The diagonal is x + 8. x 2 + ( x + 7 ) = ( x + 8) 2

2

x 2 + x 2 + 14 x + 49 = x 2 + 16 x + 64 x 2 − 2 x − 15 = 0

( x − 5)( x + 3) = 0 x −5= 0 or x + 3 = 0 x = 5 or x = −3 The negative answer is inadmissible. The width is 5 m and the length is 12 m. Chapter 6 Section 2

Question 20

Page 281

P = 1125 ( t − 1) − 4500 2

0 = 1125 ( t 2 − 2t + 1) − 4500 0 = 1125t 2 − 2250t + 1125 − 4500 0 = 1125t 2 − 2250t − 3375 0 = 1125 ( t 2 − 2t − 3) 0 = 1125 ( t − 3)( t + 1)

t −3= 0

or t + 1 = 0

t = 3 or t = −1 The positive t-intercept is 3. Ralph will lose money for the first two years, break even in the third year, and make a profit in the fourth and fifth years.

MHR • Principles of Mathematics 10 Solutions

33

Chapter 6 Section 2 a)

Question 21

Page 281

y2 + 3y + 2 = 0

( y + 1)( y + 2 ) = 0 y +1 = 0

or y + 2 = 0

y = −1 or

y = −2

The coefficients are the same for the two equations, but the solutions for y 2 + 3 yy + 2 x 2 = 0 are y = –x and y = –2x. b) i) y 2 + 3 yx + 2 x 2 = 0

( y + x )( y + 2 x ) = 0 y+x=0 or y + 2 x = 0 y = − x or y = −2 x 5 y 2 − 6 yx − 8 x 2 = 0

ii)

5 y 2 − 10 yx + 4 yx − 8 x 2 = 0

(5 y

2

) (

)

− 10 yx + 4 yx − 8 x 2 = 0

5 y ( y − 2x ) + 4x ( y − 2x ) = 0

( y − 2 x )( 5 y + 4 x ) = 0 y − 2 x = 0 or 5y + 4 x = 0 y = 2 x or

iii)

4 y=− x 5

1 2 1 1 y − yx + x 2 = 0 9 3 4 2 4 y − 12 yx + 9 x 2 = 0

( 2 y − 3x )

2

=0

2 y − 3x = 0 y=

3 x 2

34 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 3

Chapter 6 Section 3

Question 1

Page 288

Factor and solve the corresponding quadratic equation by letting y = 0. a) y = x 2 + 5 x + 6 b) y = x 2 − 11x + 28 0 = ( x + 2 )( x + 3) x + 2 = 0 or x + 3 = 0

0 = ( x − 7 )( x − 4 ) x − 7 = 0 or x − 4 = 0

x = −2 or x = −3 The x-intercepts are –2 and –3.

x = 7 or x=4 The x-intercepts are 7 and 4.

c) y = x 2 + 9 x

d) y = x 2 + 5 x − 24

0 = x ( x + 9) x = 0 or x + 3 = 0 x = −3 The x-intercepts are 0 and –9.

0 = ( x + 8 )( x − 3) x + 8 = 0 or x − 3 = 0 x = −8 or x=3 The x-intercepts are –8 and 3.

e) y = x 2 − 2 x − 8

f) y = x 2 + 9 x − 36

0 = ( x − 4 )( x + 2 ) x − 4 = 0 or x + 2 = 0 x = 4 or x = −2 The x-intercepts are 4 and –2.

0 = ( x + 12 )( x − 3) x + 12 = 0 or x − 3 = 0 x = −12 or x=3 The x-intercepts are –12 and 3.

MHR • Principles of Mathematics 10 Solutions

35

Chapter 6 Section 3

Question 2

Page 288

Factor and solve the corresponding quadratic equation by letting y = 0. a) y = 4 x 2 + 20 x + 9 b) y = 8 x 2 − 6 x 0 = 2 x ( 4 x − 3)

0 = 4 x 2 + 2 x + 18 x + 9

(

)

0 = 4 x 2 + 2 x + (18 x + 9 ) 0 = 2 x ( 2 x + 1) + 9 ( 2 x + 1) 0 = ( 2 x + 1)( 2 x + 9 )

2x + 1 = 0 x=−

or 2 x + 9 = 0 1 or 2

x=−

2x = 0 9 2

c) y = 6 x 2 − 11x − 7 0 = 6 x 2 + 3 x − 14 x − 7

(

)

or 4 x − 3 = 0 3 or x= 4

x=0

d) y = 5 x 2 + 6 x − 8 0 = 5 x 2 + 10 x − 4 x − 8

(

)

0 = 6 x 2 + 3 x + ( −14 x − 7 )

0 = 5 x 2 + 10 x + ( −4 x − 8 )

0 = 3x ( 2 x + 1) − 7 ( 2 x + 1)

0 = 5x ( x + 2 ) − 4 ( x + 2 )

0 = ( 2 x + 1)( 3 x − 7 )

0 = ( x + 2 )( 5 x − 4 )

2x + 1 = 0

or 3x − 7 = 0 1 7 x = − or x= 2 3

x+2=0

e) y = 3x 2 − 13 x + 4

f) y = 4 x 2 − 20 x + 25

0 = 3x 2 − 12 x − x + 4

(

)

0 = 3 x 2 − 12 x + ( − x + 4 )

or 5 x − 4 = 0 4 x = −2 or x= 5

0 = ( 2 x − 5)

0 = 3x ( x − 4 ) − 1( x − 4 ) 0 = ( x − 4 )( 3 x − 1)

x−4=0

or 3x − 1 = 0 1 x = 4 or x= 3

2x − 5 = 0 5 x= 2

36 MHR • Principles of Mathematics 10 Solutions

2

Chapter 6 Section 3

Question 3

Page 289

a) Factor and solve the corresponding quadratic equation. y = x 2 + 9 x + 14

0 = ( x + 2 )( x + 7 ) The x-intercepts are –2 and –7. Find the x-coordinate of the vertex, and then, the ycoordinate. −2 + ( −7 ) 2 9 =− 2 2 y = x + 9 x + 14 x=

2

⎛ 9⎞ ⎛ 9⎞ = ⎜ − ⎟ + 9 ⎜ − ⎟ + 14 ⎝ 2⎠ ⎝ 2⎠ 25 =− 4

⎛ 9 25 ⎞ The vertex is ⎜ − , − ⎟ . 4 ⎠ ⎝ 2 b) Factor and solve the corresponding quadratic equation. y = x2 − 6x + 8

0 = ( x − 2 )( x − 4 ) The x-intercepts are 2 and 4. Find the x-coordinate of the vertex, and then, the ycoordinate. 2+4 2 =3

x=

y = x2 − 6x + 8 = ( 3) − 6 ( 3) + 8 2

= −1 The vertex is ( 3, −1) .

MHR • Principles of Mathematics 10 Solutions

37

c) Factor and solve the corresponding quadratic equation. y = − x2 − 4 x + 5

0 = − ( x 2 + 4 x − 5) 0 = − ( x + 5)( x − 1) The x-intercepts are –5 and 1. Find the x-coordinate of the vertex, and then, the ycoordinate. −5 + 1 2 = −2

x=

y = − x2 − 4 x + 5 = − ( −2 ) − 4 ( −2 ) + 5 2

=9 The vertex is ( −2,9 ) .

d) Factor and solve the corresponding quadratic equation. y = − x 2 − 5x

0 = − x ( x + 5) The x-intercepts are 0 and –5. Find the x-coordinate of the vertex, and then, the ycoordinate. 0 + ( −5 ) 2 5 =− 2 y = − x 2 − 5x x=

2

⎛ 5⎞ ⎛ 5⎞ = − ⎜ − ⎟ − 5⎜ − ⎟ ⎝ 2⎠ ⎝ 2⎠ 25 = 4

⎛ 5 25 ⎞ The vertex is ⎜ − , ⎟ . ⎝ 2 4 ⎠

38 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 3

Question 4

Page 289

a) Factor and solve the corresponding quadratic equation. y = x2 − 9

0 = ( x + 3)( x − 3) The x-intercepts are –3 and 3. Find the x-coordinate of the vertex, and then, the ycoordinate. −3 + 3 2 =0

x=

y = x2 − 9 = ( 0) − 9 2

= −9 The vertex is ( 0, −9 ) . b) Factor and solve the corresponding quadratic equation. y = − x 2 + 10 x − 9

0 = − ( x 2 − 10 x + 9 ) 0 = − ( x − 9 )( x − 1) The x-intercepts are 9 and 1. Find the x-coordinate of the vertex, and then, the y-coordinate. 9 +1 2 =5

x=

y = − x 2 + 10 x − 9 = − ( 5 ) + 10 ( 5 ) − 9 2

= 16 The vertex is ( 5,16 ) .

MHR • Principles of Mathematics 10 Solutions

39

c) Factor and solve the corresponding quadratic equation. y = x 2 − 12 x + 36

0 = ( x − 6)

2

The x-intercept is 6. Find the x-coordinate of the vertex, and then, the ycoordinate. 6+6 2 =6

x=

y = x 2 − 12 x + 36 = ( 6 ) − 12 ( 6 ) + 36 2

=0 The vertex is ( 6,0 ) . d) Factor and solve the corresponding quadratic equation. y = 16 − x 2

0 = ( 4 + x )( 4 − x ) The x-intercepts are –4 and 4. Find the x-coordinate of the vertex, and then, the ycoordinate. −4 + 4 2 =0

x=

y = 16 − x 2 = 16 − ( 0 )

2

= 16 The vertex is ( 0,16 ) .

40 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 3

Question 5

Page 289

a) Factor and solve the corresponding quadratic equation. y = 2 x 2 + 15 x + 7 0 = 2 x 2 + 14 x + x + 7 0 = ( 2 x 2 + 14 x ) + ( x + 7 ) 0 = 2 x ( x + 7) + ( x + 7) 0 = ( x + 7 )( 2 x + 1)

x+7=0 x = −7

or 2 x + 1 = 0 or

x=−

1 2

1 The zeros are –7 and − . 2 Find the x-coordinate of the vertex, and then, the y-coordinate. ⎛ 1⎞ −7 + ⎜ − ⎟ ⎝ 2⎠ x= 2 15 =− 4 y = 2 x 2 + 15 x + 7 2

⎛ 15 ⎞ ⎛ 15 ⎞ = 2 ⎜ − ⎟ + 15 ⎜ − ⎟ + 7 ⎝ 4⎠ ⎝ 4⎠ 169 =− 8

⎛ 15 169 ⎞ The vertex is ⎜ − , − . 8 ⎟⎠ ⎝ 4

MHR • Principles of Mathematics 10 Solutions

41

b) Factor and solve the corresponding quadratic equation. y = 12 x 2 − 16 x + 5 0 = 12 x 2 − 6 x − 10 x + 5 0 = (12 x 2 − 6 x ) + ( −10 x + 5) 0 = 6 x ( 2 x − 1) − 5 ( 2 x − 1) 0 = ( 2 x − 1)( 6 x − 5)

2x −1 = 0 1 x= 2

or 6 x − 5 = 0 5 or x= 6 1 5 The zeros are and . 2 6 Find the x-coordinate of the vertex, and then, the y-coordinate. 1 5 + x= 2 6 2 2 = 3 y = 12 x 2 − 16 x + 5 2

⎛2⎞ ⎛2⎞ = 12 ⎜ ⎟ − 16 ⎜ ⎟ + 5 ⎝3⎠ ⎝3⎠ 1 =− 3

⎛ 2 1⎞ The vertex is ⎜ , − ⎟ . ⎝ 3 3⎠

42 MHR • Principles of Mathematics 10 Solutions

c) Factor and solve the corresponding quadratic equation. y = −8 x 2 − 13x + 6

0 = − (8 x 2 + 13x − 6 )

0 = − (8 x 2 + 16 x − 3x − 6 ) 0 = − ⎡⎣ ( 8 x 2 + 16 x ) + ( −3x − 6 )⎤⎦ 0 = − ⎡⎣8 x ( x + 2 ) − 3 ( x + 2 )⎤⎦ 0 = − ( x + 2 )( 8 x − 3)

8 x − 3 = 0 or x + 2 = 0 3 x= or x = −2 8 3 The zeros are and −2 . 8 Find the x-coordinate of the vertex, and then, the y-coordinate. 3 + ( −2 ) x= 8 2 13 =− 16 y = −8 x 2 − 13x + 6 2

⎛ 13 ⎞ ⎛ 13 ⎞ = −8 ⎜ − ⎟ − 13 ⎜ − ⎟ + 6 ⎝ 16 ⎠ ⎝ 16 ⎠ 361 = 32

⎛ 13 361 ⎞ The vertex is ⎜ − , − ⎟. ⎝ 16 32 ⎠

MHR • Principles of Mathematics 10 Solutions

43

d) Factor and solve the corresponding quadratic equation. y = 8 x 2 + 17 x + 9 0 = 8x2 + 8x + 9 x + 9 0 = (8 x 2 + 8 x ) + ( 9 x + 9 ) 0 = 8 x ( x + 1) + 9 ( x + 1) 0 = ( x + 1)(8 x + 9 )

x +1 = 0

or 8 x + 9 = 0

x = −1 or

x=−

9 8

9 The zeros are –1 and − . 8 Find the x-coordinate of the vertex, and then, the y-coordinate. ⎛ 9⎞ −1 + ⎜ − ⎟ ⎝ 8⎠ x= 2 17 =− 16 y = 8 x 2 + 17 x + 9 2

⎛ 17 ⎞ ⎛ 17 ⎞ = 8 ⎜ − ⎟ + 17 ⎜ − ⎟ + 9 ⎝ 16 ⎠ ⎝ 16 ⎠ 1 =− 32

1 ⎞ ⎛ 17 The vertex is ⎜ − , − ⎟ . 16 32 ⎝ ⎠

44 MHR • Principles of Mathematics 10 Solutions

e) Factor and solve the corresponding quadratic equation. y = 16 x 2 − 24 x + 9

0 = ( 4 x − 3)

2

4x − 3 = 0 3 x= 4 The zero is

3 . 4

Find the x-coordinate of the vertex, and then, the y-coordinate. 3 3 + x= 4 4 2 3 = 4 y = 16 x 2 − 24 x + 9 2

⎛3⎞ ⎛3⎞ = 16 ⎜ ⎟ − 24 ⎜ ⎟ + 9 ⎝4⎠ ⎝4⎠ =0

⎛3 ⎞ The vertex is ⎜ ,0 ⎟ . ⎝4 ⎠

MHR • Principles of Mathematics 10 Solutions

45

f) Factor and solve the corresponding quadratic equation. y = −4 x 2 − 18 x − 8

0 = −2 ( 2 x 2 + 9 x + 4 )

0 = −2 ( 2 x 2 + 8 x + x + 4 ) 0 = −2 ⎡⎣( 2 x 2 + 8 x ) + ( x + 4 ) ⎤⎦ 0 = −2 ⎡⎣ 2 x ( x + 4 ) + ( x + 4 ) ⎤⎦ 0 = −2 ( x + 4 )( 2 x + 1) x+4=0 x = −4

or 2 x + 1 = 0 or

x=−

1 2

1 The zeros are –4 and − . 2 Find the x-coordinate of the vertex, and then, the y-coordinate. ⎛ 1⎞ −4 + ⎜ − ⎟ ⎝ 2⎠ x= 2 9 =− 4 y = −4 x 2 − 18 x − 8 2

⎛ 9⎞ ⎛ 9⎞ = −4 ⎜ − ⎟ − 18 ⎜ − ⎟ − 8 ⎝ 4⎠ ⎝ 4⎠ 49 = 4

⎛ 9 49 ⎞ The vertex is ⎜ − , ⎟ . ⎝ 4 4 ⎠

46 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 3

Question 6

Page 289

a) Since the x-intercepts are 0 and 6, the equation has the form y = ax ( x − 6 ) . To find a, substitute (x, y) = (5, 5). 2 5 = a ( 5 ) − 6a ( 5 )

5 = 25a − 30a 5 = −5a −1 = a y = −1x ( x − 6) = − x2 + 6x

b) Since the x-intercepts are –4 and 6, the equation has the form y = a ( x + 4 )( x − 6 ) . To find a, substitute (x, y) = (1, –5). −5 = a (1 + 4 )(1 − 6 ) −5 = −25a

1 =a 5 1 y = ( x + 4 )( x − 6 ) 5 1 2 = ( x − 2 x − 24 ) 5 1 2 24 = x2 − x − 5 5 5

MHR • Principles of Mathematics 10 Solutions

47

Chapter 6 Section 3

Question 7

Page 289

a) Since the x-intercepts are –8 and –2, the equation has the form y = a ( x + 8 )( x + 2 ) . To find a, substitute (x, y ) = (–7, –10). −10 = a ( −7 + 8 )( −7 + 2 ) −10 = −5a

2=a y = 2 ( x + 8 )( x + 2 )

(

= 2 x 2 + 10 x + 16

)

= 2 x 2 + 20 x + 32

b) Since the x-intercepts are –5 and –1, the equation has the form y = a ( x + 5)( x + 1) . To find a, substitute (x, y) = (–3, 3). 3 = a ( −3 + 5)( −3 + 1)

3 = −4a −

3 =a 4 3 ( x + 5)( x + 1) 4 3 = − ( x 2 + 6 x + 5) 4 3 9 15 = − x2 − x − 4 2 4

y=−

c) Since the x-intercepts are 0 and 8, the equation has the form y = ax ( x − 8 ) . To find a, substitute (x, y) = (6, 6). 6 = a ( 6 )( 6 − 8 )

6 = −12a −

1 =a 2 1 y = − x ( x − 8) 2 1 = − ( x2 − 8x ) 2 1 = − x2 + 4 x 2

48 MHR • Principles of Mathematics 10 Solutions

d) Since the x-intercepts are –4 and 5, the equation has the form y = a ( x + 4 )( x − 5) . To find a, substitute (x, y) = (1, –6). −6 = a (1 + 4 )(1 − 5)

−6 = −20a 0.3 = a y = 0.3 ( x + 4 )( x − 5)

(

= 0.3 x 2 − x − 20

)

= 0.3x − 0.3x − 6 2

Chapter 6 Section 3

Question 8

Page 290

a) h = −d 2 + 4

0 = − (d 2 − 4) 0 = − ( d + 2 )( d − 2 ) The d-intercepts are at –2 and 2. The width of the garage is 4 m. Find the vertex. −2 + 2 2 =0

d=

h = −d 2 + 4 = − ( 0) + 4 2

=4 The height of the garage is 4 m. b)

c) The relation is valid for −2 ≤ d ≤ 2 . h must be positive.

MHR • Principles of Mathematics 10 Solutions

49

Chapter 6 Section 3

Question 9

Page 290

a) y = −3x 2 + 11x + 4

0 = − ( 3x 2 − 11x − 4 )

0 = − ( 3x 2 − 12 x + x − 4 ) 0 = − ⎡⎣( 3x 2 − 12 x ) + ( x − 4 )⎤⎦ 0 = − ⎡⎣3x ( x − 4 ) + ( x − 4 )⎤⎦ 0 = − ( x − 4 )( 3x + 1) x−4=0

or 3x + 1 = 0 1 x = 4 or x=− 3 1 The zeros are 4 and − . 3 b) The relation is valid for 0 ≤ x ≤ 4 . c) The rocket has travelled 4 m horizontally when is lands on the ground. d) Find the vertex. 1 − +4 x= 3 2 11 = 6 y = −3x 2 + 11x + 4 2

⎛ 11 ⎞ ⎛ 11 ⎞ = −3 ⎜ ⎟ + 11⎜ ⎟ + 4 6 ⎝ ⎠ ⎝6⎠ = 14.08

The maximum height of the rocket is 14.08 m.

50 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 3

Question 10

Page 290

a)

b) Since the x-intercepts are –100 and 100, the equation has the form y = a ( x + 100 )( x − 100 ) . To find a, substitute (x, y) = (0, 4). 4 = a ( 0 + 100 )( 0 − 100 )

4 = −10 000a −

1 =a 2500 1 ( x + 100 )( x − 100 ) 2500 1 =− ( x 2 − 10 000) 2500 1 =− x2 + 4 2500

y=−

1 2 (80 ) + 4 2500 = 1.44 A point on the arch that is 20 m horizontally from one end is 1.44 m high. y=−

MHR • Principles of Mathematics 10 Solutions

51

Chapter 6 Section 3

Question 11

Page 290

The x-coordinate of the vertex is –3, which is 8 units from the x-intercept –11. The other xintercept will be another 8 units to the right, or 5. Since the x-intercepts are –11 and 5, the equation has the form y = a ( x + 11)( x − 5) . To find a, substitute (x, y) = (–3, 7). 7 = a ( −3 + 11)( −3 − 5) 7 = −64a 7 − =a 64 7 y = − ( x + 11)( x − 5) 64 7 = − ( x 2 + 6 x − 55) 64 7 21 385 x+ = − x2 − 64 32 64

The y-intercept is

385 . 64

52 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 3

Question 12

Page 290

a)

1 2 16 36 d + d+ 35 35 35 1 2 d − 16d − 36 0=− 35 1 2 d + 2d − 18d − 36 0=− 35 1 0 = − ⎡⎣ d 2 + 2d − (18d + 36 )⎤⎦ 35 1 0 = − ⎡⎣ d ( d + 2 ) − 18 ( d + 2 )⎤⎦ 35 1 0 = − ( d + 2 )( d − 18 ) 35

h=−

(

)

(

)

(

)

The d-intercepts are –2 and 18. b)

c) The relation is valid for 0 ≤ d ≤ 18 . The height must be positive. d) The ball was kicked at a height of

36 m. 35

e) The ball had travelled 18 m when it landed on the ground.

MHR • Principles of Mathematics 10 Solutions

53

Chapter 6 Section 3

Question 13

Page 290

a)

1 2 2 w + w 125 25 1 0=− w ( w − 10 ) 125

b) d = −

The zeros are 0 and 10. The relation is valid for 0 ≤ w ≤ 10 , since d must be positive. c) The road is 10 m wide. d) Find the vertex.

0 + 10 2 =5

w=

1 2 2 w + w 125 25 2 1 2 =− ( 5) + ( 5) 125 25 = 0.2

d =−

The maximum height of the road is 0.2 m.

Chapter 6 Section 3

Question 14

Page 290

If a parabola has only one x-intercept, then the vertex is also the x-intercept.

54 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 3

Question 15

Page 290

a) n = 10 or –10; y = x 2 + 10 x + 25 or y = x 2 − 10 x + 25

b) n = 6 or –6; y = − x 2 + 6 x − 9 or y = − x 2 − 6 x − 9

c) n = 1; y = 16 x 2 − 8 x + 1

MHR • Principles of Mathematics 10 Solutions

55

Chapter 6 Section 3

Question 16

Page 291

a) The Marble Heads: y = − x2 + 2 x + 3

0 = − ( x 2 − 2 x − 3) 0 = − ( x − 3)( x + 1)

x −3= 0 x=3

or x + 1 = 0 or

x = −1

The XY Team: y = −2 x 2 + x + 3 0 = − ( 2 x 2 − x − 3)

0 = −2 ( 2 x 2 + 2 x − 3 x − 3 ) 0 = −2 ⎡⎣( 2 x 2 + 2 x ) + ( −3x − 3)⎤⎦ 0 = −2 ⎡⎣ 2 x ( x + 1) − 3 ( x + 1)⎤⎦ 0 = −2 ( x + 1)( 2 x − 3) x +1 = 0

or 2 x − 3 = 0 3 x = −1 or x= 2

The Marble Heads' marble travels farther by 1.5 m. b) Both marbles are at a height of 3 m at a horizontal distance of 0 m.

56 MHR • Principles of Mathematics 10 Solutions

c) Find the vertices. The Marble Heads: −1 + 3 x= 2 =1

y = − x2 + 2 x + 3 = − (1) + 2 (1) + 3 2

=4 The XY Team: 3 −1 + 2 x= 2 1 = 4 y = −2 x 2 + x + 3 2

⎛1⎞ ⎛1⎞ = −2 ⎜ ⎟ + ⎜ ⎟ + 3 ⎝4⎠ ⎝4⎠ = 3.125 The Marble Heads' marble flies higher by 0.875 m.

MHR • Principles of Mathematics 10 Solutions

57

Chapter 6 Section 3

Question 17

1 2 x + 50 100 1 0=− x 2 − 5000 100 1 x + 5000 0=− 100

Page 291

a) y = −

(

)

(

)( x −

5000

)

The width of the hangar is 2 5000 , or about 141.42 m.

141.42 6.08 23.24 Six planes can fit side by side inside the hangar. b) The width of the airplane hangar decreases as the height increases. 1 2 x + 50 100 1 2 2=− x + 50 100 1 0=− x 2 − 4800 100 1 x − 4800 =− 100 y=−

(

)

(

)( x +

4800

)

The width of the hangar is 2 4800 , or about 138.56 m.

139.56 5.96 23.24 Only 5 planes can fit side by side if the wings are 2 m above the ground. Chapter 6 Section 3

Question 18

Page 291

Answers will vary. For example: a) Parabola A: y = x 2 − 8 x + 18

Parabola B: y = − x 2 − 12 x − 37 b) The parabolas do not intersect the x-axis and thus do not have x-intercepts. Therefore, the equations cannot be factored.

58 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 3

Question 19

Page 291

4 ( x + 3)( x + a ) = 4 x 2 + x + c

(

)

4 x 2 + ( 3 + a ) x + 3a = 4 x 2 + x + c 4 x 2 + 4 ( 3 + a ) x + 12a = 4 x 2 + x + c 4 (3 + a ) = 1 12 + 4a = 1 11 4 c = 12a

a=−

⎛ 11 ⎞ = 12 ⎜ − ⎟ ⎝ 4⎠ = −33 Chapter 6 Section 3

Question 20

Page 291

There are no x-intercepts for this graph. The vertex of the graph of y = x2 + 4 is (0, 4), and the graph opens upward. The expression a2 + b2 cannot be factored since the graph of any parabola of the form y = x2 + b2 does not have any x-intercepts. It cannot be solved by factoring.

Chapter 6 Section 3 a)

Question 21

Page 291

( ) ( 3 ) − 12 ( 3 ) + 27 = 0 ( 3 − 9 )(3 − 3) = 0 32 x − 12 3x + 27 = 0 x 2

x

x

x

3x = 9 or 3x = 3 x = 2 or x = 1 b)

( ) 2 − 3 ( 2 )( 2 ) + 128 = 0 ( 2 ) − 24 ( 2 ) + 128 = 0 ( 2 − 8)( 2 − 16 ) = 0

22 x − 3 2 x + 3 + 128 = 0

2x

x

x 2

3

x

x

x

2 x = 8 or 2 x = 16 x = 3 or x = 4

MHR • Principles of Mathematics 10 Solutions

59

Chapter 6 Section 3

Question 22

Page 291

a) There are 2 × 2 × 2 = 8 possible outcomes. Three of these, BBG, BGB, and GBB, are 3 favourable. The probability of exactly 2 boys is . 8 b) The possible outcomes are

JBB JBG JGB JGG

BJB BJG GJB GJG

BBJ BGJ GBJ GGJ

Half of these have 2 boys. The probability of 2 boys is

1 . 2

c) John is a boy. This eliminates some of the possible families.

60 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 4

Chapter 6 Section 4

Question 1

a) For 7x2 + 24x + 9 = 0, a = 7, b = 24, and c = 9.

x= =

−b ± b 2 − 4ac 2a −24 ± 242 − 4 ( 7 )( 9 ) 2(7)

−24 ± 324 14 −24 ± 18 = 14 =

c) For 4x2 – 12x + 9 = 0, a = 4, b = –12, and c = 9.

=

−b ± b 2 − 4ac 2a 12 ±

( −12 ) − 4 ( 4 )( 9 ) 2 ( 4)

x= =

12 ± 0 8 3 = 2

−4 ± 42 − 4 ( 2 )( −7 ) 2 ( 2)

−4 ± 72 4

−4 + 72 −4 − 72 and . 4 4

d) For 2x2 – 7x + 4 = 0, a = 2, b = –7, and c = 4. x= = =

3 . 2

−b ± b 2 − 4ac 2a

The roots are

2

=

The root is

b) For 2x2 + 4x – 7 = 0, a = 2, b = 4, and c = –7.

=

3 The roots are –3 and − . 7

x=

Page 300

−b ± b 2 − 4ac 2a 7±

( −7 )

2

− 4 ( 2 )( 4 )

2 ( 2)

7 ± 17 4

The roots are

7 + 17 7 − 17 and . 4 4

MHR • Principles of Mathematics 10 Solutions

61

e) For 3x2 + 5x – 1 = 0, a = 3, b = 5, and c = –1.

x= = =

f) For 16x2 + 24x + 9 = 0, a = 16, b = 24, and c = 9.

−b ± b 2 − 4ac 2a

x=

−5 ± 52 − 4 ( 3)( −1)

=

2 ( 3)

−5 ± 37 6

The roots are

−b ± b 2 − 4ac 2a −24 ± 242 − 4 (16 )( 9 ) 2 (16 )

−24 ± 0 32 3 =− 4 3 The root is − . 4 =

−5 + 37 −5 − 37 and . 6 6

Chapter 6 Section 4

Question 2

Page 300

a) For 3x2 + 14x + 5 = 0, a = 3, b = 14, and c = 5. x= =

−b ± b 2 − 4ac 2a −14 ± 142 − 4 ( 3)( 5 ) 2 ( 3)

−14 ± 136 6 −14 ± 2 34 = 6 =

=

−7 ± 34 3

The exact roots are

−7 + 34 −7 − 34 and . The approximate roots are –0.39 and –4.28. 3 3

62 MHR • Principles of Mathematics 10 Solutions

b) For 8x2 + 12x + 1 = 0, a = 8, b = 12, and c = 1. x= =

−b ± b 2 − 4ac 2a −12 ± 122 − 4 ( 8 )(1) 2 (8)

−12 ± 112 16 −12 ± 4 7 = 16 =

=

−3 ± 7 4

The exact roots are

−3 + 7 −3 − 7 and . The approximate roots are –0.09 and –1.41. 4 4

c) For 4x2 – 7x – 1 = 0, a = 4, b = –7, and c = –1. x= = =

−b ± b 2 − 4ac 2a 7±

( −7 )

2

− 4 ( 4 )( −1)

2 ( 4)

7 ± 65 8

The exact roots are

7 + 65 7 − 65 and . The approximate roots are 1.88 and –0.13. 8 8

d) For 10x2 – 45x – 7 = 0, a = 10, b = –45, and c = –7. x= = =

−b ± b 2 − 4ac 2a 45 ±

( −45 ) − 4 (10 )( −7 ) 2 (10 ) 2

45 ± 2305 20

The exact roots are

45 + 2305 45 − 2305 and . The approximate roots are 4.65 and –0.15. 20 20

MHR • Principles of Mathematics 10 Solutions

63

e) For –5x2 + 16x – 2 = 0, a = –5, b = 16, and c = –2. x= = =

−b ± b 2 − 4ac 2a −16 ± 162 − 4 ( −5 )( −2 ) 2 ( −5 )

−16 ± 216 −10 −16 + 216 −16 − 216 and . The approximate roots are 0.13 and 3.07. −10 −10

The exact roots are

f) For –6x2 + 17x + 5 = 0, a = –6, b = 17, and c = 5. x= =

−b ± b 2 − 4ac 2a −17 ± 17 2 − 4 ( −6 )( 5 ) 2 ( −6 )

−17 ± 409 −12 17 ± 409 = 12

=

The exact roots are

17 + 409 17 − 409 and . The approximate roots are 3.10 and –0.27. 12 12

64 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 4

Question 3

Page 300

Use the quadratic formula to solve the corresponding quadratic equation. a) For 5x2 – 14x – 3 = 0, a = 5, b = –14, and c = –3. x= =

−b ± b 2 − 4ac 2a −14 ± 142 − 4 ( 5 )( −3) 2 ( 5)

−14 ± 256 10 −14 ± 16 = 10 =

The x-intercepts are 3 and –0.2. Find the vertex. For the x-coordinate, use x = − x=−

b . 2a

−14 2 ( 5)

= 1.4 y = 5 x 2 − 14 x − 3 = 5 (1.4 ) − 14 (1.4 ) − 3 2

= −12.8 The vertex is (1.4, –12.8). The axis of symmetry is x = 1.4 .

MHR • Principles of Mathematics 10 Solutions

65

b) For 2x2 – 5x – 12 = 0, a = 2, b = –5, and c = –12. x= =

−b ± b 2 − 4ac 2a 5±

( −5 )

2

− 4 ( 2 )( −12 )

2 ( 2)

5 ± 121 10 −14 ± 11 = 10 =

The x-intercepts are 4 and –1.5. Find the vertex. For the x-coordinate, use x = − x=−

b . 2a

−5 2 (2)

= 1.25 y = 2 x 2 − 5 x − 12 = 2 (1.25) − 5 (1.25) − 12 2

= −15.125 The vertex is (1.25, –15.125). The axis of symmetry is x = 1.25 .

66 MHR • Principles of Mathematics 10 Solutions

c) y = x 2 + 10 x + 25

0 = ( x + 5)

2

x+5=0 x = −5 The x-intercept is –5. Find the vertex. −5 − 5 2 = −5

x=

y = x 2 + 10 x + 25 = ( −5 ) + 10 ( −5 ) + 25 2

=0 The vertex is ( −5,0 ) . The axis of symmetry is x = −5 .

MHR • Principles of Mathematics 10 Solutions

67

d) y = 9 x 2 − 24 x + 16

0 = ( 3x − 4 )

2

3x − 4 = 0 x=

4 3

The x-intercept is

4 . 3

Find the vertex. 4 4 + x= 3 3 2 4 = 3 y = 9 x 2 − 24 x + 16 2

⎛4⎞ ⎛4⎞ = 9 ⎜ ⎟ − 24 ⎜ ⎟ + 16 ⎝3⎠ ⎝3⎠ =0

⎛4 ⎞ The vertex is ⎜ ,0 ⎟ . ⎝3 ⎠ The axis of symmetry is x =

4 . 3

68 MHR • Principles of Mathematics 10 Solutions

e) For x2 – 2x + 3 = 0, a = 1, b = –2, and c = 3. x= = =

−b ± b 2 − 4ac 2a 2±

( −2 ) − 4 (1)( 3) 2 (1) 2

2 ± −8 2

There are no x-intercepts. Find the vertex. For the x-coordinate, use x = − x=−

b . 2a

−2 2 (1)

=1 y = x2 − 2 x + 3 = 12 − 2 (1) + 3 =2 The vertex is (1,2 ) . The axis of symmetry is x = 1 .

MHR • Principles of Mathematics 10 Solutions

69

f) For –x2 – 3x – 3 = 0, a = –1, b = –3, and c = –3. x= = =

−b ± b 2 − 4ac 2a 3±

( −3)

2

− 4 ( −1)( −3)

2 ( −1)

3 ± −3 −2

There are no x-intercepts. Find the vertex. For the x-coordinate, use x = − x=−

b . 2a

−3 2 ( −1)

= −1.5 y = − x 2 − 3x − 3 = − ( −1.5) − 3 ( −1.5) − 3 2

= −0.75 The vertex is ( −1.5, −0.75 ) . The axis of symmetry is x = −1.5 .

Chapter 6 Section 4

Question 4

Page 300

a) The vertex is at (3, 2). The graph opens upward. There are no x-intercepts. b) The vertex is at (2,4). The graph opens downward. There are two x-intercepts. c) The vertex is at (–4, –5). The graph opens upward. There are two x-intercepts. d) The vertex is at (–3, –1). The graph opens downward. There are no x-intercepts. e) The vertex is at (5, 0). The graph opens upward. There is one x-intercept.

70 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 4

Question 5

Page 300

a) y = ( x − 3) + 2 2

= x 2 − 6 x + 11 Let y = 0 and use the quadratic formula with a = 1, b = –6, and c = 11. x= = =

−b ± b 2 − 4ac 2a 6±

( −6 )

2

− 4 (1)(11)

2 (1)

6 ± −8 2

There are no x-intercepts.

b) y = − ( x − 2 ) + 4 2

= − x2 + 4 x Factor and solve the corresponding quadratic equation. 0 = − x2 + 4 x 0 = − x ( x − 4) The x-intercepts are 0 and 4.

c) y = 2 ( x + 4 ) − 5 2

= 2 x 2 + 16 x + 27 Let y = 0 and use the quadratic formula with a = 2, b = 16, and c = 27. x= = =

−b ± b2 − 4ac 2a −16 ±

(16 ) − 4 ( 2 )( 27 ) 2 (2) 2

−16 ± 40 4

The x-intercepts are

−16 ± 40 . 4

MHR • Principles of Mathematics 10 Solutions

71

d) y = −2 ( x + 3) − 1 2

= −2 x 2 − 12 x − 19 Let y = 0 and use the quadratic formula with a = –2, b = –12, and c = –19. x= = =

−b ± b 2 − 4ac 2a 12 ±

( −12 )

2

− 4 ( −2 )( −19 )

2 ( −2 )

12 ± −8 −4

There are no x-intercepts.

e) y = ( x − 5) Factor and solve the corresponding quadratic equation. 2

0 = ( x − 5)

2

x −5= 0 x =5 The x-intercept is 5.

72 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 4

Question 6

Page 301

a) Let h = 0 and use the quadratic formula with a = –0.1, b = 1, and c = 0.5. d= =

−b ± b2 − 4ac 2a −1 ±

( −1)

2

− 4 ( −0.1)( 0.5)

2 ( −0.1)

−1 ± 1.2 −0.2 1 ± 1.2 = 0.2 So, d 10.5 or d –0.5. The positive root is 10.5. The ball has travelled 10.5 m horizontally when it lands on the ground. =

b) 2.6 = −0.1d 2 + d + 0.5 0 = −0.1d 2 + d − 2.1 Use the quadratic formula with a = –0.1, b = 1, and c = –2.1. d= =

−b ± b2 − 4ac 2a −1 ±

( −1)

2

− 4 ( −0.1)( −2.1)

2 ( −0.1)

−1 ± 0.16 −0.2 −1 ± 0.4 = −0.2 So, d = 3 or d = 7. The ball is at a height of 2.6 m at a horizontal distance of 3 m or 7 m. =

MHR • Principles of Mathematics 10 Solutions

73

Chapter 6 Section 4

Question 7

Page 301

a) Let h = 0 and use the quadratic formula with a = –4.9, b = 8.4, and c = 1.5. t= =

−b ± b2 − 4ac 2a −8.4 ±

(8.4 ) − 4 ( −4.9 )(1.5) 2 ( −4.9 ) 2

−8.4 ± 99.96 −0.2 8.4 ± 99.96 = 9.8 So, t –0.2 or t 1.9. The positive root is 1.9. The ball lands on the ground after 1.9 s. =

b) For the t-coordinate of the vertex, use t = −

b . 2a

8.4 2( −4.9) 0.86

t=−

h = −4.9t 2 + 8.4t + 1.5 = −4.9 ( 0.86 ) + 8.4 ( 0.86 ) + 1.5 2

5 The ball reaches a height of 5 m. Chapter 6 Section 4

Question 8

Page 301

74 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 4

Question 9

Page 301

7 x 2 − 12 x = 9

a)

7 x 2 − 12 x − 9 = 0 Use the quadratic formula with a = 7, b = –12, and c = –9. x= = =

−b ± b 2 − 4ac 2a 12 ±

( −12 ) − 4 ( 7 )( −9 ) 2(7) 2

12 ± 396 14

x 2.28 or x –0.56. 4 x 2 = 12 − 13x

b)

4 x 2 + 13x − 12 = 0 Use the quadratic formula with a = 4, b = 13, and c = –12. x= =

−b ± b2 − 4ac 2a −13 ±

(13)

2

− 4 ( 4 )( −12 )

2 (4)

−13 ± 361 8 −13 ± 19 = 8 =

x = 0.75 or x = –4. 4 x 2 = 2.8 x + 4.8

c)

4 x 2 − 2.8 x − 4.8 = 0 Use the quadratic formula with a = 4, b = –2.8, and c = –4.8. x= =

−b ± b2 − 4ac 2a 2.8 ±

( −2.8) − 4 ( 4 )( −4.8) 2 (4) 2

2.8 ± 84.64 8 2.8 ± 9.2 = 8 =

x = 1.5 or x –0.8.

MHR • Principles of Mathematics 10 Solutions

75

x ( 3 x − 8 ) = −1

d)

3x 2 − 8 x + 1 = 0 Use the quadratic formula with a = 3, b = –8, and c = 1. x= = =

−b ± b 2 − 4ac 2a 8±

( −8 ) − 4 ( 3)(1) 2 ( 3) 2

8 ± 52 6

x 2.54 or x 0.13

( x − 3)

e)

2

= −2 ( x + 3)

x − 6 x + 9 = −2 x − 6 2

x 2 − 4 x + 15 = 0 Use the quadratic formula with a = 1, b = –4, and c = 15. x= = =

−b ± b 2 − 4ac 2a 4±

( −4 )

2

− 4 (1)(15 )

2 (1)

4 ± −44 2

There are no real roots.

( x + 3)

f)

2

= ( 2 x + 5 )( 2 x − 5 )

x 2 + 6 x + 9 = 4 x 2 − 25 −3x 2 + 6 x + 34 = 0 Use the quadratic formula with a = –3, b = 6, and c = 34. x= = =

−b ± b 2 − 4ac 2a −6 ±

(6)

2

− 4 ( −3)( 34 )

2 ( −3 )

−6 ± 444 −6

x –2.51 or x 4.51.

76 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 4 Question 10 Page 301 Let y = 0 and use the quadratic formula with a = –0.0044, b = 0, and c = 21.3. x= = =

−b ± b 2 − 4ac 2a 0±

(0)

2

− 4 ( −0.0044 )( 21.3)

2 ( −0.0044 )

± 0.37488 −0.0088

The x-intercepts are –69.6 and 69.6. The width of the bridge is 139.2 m. The height of the bridge is 21.3 m. Chapter 6 Section 4

Question 11

Page 301

a) 6 = −4.9t 2 + 8.1t + 3

0 = −4.9t 2 + 8.1t − 3 Use the quadratic formula with a = –4.9, b = 8.1, and c = –3. t= =

−b ± b 2 − 4ac 2a −8.1 ±

(8.1) − 4 ( −4.9 )( −3) 2 ( −4.9 ) 2

−8.1 ± 6.81 −9.8 So, t 0.56 or t 1.09. The rocket first reaches a height of 6 m above the ground at t = 0.56 s. =

b) The rocket falls to a height of 6 m above the ground at 1.09 s. c) The rocket reached its maximum height above the ground at

0.56 + 1.09 , or about 0.83 s. 2

MHR • Principles of Mathematics 10 Solutions

77

Chapter 6 Section 4

Question 12

Page 301

a) 4.2 = 0.2 x 2 − 1.6 x + 4.2

0 = 0.2 x 2 − 1.6 x 0 = 0.2 x ( x − 8 ) So, x = 0 or x = 8. The half-pipe is 8 m wide. b) 2.2 = 0.2 x 2 − 1.6 x + 4.2

0 = 0.2 x 2 − 1.6 x + 2 Use the quadratic formula with a = 0.2, b = –1.6, and c = 2. x= =

−b ± b 2 − 4ac 2a 1.6 ±

( −1.6 ) − 4 ( 0.2 )( 2 ) 2 ( 0.2 ) 2

1.6 ± 0.96 0.4 So, x 6.45 or x 1.55. A skater would have travelled 1.55 m horizontally after a drop of 2 m. =

Chapter 6 Section 4

Question 13

Page 302

The vertex is at (8, 32). Therefore, the width of the arch is 16 m. Chapter 6 Section 4

Question 14

Page 302

a) Let f = 0 and use the quadratic formula with a = 0.0048, b = –0.96, and c = 64. v= = =

−b ± b 2 − 4ac 2a 0.96 ±

( −0.96 ) − 4 ( 0.0048)( 64 ) 2 ( 0.0048 ) 2

0.96 ± −0.3072 0.0096

There are no v-intercepts. b) For the v-coordinate of the vertex, use v = −

b . 2a

−0.96 2(0.0048) = 100

v=−

The speed that minimizes fuel flow is 100 km/h.

78 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 4

Question 15

Page 302

Solutions for the Achievement Checks are shown in the Teacher Resource. Chapter 6 Section 4

Question 16

Page 302

−b ± b2 − 4ac 14 ± 140 . with x = 2a 4 b = −14

a) Compare x =

2a = 4 a=2

( −14 )

2

− 4 ( 2 ) c = 140 c=7

2 x − 14 x + 7 = 0 2

−b ± b2 − 4ac −11 ± 145 . with x = 6 2a b = 11

b) Compare x =

2a = 6 a =3

(11)

2

− 4 ( 3) c = 145 c = −2

3x + 11x − 2 = 0 2

Chapter 6 Section 4

Question 17

Page 302

a) Three line segments can be drawn joining any two of three points. b) Six line segments can be drawn joining any two of four points, 10 line segments if there are 5 points, and 15 line segments if there are 6 points. c) Following the pattern in b), if there are n points, you can draw

n ( n − 1) 2

segments.

MHR • Principles of Mathematics 10 Solutions

79

n ( n − 1)

= 1000 2 n 2 − 2n − 2000 = 0 Use the quadratic formula with a = 1, b = –2, and c = –2000. d)

n= =

−b ± b 2 − 4ac 2a 2±

( −2 )

2

− 4 (1)( −2000 )

2 (1)

2 ± 8004 2 So, n 45.7 or n –43.7. You need 46 points to have at least 1000 line segments. =

Chapter 6 Section 4

Question 18

Page 302

Answers may vary. For example: Consider the equation x 2 − 6 x + 9 = 0 . When graphed there is one x-intercept at 3. This is also the vertex. There is only one real root.

80 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 4

x 2 + y 2 = 16

Question 19

Page 302

c

y = x −9 d 2 Substitute x – 9 for y in equation c. x 2 + y 2 = 16 2

x 2 + ( x 2 − 9 ) = 16 2

x 4 − 17 x 2 + 65 = 0 Use the quadratic formula with a = 1, b = –17, and c = 65. x2 = =

−b ± b 2 − 4ac 2a 17 ±

( −17 ) − 4 (1)( 65) 2 (1) 2

17 ± 29 2 So, x 11.19 or x 5.81. x 2 = 11.19 =

x 2 = 5.81

x ±3.35

x ±2.41

y = x −9

y = x2 − 9

2

= 11.19 − 9 = 2.19

= 5.81 − 9 = −3.19

The points of intersection are ( 3.35, 2.19 ) , ( −3.35, 2.19 ) , ( 2.41, −3.19 ) , and ( −2.41, −3.19 ) .

Chapter 6 Section 4

Question 20

Page 302

The possible numbers of intersection points are 0, 1, 2, 3, and 4. Equations will vary.

MHR • Principles of Mathematics 10 Solutions

81

Chapter 6 Section 4

x =1+

Question 21

Page 302

1 x

x2 − x − 1 = 0 Use the quadratic formula with a = 1, b = –1, and c = –1. x= = =

−b ± b 2 − 4ac 2a 1±

( −1)

2

− 4 (1)( −1)

2 (1)

1± 5 2

1+ 5 1− 5 and . 2 2 ⎛ 1 − 5 ⎞⎛ 1 + 5 ⎞ 1 − 5 ⎜⎜ ⎟⎜ ⎟⎟ = ⎟⎜ 4 ⎝ 2 ⎠⎝ 2 ⎠ = −1

The roots are

The roots are negative reciprocals. Chapter 6 Section 4

Question 22

Page 303

a) Use a calculator to evaluate the continued fraction, and express the answer as a fraction:

157 . 68

b) Use a calculator to evaluate the continued fraction for several steps. The answer to three decimal places is 1.618, the golden ratio. Chapter 6 Section 4

Question 23

Page 303

a) The next three terms are 34, 55, and 89. Each term is found by adding the two preceding terms. b) The ratios of terms approach 1.618, the golden ratio.

82 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 4

a–b+c=0 a–c+d=0 a + 2b + 2d = 0 b–d=3

Question 24

Page 303

c d e f

Add equations c and d to obtain 2a − b + d = 1 . g Rearrange equation f: −b + d = −3 . Substitute into equation g. 2a − b + d = 1 2a − 3 = 1

a=2 Substitute a = 2 into equation e. a + 2b + 2d = 0 2 + 2b + 2 d = 0 h b + d = −1 Add equations f and h. f b–d=3 h b + d = −1 2b = 2 f+h b =1 Substitute b = 1 into equation f to determine that d = –2. Substitute a = 2 and b = 1 into equation c. a–b+c=0 2 −1+ c = 0

c = −1 a=2 b =1 c = −1 d = −2

MHR • Principles of Mathematics 10 Solutions

83

Chapter 6 Section 5

Chapter 6 Section 5

Question 1

Page 311

a) h = −4.9t 2 + 45t + 2 b) Let h = 0 and use the quadratic formula with a = –4.9, b = 45, and c = 2. t= =

−b ± b 2 − 4ac 2a −45 ±

( 45) − 4 ( −4.9 )( 2 ) 2 ( −4.9 ) 2

−45 ± 2064.2 −9.8 So, t –0.04 or t 9.23. The rocket would take 9.23 s to fall to Earth. =

Chapter 6 Section 5

Question 2

a) For the t-coordinate of the vertex, use t = −

t=−

Page 311

b . 2a

49 2( −4.9)

=5 2 h = −4.9 ( 5 ) + 49 ( 5 ) + 1.5 = 124

The maximum height of the firework is 124 m above the ground. b) 100 = −4.9t 2 + 49t + 1.5

0 = −4.9t 2 + 49t − 98.5 Use the quadratic formula with a = –4.9, b = 49, and c = –98.5. t= =

−b ± b 2 − 4ac 2a −49 ±

( 49 )

2

− 4 ( −4.9 )( −98.5)

2 ( −4.9 )

−49 ± 470.4 −9.8 So, t 2.79 or t 7.21. The firework is more than 100 m above the ground over the time interval 2.79 ≤ t ≤ 7.21 . =

84 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 5

Question 3

Page 312

x ( x + 16 ) = 35 x + 16 x − 35 = 0 Use the quadratic formula with a = 1, b = 16, and c = –35. 2

x= =

−b ± b 2 − 4ac 2a −16 ±

(16 )

2

− 4 (1) ( −35)

2 (1)

−16 ± 396 2 So, x 1.95 or x –17.95. The width is 1.95 cm and the length is 17.95 cm. =

Chapter 6 Section 5

Question 4

Page 312

x ( x + 1) = 3306 x + x − 3306 = 0 Use the quadratic formula with a = 1, b = 1, and c = –3306. 2

x= =

−b ± b 2 − 4ac 2a −1 ±

(1)

2

− 4 (1)( −3306 )

2 (1)

−1 ± 13 225 2 −1 ± 115 = 2 So, x = 57 or x = –58. The numbers are 57 and 58, or –57 and –58. =

MHR • Principles of Mathematics 10 Solutions

85

Chapter 6 Section 5

Question 5

Page 312

x ( x + 2 ) = 323 x + 2 x − 323 = 0 Use the quadratic formula with a = 1, b = 2, and c = –323. 2

x= =

−b ± b2 − 4ac 2a −2 ±

(2)

2

− 4 (1)( −323)

2 (1)

−2 ± 1296 2 −2 ± 36 = 2 So, x = 17 or x = –19. The numbers are 17 and 19, or –17 and –19. =

Chapter 6 Section 5

( 2 x + 3)

2

= x2 + ( x + 7 )

Question 6

Page 312

2

4 x 2 + 12 x + 9 = x 2 + x 2 + 14 x + 49 2 x 2 − 2 x − 40 = 0 x 2 − x − 20 = 0

( x − 5)( x + 4 ) = 0 So, x = 5 or x = –4. The negative root is inadmissible. The lengths of the sides are 5 cm, 12 cm, and 13 cm. Chapter 6 Section 5

Question 7

Page 312

a)

b) y = −0.05 x 2 + 0.95 x + 0.5

86 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 5

Question 8

Page 312

V = π r 2h 600 = π r 2 (12 ) 50

π

= r2

r = 4.0 The radius of the cylinder is 40 mm. Chapter 6 Section 5

Question 9

Page 312

1 b ( b + 4 ) = 20 2 b ( b + 4 ) = 40 b2 + 4b − 40 = 0 Use the quadratic formula with a = 1, b = 4, and c = –40. x= =

−b ± b2 − 4ac 2a −4 ±

(4)

− 4 (1)( −40 )

2

2 (1)

−4 ± 176 2 So, x 4.6 or x –8.6. The length of the base is 46 mm. =

Chapter 6 Section 5

Question 10

Page 312

x 2 + ( x + 1) = 365 2

x 2 + x 2 + 2 x + 1 = 365 2 x 2 + 2 x − 364 = 0 x 2 + x − 182 = 0 Use the quadratic formula with a = 1, b = 1, and c = –182. x= =

−b ± b2 − 4ac 2a −1 ±

(1)

2

− 4 (1)( −182 )

2 (1)

−1 ± 729 2 −1 ± 27 = 2 So, x = 13 or x = –14. The integers are 13 and 14 or –13 and –14. =

MHR • Principles of Mathematics 10 Solutions

87

Chapter 6 Section 5

Question 11

Page 312

Perimeter: 2 x + 2 y = 23 Area: xy = 33 Solve for y in the perimeter equation and substitute into the area equation. xy = 33 x (11.5 − x ) = 33 − x + 11.5 x − 33 = 0 Use the quadratic formula with a = –1, b = 11.5, and c = –33. 2

x= =

−b ± b2 − 4ac 2a −11.5 ±

(11.5) − 4 ( −1)( −33) 2 ( −1) 2

−11.5 ± 0.25 −2 −11.5 ± 0.5 = −2 So, x = 5.5 or x = 6. =

One dimension is 6 cm. The other dimension is

33 , or 5.5 cm. 6

Chapter 6 Section 5

Page 312

Question 12

y + 2 x = 1200 y = 1200 − 2 x xy = 180 000 x (1200 − 2 x ) = 180 000 −2 x + 1200 x − 180 000 = 0 Use the quadratic formula with a = –2, b = 1200, and c = –180 000. 2

x= = =

−b ± b 2 − 4ac 2a −1200 ±

(1200 )

2

− 4 ( −2 ) ( −180 000 )

2 ( −2 )

−1200 ± 0 −4

The width is 300 m. The length is 1200 – 2(300), or 600 m.

88 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 5

x2 + ( x + 2) = ( x + 4) 2

Question 13

Page 313

2

x 2 + x 2 + 4 x + 4 = x 2 + 8 x + 16 x 2 − 4 x − 12 = 0

( x − 6 )( x + 4 ) = 0 So, x = 6 or x = –4. The negative solution is inadmissible. The sides of the triangle measure 6 units, 8 units, and 10 units. Chapter 6 Section 5

Question 14

Page 313

x 2 + (10 x ) = 62 2

101x 2 = 36 36 x2 = 101 x 0.597

The top of the ladder must be placed no higher than 5.97 m.

MHR • Principles of Mathematics 10 Solutions

89

Chapter 6 Section 5

Question 15

Page 313

a) h = −4.9t 2 + 36.85t + 0.61 b) Answers may vary. For example: Let h = 0 and use the quadratic formula with a = –4.9, b = 36.85, and c = 0.61. t= =

−b ± b 2 − 4ac 2a −36.85 ±

( 36.85) − 4 ( −4.9 )( 0.61) 2 ( −4.9 ) 2

−36.85 ± 1369.88 −9.8 So, t –0.02 or t 7.54. The rocket was in the air 7.54 s. =

For the t-coordinate of the vertex, use t = −

b . 2a

36.85 2( −4.9) 3.76 h = −4.9t 2 + 36.85t + 0.61

t=−

= −4.9 ( 3.76 ) + 36.85 ( 3.76 ) + 0.61 2

69.89

The maximum height reached was 69.89 m the time is 3.76 s. c)

90 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 5

Question 16

Page 313

a) h = −4.9t 2 + 15t + 1 b) h = −4.9t 2 + 15t + 1 = −4.9 (1) + 15 (1) + 1 2

= 11.1

The height of the ball after 1 s is 11.1 m. c) Let h = 0 and use the quadratic formula with a = –4.9, b = 15, and c = 1. t= =

−b ± b 2 − 4ac 2a −15 ±

(15) − 4 ( −4.9 )(1) 2 ( −4.9 ) 2

−15 ± 244.6 −9.8 So, t –0.1 or t 3.1. The ball lands after 3.1 s. =

d) For the t-coordinate of the vertex, use t = −

b . 2a

15 2( −4.9) 1.5 h = −4.9t 2 + 15t + 1

t=−

= −4.9 (1.5) + 15 (1.5) + 1 2

12.5

The ball reached a maximum height of 12.5 m after 1.5 s.

MHR • Principles of Mathematics 10 Solutions

91

e) h = −0.81t 2 + 15t + 1 = −0.81(1) + 15 (1) + 1 2

15.2 The height of the ball after 1 s is 15.2 m.

Let h = 0 and use the quadratic formula with a = –0.81, b = 15, and c = 1. t= =

−b ± b 2 − 4ac 2a −15 ±

(15) − 4 ( −0.81)(1) 2 ( −0.81) 2

−15 ± 228.24 −1.62 So, t –0.1 or t 3.1. The ball lands after 18.6 s. =

For the t-coordinate of the vertex, use t = −

b . 2a

15 2( −0.81) 9.3 h = −0.81t 2 + 15t + 1

t=−

= −0.81( 9.3) + 15 ( 9.3) + 1 2

70.4

The ball reached a maximum height of 70.4 m after 9.3 s.

92 MHR • Principles of Mathematics 10 Solutions

f) h = −11.55t 2 + 15t + 1 = −11.55 (1) + 15 (1) + 1 2

4.5 The height of the ball after 1 s is 4.5 m.

Let h = 0 and use the quadratic formula with a = –11.55, b = 15, and c = 1. t= =

−b ± b 2 − 4ac 2a −15 ±

(15) − 4 ( −11.55)(1) 2 ( −11.55) 2

−15 ± 271.2 −23.1 So, t –0.1 or t 1.4. The ball lands after 1.4 s. =

For the t-coordinate of the vertex, use t = −

b . 2a

15 2( −11.55) 0.65 h = −11.55t 2 + 15t + 1

t=−

= −11.55 ( 0.65 ) + 15 ( 0.65 ) + 1 2

= 5.9

The ball reached a maximum height of 5.9 m after 0.6 s.

MHR • Principles of Mathematics 10 Solutions

93

Chapter 6 Section 5

Question 17

Page 313

a) R = (10 − 0.5 x )( 30 + 2 x ) , where x represents the number of price reductions. b)

R = (10 − 0.5 x )( 30 + 2 x ) 150 = 300 + 5 x − x 2 0 = 150 + 5 x − x 2

0 = (15 − x )(10 + x ) So, x = 15 or x = –10. The revenue will be \$150 after 15 price reductions. The price is 10 − 15 ( 0.5) , or \$2.50.

c) R = (10 − 0.5 x )( 30 + 2 x ) 0 = 300 + 5 x − x 2 0 = ( 20 − x )(15 + x ) So, x = 20 or x = –15.

The x-intercepts are –15 and 20. The x-coordinate of the vertex is

−15 + 20 , or 2.5. 2

The maximum revenue occurs after 2.5 price reductions. The price is 10 − 2.5 ( 0.5) , or \$8.75. d) Her revenue will be \$0 at a price of \$0. e)

Part b) is represented by the point (15, 150). Part c) is represented by the vertex (2.5, 306.25), which is the maximum point. Part d) is represented by the x-intercept 20.

94 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 5

Question 18

Page 313

( 30 + 2 x )( 20 + 2 x ) = 1064 600 + 100 x + 4 x 2 = 1064 4 x 2 + 100 x − 464 = 0 x 2 + 25 x − 116 = 0

( x − 4 )( x + 29 ) = 0 So, x = 4 or x = –29. The negative answer is inadmissible. The new frame measures 28 cm by 38 cm. Chapter 6 Section 5

Question 19

Page 313

( 24 + 2 x )(15 + 2 x ) = 1.5 (15)( 24 ) 360 + 78 x + 4 x 2 = 540 4 x 2 + 78 x − 180 = 0 2 x 2 + 39 x − 90 = 0 Use the quadratic formula with a = 2, b = 39, and c = –90. x= =

−b ± b 2 − 4ac 2a −39 ±

( 39 )

2

− 4 ( 2 )( −90 )

2 (2)

−39 ± 2241 4 So, x 2.1 or x –21.6. The negative answer is inadmissible. The new dimensions are 19.2 m and 28.2 m. =

MHR • Principles of Mathematics 10 Solutions

95

Chapter 6 Section 5

Question 20

Page 313

w ( 3w + 2 ) = 1496 3w + 2 w − 1496 = 0 Use the quadratic formula with a = 3, b = 2, and c = –1496. 2

w= =

−b ± b 2 − 4ac 2a −2 ±

(2)

2

− 4 ( 3)( −1496 )

2 ( 3)

−2 ± 17 956 6 −2 ± 134 = 6 So, w = 22 or w –22.7. The negative answer is inadmissible. The width of the field is 22 m and the length is 68 m. =

96 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 5

Question 21

Page 314

a)

b) ( 40 − 2 x )( 50 − 2 x ) = 875 2000 − 180 x + 4 x 2 = 875 4 x 2 − 180 x + 1125 = 0 Use the quadratic formula with a = 4, b = –180, and c = 1125. x= =

−b ± b2 − 4ac 2a 180 ±

( −180 ) − 4 ( 4 )(1125) 2 (4) 2

180 ± 14 400 8 180 ± 120 = 8 So, x = 37.5 or x = 7.5. The side length of the squares being removed is 7.5 cm. =

c) V = 875 × x

= 875 ( 7.5 ) = 6562.5 The volume of the box is 6562.5 cm2.

MHR • Principles of Mathematics 10 Solutions

97

Chapter 6 Section 5

Question 22

Page 314

( 21 − 2 x )(15 − 2 x ) = 216 315 − 72 x + 4 x 2 = 216 4 x 2 − 72 x + 99 = 0 Use the quadratic formula with a = 4, b = –72, and c = 99.

x= =

−b ± b2 − 4ac 2a 72 ±

( −72 ) − 4 ( 4 )( 99 ) 2 (4) 2

72 ± 3600 8 72 ± 60 = 8 So, x = 79.5 or x = 1.5. The width of the cut is 1.5 cm. =

Chapter 6 Section 5

Question 23

Page 314

Question 24

Page 314

( 20 − x )(16 − x ) = 0.6 ( 20 )(16 ) 320 − 36 x + x 2 = 192 x 2 − 36 x + 128 = 0

( x − 4 )( x − 32 ) = 0 So, x = 4 or x = 32. The width of the cut is 4 cm. Chapter 6 Section 5

1 2 2 300 − 35 x + x = 150

( 20 − x )(15 − x ) = ( 20 )(15) x 2 − 35 x + 150 = 0

( x − 5)( x − 30 ) = 0 So, x = 5 or x = 30. The reduction is 5 m. The dimensions of the fenced-in area are 10 m by 15 m.

98 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 5

Question 25

Page 314

Answers will vary. For example: Set the vertex 23 cm below the origin. The parabola opens upward. y = ax 2 − 23 0 = a ( 300 ) − 23 2

23 90 000 23 y= x 2 − 23 90 000

a=

Chapter 6 Section 5

Question 26

Page 314

a)

b) y = 0.008 x 2 − 0.383x + 8.726 c) Using the equation from part b), the stopping distance for a speed of 110 km/h is 63.4 m. d) Use the equation from part b) and the quadratic formula. A speed of 81 km/h results in a stopping distance of 30 m. A speed of 111 km/h results in a stopping distance of 65 m. A speed of 180 km/h results in a stopping distance of 200 m. e) Answers may vary. For example: The model does not make sense for speeds less than 24.5 km/h because the stopping distances should be less when the car is going slower.

MHR • Principles of Mathematics 10 Solutions

99

Chapter 6 Section 5

Question 27

Page 314

a) x 2 + 2 x + 7 = x 2 − 4 x − 1 6 x = −7

There is one point of intersection because the resulting equation is linear. b) 3x 2 − 12 x + 16 = −2 x 2 − 4 x + 3

5 x 2 − 8 x + 13 = 0 Use the quadratic formula. x= =

−b ± b 2 − 4ac 2a 8±

( −8 )

2

− 4 ( 5)(13)

2 ( 5)

8 ± −196 10 There are no points of intersection because the resulting quadratic equation has no real roots. =

x 2 − 6 x + 10 = 5 x 2 − 30 x + 46

c)

−4 x 2 + 24 x − 36 = 0 Use the quadratic formula. x= =

−b ± b2 − 4ac 2a −24 ±

( 24 )

2

− 4 ( −4 )( −36 )

2 ( −4 )

−24 ± 0 −8 There is one point of intersection because the resulting quadratic equation has two equal real roots. =

Chapter 6 Section 5

Question 28

Page 315

a) WC = 0.0032w2 − 0.425w + 6 , where WC represents the wind chill temperature and w represents the wind speed. b) The QuadReg operation results in a linear relation, since the coefficient of the x2-term is 0. WC = t − 13 , where WC represents the wind chill temperature and t represents the air temperature. c) Answers may vary. For example: The wind chill model from part b) is very good, because the data follow a linear model exactly. The model from part a) is quite good because the result for w = 60 is very close to the actual result.

100 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Section 5

Question 29

Page 315

Answers will vary. For example: If the initial speed is in a diagonal direction, only the component that is vertical is affected by gravity. Chapter 6 Section 5 Question 30 Solve two equations with two unknowns.

Page 315

Using the maximum height reached, the equation will be of the form y = a ( x − h ) + 14 . Use the two points (0, 4) and (24, 0). 2 4 = a ( 0 − h ) + 14 2 0 = a ( 24 − h ) + 14 2 −10 = a ( 0 − h ) 2 −14 = a ( 24 − h ) −10 = ah 2 −10 =a c −14 = a ( 576 − 48h + h 2 ) d 2 h 2

Substitute the expression for a from equation c into equation d. −14 = a ( 576 − 48h + h 2 ) ⎛ −10 ⎞ −14 = ⎜ 2 ⎟ ( 576 − 48h + h 2 ) ⎝ h ⎠ 2 −14h = −5760 + 480h − 10h 2 −4h 2 − 480h + 5760 = 0 h 2 + 120h − 1440 = 0 Use the quadratic formula. h= =

−b ± b2 − 4ac 2a −120 ±

( −120 ) − 4 (1)( −1440 ) 2 (1) 2

−120 ± 20 160 2 So, h –131 or h 11. =

Substitute h = 11 into equation c. −10 =a 112 10 a=− 121

MHR • Principles of Mathematics 10 Solutions

101

Now , use the equation y = −

10 2 ( x − 11) + 14 with h = 10. 121

10 2 ( x − 11) + 14 121 1210 = −10 x 2 + 220 x − 1210 + 1694 10 = −

10 x 2 − 220 x + 726 = 0 Use the quadratic formula. x= =

−b ± b 2 − 4ac 2a 220 ±

( −220 ) − 4 (10 )( 726 ) 2 (10 ) 2

220 ± 19 360 20 So, x 4 or x 18. The pumpkin was at a height of 10 m at horizontal distances of 4 m and 18 m. =

Chapter 6 Section 5

Question 31

Page 315

Use a calculator to verify the formula. f1 = 1 f2 = 1 f3 = 2 f4 = 3 f5 = 5

102 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Review Chapter 6 Review

Question 1

Page 316

a) y = x 2 + 8 x − 7

= ( x2 + 8x ) − 7

= ( x 2 + 8 x + 42 − 42 ) − 7 = ( x 2 + 8 x + 42 ) − 42 − 7 = ( x + 4 ) − 23 2

MHR • Principles of Mathematics 10 Solutions

103

b) y = x 2 + 2 x + 7

= ( x2 + 2x ) + 7

= ( x 2 + 2 x + 12 − 12 ) + 7 = ( x 2 + 2 x + 12 ) − 12 + 7 = ( x + 1) + 6 2

104 MHR • Principles of Mathematics 10 Solutions

c) y = x 2 + 4 x + 6

= ( x2 + 4 x ) + 6

= ( x 2 + 4 x + 22 − 22 ) + 6 = ( x 2 + 4 x + 22 ) − 22 + 6 = ( x + 2) + 2 2

MHR • Principles of Mathematics 10 Solutions

105

d) y = x 2 + 6 x − 3

= ( x2 + 6x ) − 3

= ( x 2 + 6 x + 32 − 32 ) − 3 = ( x 2 + 6 x + 32 ) − 32 − 3 = ( x + 3) − 12 2

106 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Review

Question 2

Page 316

a) y = x 2 + 12 x + 30

= ( x 2 + 12 x ) + 30

= ( x 2 + 12 x + 62 − 62 ) + 30 = ( x 2 + 12 x + 62 ) − 62 + 30 = ( x + 6) − 6 2

The vertex is (–6, –6).

b) y = x 2 − 14 x + 50

(

)

= x 2 − 14 x + 50

( = (x

)

= x 2 − 14 x + ( −7 ) − ( −7 ) + 50 2

2

2

)

− 14 x + ( −7 ) − ( −7 ) + 50 2

2

= ( x − 7) + 1 2

The vertex is (7, 1).

c) y = − x 2 + 6 x − 7

(

)

= − x2 − 6x − 7

( = −(x

)

= − x 2 − 6 x + ( −3) − ( −3) − 7 2

2

2

)

− 6 x + ( −3) − ( −1)( −3) − 7 2

2

= − ( x − 3) + 2 2

The vertex is (3, 2).

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107

d) y = 5 x 2 − 40 x + 76

(

)

= 5 x 2 − 8 x + 76

( = 5( x

)

= 5 x 2 − 8 x + ( −4 ) − ( −4 ) + 76 2

2

)

− 8 x + ( −4 ) − 5 ( −4 ) + 76 2

2

2

= 5( x − 4) − 4 2

The vertex is (4, –4).

Chapter 6 Review

Question 3

Page 316

a) The minimum point is at (0.6, 7.3). b) The maximum point is at (0.2, –200.1). c) The minimum point is at (0.2, 0.6). Chapter 6 Review

Question 4

Page 316

a) x 2 + 10 x + 21 = 0

( x + 3)( x + 7 ) = 0 x + 3 = 0 or x + 7 = 0 x = −3 or x = −7 Check by substituting both solutions in the original equation. For x = –3: For x = –7: R.S. = 0 L.S. = x 2 + 10 x + 21 L.S. = x 2 + 10 x + 21 = ( −3) + 10 ( −3) + 21

= ( −7 ) + 10 ( −7 ) + 21

=0

=0

2

2

L.S. = R.S.

The roots are –3 and –7.

108 MHR • Principles of Mathematics 10 Solutions

L.S. = R.S.

R.S. = 0

m 2 + 8m − 20 = 0

b)

( m − 2 )( m + 10 ) = 0 m−2=0 m=2

or m + 10 = 0 or m = −10

Check by substituting both solutions in the original equation. For m = 2: For m = –10: R.S. = 0 L.S. = m 2 + 8m − 20 L.S. = m 2 + 8m − 20

R.S. = 0

= ( 2 ) + 8 ( 2 ) − 20

= ( −10 ) + 8 ( −10 ) − 20

=0

=0

2

2

L.S. = R.S. The roots are 2 and –10.

L.S. = R.S.

c)

6 y 2 + 21 y + 9 = 0

Divide both sides by 3.

2y + 7y + 3 = 0 2

2y + 6y + y + 3 = 0 2

(2 y

2

+ 6 y ) + ( y + 3) = 0

2 y ( y + 3 ) + 1 ( y + 3) = 0

( y + 3)( 2 y + 1) = 0 y+3=0

or 2 y + 1 = 0

y = −3 or

y=−

1 2

Check by substituting both solutions in the original equation. 1 For y = –3: For y = − : 2 R.S. = 0 L.S. = 6 y 2 + 21y + 9 L.S. = 6 y 2 + 21 y + 9 = 6 ( −3) + 21( −3) + 9 2

=0

L.S. = R.S.

The roots are –3 and –

R.S. = 0

2

⎛ 1⎞ ⎛ 1⎞ = 6 ⎜ − ⎟ + 21⎜ − ⎟ + 9 ⎝ 2⎠ ⎝ 2⎠ =0 L.S. = R.S.

1 . 2

MHR • Principles of Mathematics 10 Solutions

109

5n 2 + 13n − 6 = 0

d)

5n 2 + 15n − 2n − 6 = 0

(5n

2

)

+ 15n + ( −2n − 6 ) = 0

5n ( n + 3) − 2 ( n + 3) = 0

( n + 3)( 5n − 2 ) = 0 n+3=0

or 5n − 2 = 0 2 n = −3 or n= 5

Check by substituting both solutions in the original equation. 2 For n = –3: For n = : 5 2 R.S. = 0 L.S. = 5n 2 + 13n − 6 L.S. = 5n + 13n − 6 = 5 ( −3) + 13 ( −3) − 6 2

=0

⎛2⎞ ⎛2⎞ = 5 ⎜ ⎟ + 13 ⎜ ⎟ − 6 ⎝5⎠ ⎝5⎠ =0

L.S. = R.S.

The roots are –3 and

2

2 . 5

110 MHR • Principles of Mathematics 10 Solutions

L.S. = R.S.

R.S. = 0

Chapter 6 Review

Question 5

Page 316

y2 = 8 y + 9

a)

y2 − 8 y − 9 = 0

( y + 1)( y − 9 ) = 0 y + 1 = 0 or y − 9 = 0 y = −1 or y=9

x 2 − 8 x = −7

b)

x2 − 8x + 7 = 0

( x − 1)( x − 7 ) = 0 x −1 = 0

or x − 7 = 0

x =1

or

x=7

3m 2 = −10m − 7

c)

3m 2 + 10m + 7 = 0 3m 2 + 3m + 7 m + 7 = 0

( 3m

2

)

+ 3m + ( 7 m + 7 ) = 0

3m ( m + 1) + 7 ( m + 1) = 0

( m + 1)( 3m + 7 ) = 0 m +1 = 0

or 3m + 7 = 0

m = −1 or d)

m=−

7 3

30 x − 25 x 2 = 9 25 x 2 − 30 x + 9 = 0

( 5 x − 3)

2

=0

5x − 3 = 0 3 x= 5

MHR • Principles of Mathematics 10 Solutions

111

8k 2 = −5 + 14k

e)

8k 2 − 14k + 5 = 0 8k 2 − 4k − 10k + 5 = 0

(8k

2

)

− 4k + ( −10k + 5) = 0

4k ( 2k − 1) − 5 ( 2k − 1) = 0

( 2k − 1)( 4k − 5) = 0 2k − 1 = 0 or 4k − 5 = 0 1 5 k= or k= 2 4 3 x 2 + 2 = −5 x

f)

3x 2 − 5 x + 2 = 0 3x 2 + 3x + 2 x + 2 = 0

( 3x

2

)

+ 3x + ( 2 x + 2 ) = 0

3 x ( x + 1) + 2 ( x + 1) = 0

( x + 1)( 3x + 2 ) = 0 x +1 = 0

or 3x + 2 = 0

x = −1 or Chapter 6 Review

( 3x + 1)

2

= x 2 + ( 3x − 1)

x=−

2 3

Question 6

Page 316

2

9 x2 + 6x + 1 = x2 + 9 x2 − 6x + 1 − x 2 + 12 x = 0 x 2 − 12 x = 0 x ( x − 12 ) = 0 x=0 or x − 12 = 0

or x = 12 The side lengths are 12 cm, 35 cm, and 37 cm.

112 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Review

Question 7

Page 316

a) Factor and solve the corresponding quadratic equation. y = x 2 + 8 x + 12

0 = ( x + 2 )( x + 6 ) The x-intercepts are –2 and –6. Find the x-coordinate of the vertex, and then, the ycoordinate. −2 − 6 x= 2 = −4 y = x 2 + 8 x + 12 = ( −4 ) + 8 ( −4 ) + 12 2

= −4 The vertex is ( −4, −4 ) . b) Factor and solve the corresponding quadratic equation. y = x2 − 4 x − 5

0 = ( x + 1)( x − 5) The x-intercepts are –1 and 5. Find the x-coordinate of the vertex, and then, the ycoordinate. −1 + 5 x= 2 =2 y = x2 − 4 x − 5 = ( 2) − 4 ( 2) − 5 2

= −9 The vertex is ( 2, −9 ) .

MHR • Principles of Mathematics 10 Solutions

113

c) Factor and solve the corresponding quadratic equation. y = − x 2 − 6 x + 27

0 = − ( x 2 + 6 x − 27 ) 0 = − ( x + 9 )( x − 3) The x-intercepts are –9 and 3. Find the x-coordinate of the vertex, and then, the ycoordinate. −9 + 3 x= 2 = −3 y = − x 2 − 6 x + 27 = − ( −3) − 6 ( −3) + 27 2

= 36 The vertex is ( −3,36 ) .

114 MHR • Principles of Mathematics 10 Solutions

d) Factor and solve the corresponding quadratic equation. y = 3 x 2 + 10 x + 8 0 = 3x 2 + 6 x + 4 x + 8

(

)

0 = 3x 2 + 6 x + ( 4 x + 8 ) 0 = 3x ( x + 2 ) + 4 ( x + 2 ) 0 = ( x + 2 )( 3 x + 4 ) x + 2 = 0 or 3x + 4 = 0

x = −2

or

x=−

4 3

The x-intercepts are –2 and –

4 . 3

Find the x-coordinate of the vertex, and then, the y-coordinate. 4 −2 − 3 x= 2 5 =− 3 y = 3 x 2 + 10 x + 8 2

⎛ 5⎞ ⎛ 5⎞ = 3 ⎜ − ⎟ + 10 ⎜ − ⎟ + 8 ⎝ 3⎠ ⎝ 3⎠ 1 =− 3

⎛ 5 1⎞ The vertex is ⎜ − , − ⎟ . ⎝ 3 3⎠

MHR • Principles of Mathematics 10 Solutions

115

e) Factor and solve the corresponding quadratic equation. y = − x 2 − 3x

0 = − x ( x + 3) The x-intercepts are 0 and –3. Find the x-coordinate of the vertex, and then, the ycoordinate. 0−3 x= 2 3 =− 2 y = − x 2 − 3x + 12 2

⎛ 3⎞ ⎛ 3⎞ = − ⎜ − ⎟ − 3⎜ − ⎟ ⎝ 2⎠ ⎝ 2⎠ 9 = 4

⎛ 3 9⎞ The vertex is ⎜ − , ⎟ . ⎝ 2 4⎠ f) Factor and solve the corresponding quadratic equation. y = x2 − 4

0 = ( x + 2 )( x − 2 ) The x-intercepts are –2 and 2. Find the x-coordinate of the vertex, and then, the ycoordinate. −2 + 2 x= 2 =0 y = x2 − 4 = ( 0) − 4 2

= −4 The vertex is ( 0, −4 ) . Chapter 6 Review

Question 8

Page 316

If two different quadratic relations have the same zeros, they will have the same axis of symmetry because it will pass through the midpoint of the line segment connecting the x-intercepts. However, the vertex can be different because vertical stretching or compressing will change the y-coordinate of the vertex but not the zeros.

116 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Review

Question 9

Page 316

a) k = 20 or –20; y = x 2 + 20 x + 100 or y = x 2 − 20 x + 100 b) k = –49; y = −4 x 2 + 28 x − 49 Chapter 6 Review

Question 10

Page 316

a) For –3x2 – 2x + 5 = 0, a = –3, b = –2, and c = 5.

x= =

−b ± b2 − 4ac 2a 2±

( −2 ) − 4 ( −3)(5) 2 ( −3) 2

2 ± 64 −6 2±8 = −6 =

5 The roots are 1 and − . 3 b) For 9x2 – 8x – 3 = 0, a = 9, b = –8, and c = –3.

x= =

−b ± b 2 − 4ac 2a 8±

( −8)

2

− 4 ( 9 )( −3)

2 (9)

8 ± 172 18 8 ± 2 43 = 18 4 ± 43 = 9 =

The roots are

4 + 43 4 − 43 and . 9 9

MHR • Principles of Mathematics 10 Solutions

117

c) For 5x2 + 7x + 1 = 0, a = 5, b = 7, and c = 1. x= = =

−b ± b2 − 4ac 2a −7 ±

( 7 ) − 4 ( 5)(1) 2 ( 5) 2

−7 ± 29 10

The roots are

−7 + 29 −7 − 29 and . 10 10

d) For 25x2 + 90x + 81 = 0, a = 25, b = 90, and c = 81.

x= =

−b ± b 2 − 4ac 2a −90 ±

( 90 ) − 4 ( 25)(81) 2 ( 25 ) 2

−90 ± 0 50 9 =− 5 =

9 The root is − . 5 Chapter 6 Review

Question 11

Page 317

d = 0.0034 s 2 + 0.004 s − 0.3 125 = 0.0034 s 2 + 0.004 s − 0.3 0 = 0.0034 s 2 + 0.004 s − 125.3 Use the quadratic formula with a = 0.0034, b = 0.004, and c = –125.3. s= =

−b ± b2 − 4ac 2a −0.004 ±

( 0.004 ) − 4 ( 0.0034 )( −125.3) 2 ( 0.0034 ) 2

−0.004 ± 1.704 0.0068 So, s 191.4 or s –192.6. The speed required for the ball to fly 125 m is 191.4 km/h. =

118 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Review

Question 12

Page 317

a) h = −4.9t 2 + 4t + 3 b) h = −4.9t 2 + 4t + 3

0 = −4.9t 2 + 4t + 3 Use the quadratic formula with a = –4.9, b = 4, and c = 3. t= =

−b ± b 2 − 4ac 2a −4 ±

( 4 ) − 4 ( −4.9 )( 3) 2 ( −4.9 ) 2

−4 ± 74.8 −9.8 So, t –0.47 or t 1.29. The diver enters the water after 1.29 s. =

h = −4.9t 2 + 4t + 3

c)

3.5 = −4.9t 2 + 4t + 3 0 = −4.9t 2 + 4t − 0.5 Use the quadratic formula with a = –4.9, b = 4, and c = –0.5. t= =

−b ± b 2 − 4ac 2a −4 ±

(4)

2

− 4 ( −4.9 )( −0.5)

2 ( −4.9 )

−4 ± 6.2 −9.8 So, t 0.15 or t 0.66. =

The height of the diver is greater than 3.5 m above the water over the interval 0.15 ≤ t ≤ 0.66 .

MHR • Principles of Mathematics 10 Solutions

119

Chapter 6 Review

Question 13

Page 317

( 5 + 2 x )( 7 + 2 x ) = 2 ( 5 )( 7 ) 35 + 24 x + 4 x 2 = 70 4 x 2 + 24 x − 35 = 0 Use the quadratic formula with a = 4, b = 24, and c = –35.

x= =

−b ± b2 − 4ac 2a −24 ±

( 24 )

2

− 4 ( 4 )( −35)

2 (4)

−24 ± 1136 8 = 1.2 So, x 1.2 or x –7.2. Each dimension of the garden should be extended by 2.4 m. =

Chapter 6 Review

Question 14

Page 317

a) R = ( 6 + x )( 4 − 0.25 x ) , where R is the revenue in dollars, and x is the number of price reductions. b) R = ( 6 + x )( 4 − 0.25 x ) 30 = 24 + 2.5 x − 0.25 x 2 0 = −6 + 2.5 x − 0.25 x 2 Use the quadratic formula with a = –0.25, b = 2.5, and c = –6.

x= =

−b ± b2 − 4ac 2a −2.5 ±

( 2.5) − 4 ( −0.25)( −6 ) 2 ( −0.25) 2

−2.5 ± 0.25 −0.5 −2.5 ± 0.5 = −0.5 So, x = 4 or x = 6. Either 4 or 6 price reductions will result in revenue of \$30 per customer. =

120 MHR • Principles of Mathematics 10 Solutions

c) For the x-coordinate of the vertex, use x = −

x=−

b . 2a

2.5 2( −0.25)

=5 R = ( 6 + x )( 4 − 0.25 x ) = ( 6 + 5 )( 4 − 0.25 × 5 ) = 30.25 The maximum predicted revenue per customer is \$30.25 at 5 price reductions.

MHR • Principles of Mathematics 10 Solutions

121

Chapter 6 Chapter Test Chapter 6 Chapter Test

Question 1

Page 318

a) y = x 2 + 6 x + 4

= ( x2 + 6x ) + 4

= ( x 2 + 6 x + 32 − 32 ) + 4 = ( x 2 + 6 x + 32 ) − 32 + 4 = ( x + 3) − 5 2

The vertex is (–3, –5). b) y = − x 2 + 8 x − 3

(

)

= − x2 − 8x − 3

( = −(x

)

= − x 2 − 8 x + ( −4 ) − ( −4 ) − 3 2

2

2

)

− 8 x + 16 ( −4 ) − ( −1)( −4 ) − 3 2

2

= − ( x − 4 ) + 13 2

The vertex is (4, 13).

c) y = 3x 2 + 24 x + 10

= 3 ( x 2 + 8 x ) + 10

= 3 ( x 2 + 8 x + 42 − 42 ) + 10

= 3 ( x 2 + 8 x + 4 2 ) − 3 ( 4 2 ) + 10 = 3 ( x + 4 ) − 38 2

The vertex is (–4, –38).

122 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapter Test

Question 2

Page 318

x2 − 5x + 4 = 0

a)

( x − 1)( x − 4 ) = 0 x − 1 = 0 or x − 4 = 0 x = 1 or x=4 9 y2 −1 = 0

b)

( 3 y + 1)( 3 y − 1) = 0 3y +1 = 0 y=−

or 3 y − 1 = 0 1 or 3

y=

1 3

x 2 = 3 x + 10

c)

x 2 − 3x − 10 = 0

( x − 5)( x + 2 ) = 0 x − 5 = 0 or x + 2 = 0 x = 5 or x = −2 d) 9b 2 − 12b + 4 = 0

( 3b − 2 )

2

=0

3b − 2 = 0 2 b= 3 e)

3x 2 + 13x = 10 3x 2 + 13x − 10 = 0 3x 2 + 15 x − 2 x − 10 = 0

( 3x

2

+ 15 x ) + ( −2 x − 10 ) = 0

3 x ( x + 5) − 2 ( x + 5) = 0

( x + 5)( 3x − 2 ) = 0 x+5= 0

or 3 x − 2 = 0 2 x = −5 or x= 3

MHR • Principles of Mathematics 10 Solutions

123

f) 6m 2 + 30m = 0

6m ( m + 5) = 0 m + 5 = 0 or 6m = 0 m = −5 or m = 0 g)

5 x 2 = 10 x 5 x 2 − 10 x = 0 5x ( x − 2 ) = 0 x − 2 = 0 or 5x = 0 x = 2 or x=0

h)

4d + 1 = −4d 2 4d 2 + 4d + 1 = 0

( 2d + 1)

2

=0

2d + 1 = 0 d =−

1 2

Chapter 6 Chapter Test

Question 3

Page 318

Answers may vary. For example: Use factoring to find the x-intercepts. Then, find the mean of the x-intercepts to find the x-coordinate of the vertex. Next, substitute this value into the relation to find the corresponding y-coordinate of the vertex. Examples will vary.

124 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapter Test Question 4 Page 318 Factor and solve the corresponding quadratic equation. a) y = x 2 + 2 x − 35

0 = ( x + 7 )( x − 5) The x-intercepts are –7 and 5. Find the x-coordinate of the vertex, and then, the ycoordinate. −7 + 5 x= 2 = −1 y = x 2 + 2 x − 35 = ( −1) + 2 ( −1) − 35 2

= −36 The axis of symmetry is x = –1. The vertex is (–1, –36). b) y = − x 2 − x + 20

0 = − ( x 2 + x − 20 ) 0 = − ( x + 5)( x − 4 ) The x-intercepts are –5 and 4. Find the x-coordinate of the vertex, and then, the ycoordinate. −5 + 4 x= 2 = −0.5 y = − x 2 − x + 20 = − ( −0.5) − ( −0.5) + 20 2

= 20.25 The axis of symmetry is x = –0.5. The vertex is (–0.5, 20.25).

MHR • Principles of Mathematics 10 Solutions

125

c) y = −3x 2 − 6 x

0 = −3x ( x + 2 ) The x-intercepts are 0 and –2. Find the x-coordinate of the vertex, and then, the ycoordinate. −2 + 0 x= 2 = −1 y = −3x 2 − 6 x = −3 ( −1) − 6 ( −1) 2

=3 The axis of symmetry is x = –1. The vertex is (–1, 3). d) y = x 2 − 10 x + 25

0 = ( x − 5) The x-intercept is 5. 2

Find the x-coordinate of the vertex, and then, the ycoordinate. 5+5 x= 2 =5 y = x 2 − 10 x + 25 = ( 5) − 10 ( 5) + 25 2

=0 The axis of symmetry is x = 5. The vertex is (5, 0).

126 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapter Test

Question 5

Page 318

a) For 4x2 – 11x – 3 = 0, a = 4, b = –11, and c = –3. −b ± b 2 − 4ac 2a

x= =

11 ±

( −11) − 4 ( 4 )( −3) 2 ( 4) 2

11 ± 169 8 11 ± 13 = 8 =

1 The roots are 3 and − . 4 x2 + 5x = 7

b)

x2 + 5x − 7 = 0 For x2 + 5x – 7 = 0, a = 1, b = 5, and c = –7. x= = =

−b ± b 2 − 4ac 2a −5 ±

( 5)

2

− 4 (1)( −7 )

2 (1)

−5 ± 53 2

The roots are

−5 + 53 −5 − 53 and . 2 2

c) 9 x 2 = 30 x − 25

9 x 2 − 30 x + 25 = 0 For 9x2 – 30x + 25 = 0, a = 9, b = –30, and c = 25. x= =

−b ± b 2 − 4ac 2a 30 ±

( −30 ) − 4 ( 9 )( 25 ) 2 (9) 2

30 ± 0 18 5 = 3 5 The root is . 3 =

MHR • Principles of Mathematics 10 Solutions

127

d) For 7k2 – 9k + 3 = 0, a = 7, b = –9, and c = 3. k= = =

−b ± b 2 − 4ac 2a 9±

( −9 )

2

− 4 ( 7 )( 3)

2(7)

9 ± −3 14

There are no real roots. 4s 2 − 9s = −3

e)

4s 2 − 9s + 3 = 0 For 4s2 – 9s + 3 = 0, a = 4, b = –9, and c = 3. s= = =

−b ± b 2 − 4ac 2a 9±

( −9 ) − 4 ( 4 )( 3) 2 (4) 2

9 ± 33 8

The roots are

9 + 33 9 − 33 3 and . 8 8

3t 2 − 7 = t

f)

3t 2 − t − 7 = 0 For 3t2 – t – 7 = 0, a = 3, b = –1, and c = –7. t= = =

−b ± b 2 − 4ac 2a 1±

( −1)

2

− 4 ( 3)( −7 )

2 ( 3)

1 ± 85 6

The roots are

1 + 85 1 − 85 and . 6 6

128 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapter Test

Question 6

Page 318

a) Use the quadratic formula with a = 3, b = 12, and c = 6.

x= =

−b ± b2 − 4ac 2a

(12 ) − 4 ( 3)( 6 ) 2 ( 3) 2

−12 ±

−12 ± 72 6 −12 ± 3 8 = 6 −4 ± 8 = 2 −4 + 8 −4 − 8 The roots are and . 2 2 =

b) Use the quadratic formula with a = 1, b = –8, and c = 3. x= = =

−b ± b 2 − 4ac 2a 8±

( −8 )

2

− 4 (1)( 3)

2 (1)

8 ± 52 2 8 + 52 8 − 52 and . 2 2

The roots are

c) Use the quadratic formula with a = 4, b = 0, and c = –10. m= = =

−b ± b 2 − 4ac 2a 0±

( 0)

2

− 4 ( 4 )( −10 )

2 ( 4)

± 160 8

The roots are

+ 160 − 160 and . 8 8

MHR • Principles of Mathematics 10 Solutions

129

−5 x 2 + 10 x = 5

d)

−5 x 2 + 10 x − 5 = 0

(

)

−5 x 2 − 2 x + 1 = 0 −5 ( x − 1) = 0 2

x −1 = 0 x =1 The root is 1.

( k − 5)

e)

2

= 16

k 2 − 10k + 25 = 16 k 2 − 10k + 9 = 0

( k − 1)( k − 9 ) = 0 The roots are 1 and 9. f)

x2 1 +x+ =0 2 2 2 x + 2x + 1 = 0

( x + 1)

2

=0

The root is –1. 2 ( m − 1) = ( m + 2 )( m + 1) 2

g)

2m 2 − 4m + 2 = m 2 + 3m + 2 m2 − 7m = 0 m (m − 7) = 0 The roots are 0 and 7. h) ( 5 x + 2 )( 3 x − 1) = 4 x 2 + 5 15 x 2 + x − 2 = 4 x 2 + 5 11x 2 + x − 7 = 0 Use the quadratic formula with a = 11, b = 1, and c = –7. x= =

−b ± b 2 − 4ac 2a −1 ±

(1)

2

− 4 (11)( −7 )

2 (11)

−1 ± 309 22 −1 + 309 −1 − 309 The roots are and . 22 22 =

130 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapter Test

Question 7

Page 318

Answers may vary. For example: The axis of symmetry is the same for both because they have the same value for a and b. Chapter 6 Chapter Test

Question 8

Page 318

Find d-coordinate of the vertex. Then, find the h-coordinate. b d =− 2a 20 =− 2 ( −5) =2 h = −5d 2 + 20d + 1 = −5 ( 2 ) + 20 ( 2 ) + 1 2

= 21

The maximum height of the firework is 21 m. Chapter 6 Chapter Test

Question 9

Page 318

a) There will be two x-intercepts. The parabola opens downward and the vertex is above the xaxis. b) Let y = 0, subtract 18 from both sides, and then divide both sides by –2. Next, take the square root of both sides, keeping both roots. Then, subtract 1 from both sides. Finally, simplify the results to find the x-intercepts, –4 and 2. Chapter 6 Chapter Test

Question 10

Page 318

a) ( x − 5 )( x + 3) = 0 x 2 − 2 x − 15 = 0

1 ⎞⎛ 3⎞ ⎛ b) ⎜ x − ⎟⎜ x − ⎟ = 0 2 5 ⎝ ⎠⎝ ⎠ ( 2 x − 1)( 5 x − 3) = 0 10 x 2 − 11x + 3 = 0

MHR • Principles of Mathematics 10 Solutions

131

Chapter 6 Chapter Test

Question 11

Page 318

a) h = −d 2 + 4d

0 = −d ( d − 4 ) The d-intercepts are 0 and 4. The shed is 4 m wide. The d-coordinate of the vertex is 2. h = − d 2 + 4d = − ( 2) + 4 ( 2) 2

=4

The height of the shed is 4 m. b)

c) The relation is valid for values 0 ≤ d ≤ 4 . h must be positive. Chapter 6 Chapter Test

Question 12

Page 318

Find t-coordinate of the vertex. Then, find the C-coordinate.

b 2a −96 =− 2 ( 3)

t=−

= 16 C = 3t 2 − 96t + 1014 = 3 (16 ) − 96 (16 ) + 1014 2

= 246

The minimum cost of operating the machine is \$246 for 16 h of operation.

132 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapter Test

Question 13

Page 319

1 ( 2 x + 1)( 6 x − 3) = 240 2 12 x 2 − 3 = 480 12 x 2 = 483 483 x2 = 12 x 6.34 For an area of 240 m2, the value of x is 6.34 m. Chapter 6 Chapter Test

Question 14

Page 319

d = 0.0052s 2 + 0.13s 20 = 0.0052s 2 + 0.13s 0 = 0.0052s 2 + 0.13s − 20 Use the quadratic formula with a = 0.0052, b = 0.13, and c = –20. s= =

−b ± b 2 − 4ac 2a −0.13 ±

( 0.13) − 4 ( 0.0052 )( −20 ) 2 ( 0.0052 ) 2

−0.13 ± 0.4329 0.0104 So, s 50.8 or s –75.8. The speed for a stopping distance of 20 m is 50.8 km/h. =

MHR • Principles of Mathematics 10 Solutions

133

Chapter 6 Chapter Test

Question 15

Page 319

a) Let d = 0 and use the quadratic formula with a = –1, b = 3, and c = 4.

d= =

−b ± b2 − 4ac 2a −3 ±

( 3)

2

− 4 ( −1)( 4 )

2 ( −1)

−3 ± 25 −2 −3 ± 5 = −2

=

The zeros are –1 and 4. b)

c) The relation is valid for 0 ≤ d ≤ 4 . h must be positive. d) Van hits the water at 4 m. e) The d-coordinate of the vertex is

−1 + 4 , or 1.5. 2

h = − d 2 + 3d + 4 = − (1.5 ) + 3 (1.5 ) + 4 2

= 6.25

Van's maximum height is 6.25 m.

134 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapter Test

Question 16

Page 319

a) Find the t-coordinate of the vertex. Then, find the h-coordinate.

t=− =−

b 2a 14 2 ( −4.9 )

1.4 h = −4.9t 2 + 14t + 2.5 = −4.9 (1.4 ) + 14 (1.4 ) + 2.5 2

12.5

The maximum height of 12.5 m occurs at 1.4 s.

h = −4.9t 2 + 14t + 2.5

b)

0.5 = −4.9t 2 + 14t + 2.5 0 = −4.9t 2 + 14t + 2 Use the quadratic formula with a = –4.9, b = 14, and c = 2. d= =

−b ± b2 − 4ac 2a −14 ±

(14 ) − 4 ( −4.9 )( 3) 2 ( −4.9 ) 2

−14 ± 235.2 −9.8 So, d 0 or d 3. It takes the ball 3 s to reach the player. =

MHR • Principles of Mathematics 10 Solutions

135

Chapter 6 Chapter Test Question 17 Let x and y represent the legs of the triangle. x + y = 23 y = 23 − x

Page 319

x 2 + y 2 = 172 x 2 + ( 23 − x ) = 289 2

x 2 + 529 − 46 x + x 2 = 289 2 x 2 − 46 x + 240 = 0 x 2 − 23x + 120 = 0

( x − 8 )( x − 15) = 0 So, x = 8 or x = 15. One leg measures 8 cm, and the other measures 15 cm.

Chapter 6 Chapter Test

Question 18

Page 319

Let the side length of the base of the box be represented by x. 8 x 2 = 512 x 2 = 64 x =8

The dimensions of the original piece of cardboard were 24 cm by 24 cm.

Chapter 6 Chapter Test

Question 19

Page 319

Solutions for the Achievement Checks are shown in the Teacher Resource.

136 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapters 4 to 6 Review Chapter 6 Chapters 4 to 6 Review

Question 1

Page 320

a) x

y

–2 –1 0 1 2

9 7 5 3 1

First Differences –2 –2 –2 –2

The first differences are constant. The relation is linear.

b) x

y

–2 –1 0 1 2

–3 3 5 3 -3

First Differences

Second Differences

6 2 -2 -6

–4 –4 –4

The second differences are constant. The relation is quadratic.

MHR • Principles of Mathematics 10 Solutions

137

Chapter 6 Chapters 4 to 6 Review

Question 2

Page 320

a) The graph of y = x 2 + 2 is the graph of y = x 2 translated 2 units upward.

b) The graph of y = ( x + 3) is the graph of y = x 2 translated 3 units to the left. 2

1 c) The graph of y = x 2 + 2 is the graph of y = − x 2 vertically 4 1 compressed by a factor of and reflected in the x-axis. 4

Chapter 6 Chapters 4 to 6 Review

Question 3

Page 320

a)

b) Her maximum height above the water was 11 m. c) Using a graphing calculator, the x-intercept is about 4.3. Diane had travelled 4.3 m horizontally when she entered the water.

138 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapters 4 to 6 Review

Question 4

Page 320

The x-intercepts are –3 and 5, so the equation is of the form y = a ( x + 3)( x − 5) . Substitute (x, y) = (1, –4) and solve for a. −4 = a (1 + 3)(1 − 5 ) −4 = −16a 1 =a 4

An equation for the parabola is y =

1 ( x + 3)( x − 5) . 4

Chapter 6 Chapters 4 to 6 Review

Question 5

1 24 1 = 16

Page 320

b) ( −3) =

1

−2

a) 2 −4 =

=

( −3 ) 1 9

1 8

c) 250 = 1

d) 8−1 =

e) ( −1) = 1

1 ⎛ 3⎞ f) ⎜ ⎟ = 3 4 ⎝ ⎠ ⎛ 3⎞ ⎜ ⎟ ⎝4⎠ 1 = 27 64 64 = 27

−3

12

Chapter 6 Chapters 4 to 6 Review

2

Question 6

Page 320

a) 20.8 years = 4 × 5.2 years 4 ⎛1⎞ 20 ⎜ ⎟ = 1.25 ⎝2⎠ After 20.8 years, 1.25 g of Cobalt-60 remains.

b) 36.4 years = 7 × 5.2 years 7 ⎛1⎞ 20 ⎜ ⎟ = 0.156 25 ⎝2⎠ After 36.4 years, 0.156 25 g of Cobalt-60 remains.

MHR • Principles of Mathematics 10 Solutions

139

Chapter 6 Chapters 4 to 6 Review

Question 7

Page 320

Question 8

Page 320

A = 3 x ( 2 x + 1) − x 2 = 6 x 2 + 3x − x 2 = 5 x 2 + 3x

Chapter 6 Chapters 4 to 6 Review

b) ( h + 5 ) = h 2 + 10h + 25

a) ( n + 3)( n − 3) = n 2 − 3n + 3n − 9

2

= n2 − 9

c) ( d − 4 )( d − 2 ) = d 2 − 2d − 4d + 8

d) ( m + 3)( m + 7 ) = m 2 + 7 m + 3m + 21

= d 2 − 6d + 8

= m 2 + 10m + 21

f) ( x − 7 ) = x 2 − 14 x + 49

e) ( 3t − 5 )( 3t + 5 ) = 9t 2 − 25

2

Chapter 6 Chapters 4 to 6 Review

( = x (6x

Question 9

a) x ( 3x + 1)( 2 x − 5 ) = x 6 x 2 − 15 x + 2 x − 5 2

− 13 x − 5

)

Page 320

)

= 6 x − 13 x − 5 x 3

2

b) ( 2k + 3) − ( k + 2 )( k − 2 ) = 4k 2 + 12k + 9 − k 2 + 4 2

= 3k 2 + 12k + 13

(

)

c) 5 ( y − 4 )( 3 y + 1) + ( 3 y − 4 ) = 5 3 y 2 − 11y − 4 + 9 y 2 − 24 y + 16 2

= 15 y 2 − 55 y − 20 + 9 y 2 − 24 y + 16 = 24 y 2 − 79 y − 4

( = 3 ( 6a

d) 3 ( 2a + 3b )( 3a − 2b ) = 3 6a 2 − 4ab + 9ab − 6b 2 2

+ 5ab − 6b 2

= 18a + 15ab − 18b 2

)

)

2

140 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapters 4 to 6 Review

Question 10

Page 320

Chapter 6 Chapters 4 to 6 Review

Question 11

Page 321

a) y 2 + 12 y + 27 = ( y + 3)( y + 9 )

b) x 2 + 2 x − 3 = ( x + 3)( x − 1)

c) n 2 + 22n + 21 = ( n + 21)( n + 1)

d) p 2 − 8 p + 15 = ( p − 5 )( p − 3)

e) x 2 + 2 x − 15 = ( x + 5 )( x − 3)

f) k 2 + 5k + 24 cannot be factored

Chapter 6 Chapters 4 to 6 Review a) p 2 + 12 p + 36 = ( p + 6 )

Question 12

Page 321

b) 9d 2 + 6d + 1 = ( 3d − 1)

2

2

c) x 2 − 49 = ( x + 7 )( x − 7 )

d) 4a 2 − 20a + 25 = ( 2a − 5 )

e) 8t 2 − 18 = 2 ( 4t 2 − 9 )

f) a 2 − 4b 2 = ( a + 2b )( a − 2b )

2

= 2 ( 2t + 3)( 2t − 3)

Chapter 6 Chapters 4 to 6 Review

Question 13

Page 321

a) k = 11; m 2 + 11m + 10 = ( m + 1)( m + 10 ) k = −11; m 2 − 11m + 10 = ( m − 1)( m − 10 ) k = 7; m 2 + 7m + 10 = ( m + 2 )( m + 5) k = −7; m 2 − 7m + 10 = ( m − 2 )( m − 5)

b) k = 12; 9a 2 − 12a + 4 = ( 3a − 2 )

2

k = −12; 9a 2 + 12a + 4 = ( 3a + 2 )

2

MHR • Principles of Mathematics 10 Solutions

141

Chapter 6 Chapters 4 to 6 Review

(

Question 14

Page 321

)

A = π 4 x 2 + 36 x + 81 = π ( 2x + 9)

2

The radius of the circle is 2 x + 9 . The diameter is 4x + 18.

142 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapters 4 to 6 Review

Question 15

Page 321

a) y = x 2 + 6 x − 16

( = (x = (x

)

= x 2 + 6 x − 16

)

2

+ 6 x + 32 − 32 − 16

2

+ 6 x + 32 − 32 − 16

)

= ( x + 3) − 25 2

b) y = x 2 − 8 x + 7

(

)

= x2 − 8x + 7

( = (x

)

= x 2 − 8 x + ( −4 ) − ( −4 ) + 7 2

2

2

)

− 8 x + ( −4 ) − ( −4 ) + 7 2

2

= ( x − 4) − 9 2

c) y = x 2 + 4 x + 10

( = (x = (x

)

= x 2 + 4 x + 10

)

2

+ 4 x + 22 − 2 2 + 10

2

+ 4 x + 22 − 2 2 + 10

)

= ( x + 2) + 6 2

MHR • Principles of Mathematics 10 Solutions

143

d) y = − x 2 + 6 x − 8

(

)

= − x2 − 6x − 8

( = −(x

)

= − x 2 − 6 x + ( −3) − ( −3) − 8 2

2

2

)

− 6 x + ( −3) − ( −1)( −3) − 8 2

2

= − ( x − 3) + 1 2

e) y = 2 x 2 + 8 x + 5

( = 2( x = 2( x

)

= 2 x2 + 4 x + 5

) ) − 2 ( −2 ) + 5

2

+ 4 x + 22 − 22 + 5

2

+ 4 x + 22

2

= 2 ( x + 2) − 3 2

f) y = −2 x 2 − 12 x − 7

( = −2 ( x = −2 ( x

)

= −2 x 2 + 6 x − 7

) − 3 ) − ( −2 ) ( 3 ) − 7

2

+ 6 x + 32 − 32 − 7

2

+ 6x + 3

2

2

2

= −2 ( x + 3) + 11 2

144 MHR • Principles of Mathematics 10 Solutions

Chapter 6 Chapters 4 to 6 Review a)

Question 16

Page 321

x 2 − 14 x + 24 = 0

( x − 2 )( x − 12 ) = 0 x−2=0

or x − 12 = 0

x=2

or

x = 12

Check by substituting both solutions in the original equation. For x = 2: For x = 12: 2 R.S. = 0 L.S. = x − 14 x + 24 L.S. = x 2 − 14 x + 24

R.S. = 0

= ( 2 ) − 14 ( 2 ) + 24

= (12 ) − 14 (12 ) + 24

=0

=0

2

2

L.S. = R.S.

L.S. = R.S.

The roots are 2 and 12.

b)

n 2 + 4n − 21 = 0

( n + 7 )( n − 3) = 0 n+7=0 n = −7

or n − 3 = 0 or

n=3

Check by substituting both solutions in the original equation. For n = –7: For n = 3: R.S. = 0 L.S. = n 2 + 4n − 21 L.S. = n 2 + 4n − 21 = ( −7 ) + 4 ( −7 ) − 21

= ( 3 ) + 4 ( 3 ) − 21

=0

=0

2

L.S. = R.S.

R.S. = 0

2

L.S. = R.S.

The roots are –7 and 3.

MHR • Principles of Mathematics 10 Solutions

145

m 2 − 16 = 0

c)

( m + 4 )( m − 4 ) = 0 m+4=0

or m − 4 = 0

m = −4 or

m=4

Check by substituting both solutions in the original equation. For m = –4: For m = 4: R.S. = 0 L.S. = m 2 − 16 L.S. = m 2 − 16 = ( 4 ) − 16

R.S. = 0

= ( −4 ) − 16

2

2

=0

=0

L.S. = R.S.

L.S. = R.S.

The roots are 4 and –4. 2 y2 + 5 y + 2 = 0

d)

2 y2 + 4 y + y + 2 = 0

(2y

)

+ 4 y + ( y + 2) = 0

2

2 y ( y + 2) + ( y + 2) = 0

( y + 2 )( 2 y + 1) = 0 y+2=0 y = −2

or 2 y + 1 = 0 or

y=−

1 2

Check by substituting both solutions in the original equation. 1 For y = –2: For y = − : 2 2 2 R.S. = 0 L.S. = 2 y + 5 y + 2 L.S. = 2 y + 5 y + 2 = 2 ( −2 ) + 5 ( −2 ) + 2 2

=0

L.S. = R.S.

2

⎛ 1⎞ ⎛ 1⎞ = 2⎜ − ⎟ + 5⎜ − ⎟ + 2 ⎝ 2⎠ ⎝ 2⎠ =0 L.S. = R.S.

1 The roots are –2 and − . 2

146 MHR • Principles of Mathematics 10 Solutions

R.S. = 0

7t 2 = 70t − 175

e)

7t 2 − 70t + 175 = 0

(

)

7 t 2 − 10t + 25 = 0 7 ( t − 5) = 0 2

t −5= 0 t =5 Check by substituting the solution in the original equation.

R.S. = 70t − 175

L.S. = 7t 2

= 7 ( 5)

= 70 ( 5 ) − 175

2

= 175

= 175 L.S. = R.S.

The root is 5.

Chapter 6 Chapters 4 to 6 Review Question 17 Page 321 Factor and solve the corresponding quadratic equation. a) y = x 2 − 3 x + 2 0 = ( x − 2 )( x − 1) The x-intercepts are 2 and 1. Find the x-coordinate of the vertex, and then, the ycoordinate. 1+ 2 x= 2 = 1.5 y = x 2 − 3x + 2 = (1.5 ) − 3 (1.5 ) + 2 2

= −0.25 The vertex is (1.5, −0.25 ) .

MHR • Principles of Mathematics 10 Solutions

147

b) y = x 2 − 16 0 = ( x + 4 )( x − 4 ) The x-intercepts are –4 and 4. Find the x-coordinate of the vertex, and then, the ycoordinate. −4 + 4 x= 2 =0 y = x 2 − 16 = ( 0 ) − 16 2

= −16 The vertex is ( 0, −16 ) .

c) y = 2 x 2 − 5 x − 12 0 = 2 x 2 − 8 x + 3 x − 12

(

)

0 = 2 x 2 − 8 x + ( 3 x − 12 ) 0 = 2 x ( x − 4 ) + 3( x − 4 ) 0 = ( x − 4 )( 2 x + 3)

The x-intercepts are 4 and −1.5 . Find the x-coordinate of the vertex, and then, the y-coordinate. −1.5 + 4 x= 2 = 1.25 y = 2 x 2 − 5 x − 12 = 2 (1.25 ) − 5 (1.25 ) − 12 2

= −15.125 The vertex is (1.25, −15.125 ) .

148 MHR • Principles of Mathematics 10 Solutions

d) y = − x 2 − 7 x − 12

(

0 = − x 2 + 7 x + 12

)

0 = − ( x + 3)( x + 4 ) The x-intercepts are –3 and –4. Find the x-coordinate of the vertex, and then, the ycoordinate. −3 − 4 x= 2 = −3.5 y = − x 2 − 7 x − 12 = − ( −3.5 ) − 7 ( −3.5 ) − 12 2

= 0.25 The vertex is ( −3.5,0.25) .

e) y = −3x 2 + 6 x 0 = −3x ( x − 2 ) The x-intercepts are 0 and 2. Find the x-coordinate of the vertex, and then, the y-coordinate. 0+2 x= 2 =1 y = −3x 2 + 6 x = −3 (1) + 6 (1) 2

=3 The vertex is (1,3) .

MHR • Principles of Mathematics 10 Solutions

149

Chapter 6 Chapters 4 to 6 Review

Question 18

Page 321

The equation is of the form y = ax ( x + 6 ) . Substitute (x, y) = –3, 4) and solve for a. 4 = a ( −3)( −3 + 6 ) 4 = −9a −

4 =a 9 4 y = − x ( x + 6) 9 4 8 = − x2 − x 9 3

Chapter 6 Chapters 4 to 6 Review

Question 19

Page 321

a) Use the quadratic formula with a = 1, b = –6, and c = 1. x= = =

−b ± b 2 − 4ac 2a 6±

( −6 ) − 4 (1)(1) 2 (1) 2

6 ± 32 2

The roots are

6 + 32 6 − 32 and . 2 2

b) Use the quadratic formula with a = 3, b = –5, and c = 1. x= = =

−b ± b 2 − 4ac 2a 5±

( −5 ) − 4 ( 3)(1) 2 ( 3) 2

5 ± 13 6

The roots are

5 + 13 5 − 13 and . 6 6

150 MHR • Principles of Mathematics 10 Solutions

c) Use the quadratic formula with a = 3.2, b = –5.6, and c = –7.1. x= =

−b ± b 2 − 4ac 2a 5.6 ±

( −5.6 ) − 4 ( 3.2 )( −7.1) 2 ( 3.2 ) 2

5.6 ± 122.24 6.4 5.6 + 122.24 5.6 − 122.24 The roots are and . 6.4 6.4 =

d) Use the quadratic formula with a = 2, b = 5, and c = –9. x= = =

−b ± b 2 − 4ac 2a −5 ±

( 5)

2

− 4 ( 2 )( −9 )

2 (2)

−5 ± 97 4

The roots are

−5 + 97 −5 − 97 and . 4 4

e) Use the quadratic formula with a = 3, b = –8, and c = 3. x= = =

−b ± b 2 − 4ac 2a 8±

( −8)

2

− 4 ( 3)( 3)

2 ( 3)

8 ± 28 6

The roots are

8 + 28 8 − 28 and . 6 6

MHR • Principles of Mathematics 10 Solutions

151

Chapter 6 Chapters 4 to 6 Review

Question 20

Page 321

Perimeter: 2 x + 2 y = 8 Area: xy = 2 Rearrange the perimeter equation to obtain an expression for y. Then, substitute the expression into the area equation. 2x + 2 y = 8 x+ y =4 y = 4− x xy = 2 x (4 − x ) = 2 − x2 + 4 x − 2 = 0 x2 − 4 x + 2 = 0 Use the quadratic formula with a = 1, b = –4, and c = 2. x= =

−b ± b 2 − 4ac 2a 4±

( −4 )

2

− 4 (1)( 2 )

2 (1)

4± 8 2 So, x 3.4 or x 0.6. =

The length is 3.4 m and the width is 0.6 m.

Chapter 6 Chapters 4 to 6 Review

Question 21

Page 321

R = ( 20 + x )( 500 − 20 x ) = 10 000 + 100 x − 20 x 2 Calculate the x-coordinate of the vertex. b 2a 100 =− 2 ( −20 )

x=−

= 2.5 The fare that will maximize the revenue is \$22.50.

152 MHR • Principles of Mathematics 10 Solutions