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Chapter 6 Quadratic Equations Chapter 6 Get Ready Chapter 6 Get Ready Question 1 Page 262 a) b) c) d)
MHR • Principles of Mathematics 10 Solutions 1
Chapter 6 Get Ready Question 2 Page 262 a) b) c) d)
6
2 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Get Ready 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. Chapter 6 Get Ready Question 4 Page 262 a) 25 16 9
3± − = ±
= ± b) 2 23 4 9 16
255
± + = ± +
= ±= ±
c) 26 11 36 11
255
± − = ± −
= ±= ±
d) ( )28 3 5 64 15
497
± − = ± −
= ±= ±
Chapter 6 Get Ready Question 5 Page 262 a) ( )
( )(
2 22 10 12 2 5 6
2 3 2
x x x x
x x
+ + = + +
= + + ) b) ( )( )2 16 28 14 2x x x x+ + = + +
c) ( )( )2 16 63 9 7x x x x− + = − − d) ( )
( )(
2 24 36 4 9
4 3 3
x x
x x
− = −
)= − +
e) ( )( )281 49 9 7 9 7x x x− = − + f) ( )
( )(
2 23 6 24 3 2 8
3 4 2
x x x x
x x
− − = − −
)= − +
g) h) ( 2216 8 1 4 1x x x− + = − )
( ) ( )( ) ( )
( )( )
2 2
2
12 19 4 12 16 3 4
12 16 3 4
4 3 4 3 4
3 4 4 1
x x x x x
x x x
x x x
x x
+ + = + + +
= + + +
= + + +
= + +
MHR • Principles of Mathematics 10 Solutions 3
Chapter 6 Get Ready Question 6 Page 262 a) b) ( )(2 4 32 8 4x x x x− − = − + ) ( )
( )
2 2
2
2 12 18 2 6 9
2 3
x x x x
x
+ + = + +
= +
c) ( )( )264 1 8 1 8 1x x x− = − + d) ( )22144 312 169 12 13x x x− + = − e) f) ( 2249 70 25 7 5x x x+ + = + )
( ) ( )( ) ( )
( )( )
2 2
2
2 9 5 2 10 5
2 10 5
2 5 5
5 2 1
x x x x x
x x x
x x x
x x
− − = − + −
= − + −
= − + −
= − +
g) h)
( ) ( )( ) ( )
( )( )
2 2
2
9 18 8 9 12 6 8
9 12 6 8
3 3 4 2 3 4
3 4 3 2
x x x x x
x x x
x x x
x x
− + = − − +
= − + − +
= − − −
= − −
( )
( ) ( )( ) ( )
( )( )
2 2
2
2
6 21 9 3 2 7 3
3 2 6 3
3 2 6 3
3 2 3 3
3 3 2 1
x x x x
x x x
x x x
x x x
x x
+ + = + +
⎡ ⎤= + + +⎣ ⎦⎡ ⎤= + + +⎣ ⎦⎡ ⎤= + + +⎣ ⎦
= + +
Chapter 6 Get Ready Question 7 Page 262
a) The first number is x. The second number is 12
x.
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. Chapter 6 Get Ready Question 8 Page 262 a) x + y = 100 b) 2(l + w) = 50 c) P = 8w d) ( )2
2
2
4
xy x
x
=
=
e) x + y + 3 = 12
4 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 1: Maxima and Minima Chapter 6 Section 1 Question 1 Page 270 a)
( )
2
2
2 5
1 4
y x x
x
= + +
= + +
MHR • Principles of Mathematics 10 Solutions 5
b)
( )
2
2
4 7
2 3
y x x
x
= + +
= + +
6 MHR • Principles of Mathematics 10 Solutions 6 MHR • Principles of Mathematics 10 Solutions
c)
( )
2
2
6 3
3 6
y x x
x
= + +
= + −
MHR • Principles of Mathematics 10 Solutions 7
MHR • Principles of Mathematics 10 Solutions 7
Chapter 6 Section 1 Question 2 Page 270 a) c = 9; b) c = 49; ( 22 6 9 3x x x+ + = + )
)
1
1
7
7
( )22 14 49 7x x x+ + = + c) c = 36; d) c = 25; ( 22 12 36 6x x x− + = − ( )22 10 25 5x x x− + = − e) c = 1; f) c = 1600; ( )22 2 1 1x x x+ + = + ( )22 80 1600 40x x x− + = − Chapter 6 Section 1 Question 3 Page 270 a) b)
( )( )( )( )
2
2
2 2 2
2 2 2
2
6 1
6 1
6 3 3
6 3 3
3 10
y x x
x x
x x
x x
x
= + −
= + −
= + + − −
= + + − −
= + −
( )( )( )( )
2
2
2 2 2
2 2 2
2
2 7
2 7
2 1 1
2 1 1
1 6
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + +
c) d)
( )( )( )( )
2
2
2 2 2
2 2 2
2
10 20
10 20
10 5 5 20
10 5 5 20
5 5
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + −
( )( )( )( )
2
2
2 2 2
2 2 2
2
2 1
2 1
2 1 1
2 1 1
1 2
y x x
x x
x x
x x
x
= + −
= + −
1
1
= + + − −
= + + − −
= + −
e) f)
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
6 4
6 4
6 3 3
6 3 3
3 13
y x x
x x
x x
x x
x
= − −
= − −
= − + − − − −
= − + − − − −
= − −
4
4
2
2
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
8 2
8 2
8 4 4
8 4 4
4 18
y x x
x x
x x
x x
x
= − −
= − −
= − + − − − −
= − + − − − −
= − −
g)
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
12 8
12 8
12 6 6 8
12 6 6 8
6 28
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − − +
= − −
8 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 1 Question 4 Page 270 a) b)
( )( )( )( )
2
2
2 2 2
2 2 2
2
6 2
6 2
6 3 3 2
6 3 3 2
3 7
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + −
( )( )( )( )
2
2
2 2 2
2 2 2
2
12 30
12 30
12 6 6 30
12 6 6 30
6 6
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + −
The vertex is (–3, –7). The vertex is (–6, –6). c) d)
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
8 13
8 13
8 4 4
8 4 4
4 3
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − − +
= − −
13
13
7
7
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
6 17
6 17
6 3 3 1
6 3 3 1
3 8
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − − +
= − +
The vertex is (4, –3). The vertex is (3, 8). Chapter 6 Section 1 Question 5 Page 270 a) ( )25 1y x= − + The vertex is (5, 1). Graph D b) ( )21 4y x= − + The vertex is (1, 4). Graph A c) ( )21 4y x= + + The vertex is (–1, 4). Graph B d) ( )25 1y x= + − The vertex is (–5, –1). Graph C
MHR • Principles of Mathematics 10 Solutions 9
Chapter 6 Section 1 Question 6 Page 270 a)
( )( )( )( )
2
2
2 2 2
2 2 2
2
10 20
10 20
10 5 5 20
10 5 5 20
5 5
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + − The vertex is (–5, –5). b)
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
16 60
16 60
16 8 8 60
16 8 8 60
8 4
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − − +
= − − The vertex is (8, –4).
10 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 1 Question 7 Page 270 a) b)
( )( ) ( )( )
( ) ( )( )
( )
2
2
2 22
22
2
80 100
80 100
80 40 40 100
80 1600 1 40 100
40 1500
y x x
x x
x x
x x
x
= − + −
= − − −
= − − + − − − −
= − − + − − − −
= − − +
( )( )( ) ( )( )( )
2
2
2 2 2
2 2
2
6 4
6 4
6 3 3 4
6 9 1 3
3 13
y x x
x x
x x
x x
x
= − − +
= − + +
= − + + − +
4= − + + − − +
= − + +
c) d)
( )( )( ) (( )
2
2
2 2 2
2 2
2
3 90 50
3 30 50
3 30 15 15 50
3 30 225 3 15 5
3 15 625
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + −
) 0
5
( )( ) ( )( )
( ) ( )
( )
2
2
2 22
22
2
2 16 15
2 8 15
2 8 4 4 1
2 8 16 2 4 15
2 4 17
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − +
= − −
e)
( )( ) ( )( )
( ) ( )( )
( )
2
2
2 22
22
2
7 14 3
7 2 3
7 2 1 1
7 2 1 7 1
7 1 4
y x x
x x
x x
x x
x
= − + −
= − − −
= − − + − − − −
= − − + − − − −
= − − +
3
3
MHR • Principles of Mathematics 10 Solutions 11
Chapter 6 Section 1 Question 8 Page 271 a) b)
( )( )( ) ( )( )( )
2
2
2 2 2
2 2
2
10 9
10 9
10 5 5 9
10 25 1 5 9
5 16
y x x
x x
x x
x x
x
= − − −
= − + −
= − + + − −
= − + + − − −
= − + +
( )( ) ( )( )
( ) ( )( )
( )
2
2
2 22
22
2
14 50
14 50
14 7 7 50
14 49 1 7 50
7 1
y x x
x x
x x
x x
x
= − + −
= − − −
= − − + − − − −
= − − + − − − −
= − − −
The maximum point is at (–5, 16). The maximum point is at (7, –1). c) d)
( )( )( ) (( )
2
2
2 2 2
2 2
2
2 120 75
2 60 75
2 60 30 30 75
2 60 900 2 30 7
2 30 1725
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + −
) 5
0
( )( ) ( )( )
( ) ( )
( )
2
2
2 22
22
2
3 24 10
3 8 10
3 8 4 4 1
3 8 16 3 4 10
3 4 38
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − +
= − −
The minimum point is at (–30, –1725). The minimum point is at (4, –38). e)
( )( )( ) ( )( )( )
2
2
2 2 2
2 2
2
5 200 120
5 40 120
5 40 20 20 120
5 40 400 5 20 12
5 20 1880
y x x
x x
x x
x x
x
= − − −
= − + −
= − + + − −
= − + + − − −
= − + +
0
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 13
Chapter 6 Section 1 Question 10 Page 271 a)
( )( )( ) ( )( )( )
2
2
2 2 2
2 2
2
2 6
2 6
2 1 1 6
2 25 1 1
1 5
y x x
x x
x x
x x
x
= − − −
= − + −
= − + + − −
= − + + − − −
= − + −
6
41
41
3
3
The vertex is (–1, –5). Other points will vary. Possible other points are (–2, –6) and (0, –6). b)
( )( )( ) ( )( )
2
2
2 2 2
2 2
2
4 24 41
4 6 41
4 6 3 3 41
4 6 9 4 3
4 3 5
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + + The vertex is (–3, 5). Other points will vary. Possible other points are (–2, 9) and (–4, 9). c)
( )( ) ( )( )
( ) ( )
( )
2
2
2 22
22
2
5 30 41
5 6 41
5 6 3 3
5 6 9 5 3 41
5 3 4
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − +
= − −
The vertex is (3, –4). Other points will vary. Possible other points are (2, 1) and (4, 1). d)
( )( ) ( )( )
( ) ( )( )
( )
2
2
2 22
22
2
3 12 13
3 4 13
3 4 2 2 1
3 4 4 3 2 1
3 2 1
y x x
x x
x x
x x
x
= − + −
= − − −
= − − + − − − −
= − − + − − − −
= − − −
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
MHR • Principles of Mathematics 10 Solutions 15
) 3
2
)
0
e)
( )( )( ) (( )
2
2
2 2 2
2 2
2
2 8 3
2 4 3
2 4 2 2 3
2 4 4 2 2
2 2 5
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + − 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)
( )( )( ) ( )( )( )
2
2
2 2 2
2 2
2
2 3 7
2 1.5 7
2 1.5 0.75 0.75 7
2 1.5 0.75 2 0.75 7
2 0.75 8.125
y x x
x x
x x
x x
x
= − − +
= − + +
= − + + − +
= − + + − − +
= − + + The vertex is (–0.75, 8.125). Other points will vary. Possible other points are (0, 7) and (–1, 8). b)
( )( )( ) (( )
2
2
2 2 2
2 2 2
2
3 9 11
3 3 11
3 3 1.5 1.5 11
3 3 1.5 3 1.5 11
3 1.5 4.25
y x x
x x
x x
x x
x
= − +
= − +
= − + − +
= − + − +
= − + The vertex is (1.5, 4.25). Other points will vary. Possible other points are (1, 5) and (2, 5). c)
( )( ) ( )( )
( ) ( )( )
( )
2
2
2 22
22
2
8 10
8 10
8 4 4 10
8 16 1 4 1
4 6
y x x
x x
x x
x x
x
= − + −
= − − −
= − − + − − − −
= − − + − − − −
= − − +
The vertex is (4, 6). Other points will vary. Possible other points are (3, 5) and (6, 2).
MHR • Principles of Mathematics 10 Solutions 15
16 MHR • Principles of Mathematics 10 Solutions
11
8
1
1
x= − + +
d)
( )( ) ( )( )
( ) ( )
( )
2
2
2 22
22
2
4 16 11
4 4 11
4 4 2 2
4 4 4 4 2 11
4 2 5
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − +
= − −
The vertex is (2, –5). Other points will vary. Possible other points are (1, –1) and (3, –1). e)
( )( )( ) ( )( )( )
2
2
2 2 2
2 2 2
2
5 30 48
5 6 48
5 6 3 3 48
5 6 3 5 3 4
5 3 3
y x x
x x
x x
x x
x
= − − −
= − + −
= − + + − −
= − + + − − −
= − + − 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
( )( ) ( )( )
( ) ( )( )
( )
2
2
2 22
22
2
4 1
4 1
4 2 2
4 4 1 2
2 5
y x x
x x
x x
x x
x
= − + +
= − − +
= − − + − − − +
= − − + − − − +
= − − +
The maximum height of 5 m occurs at a horizontal distance of 2 m. Chapter 6 Section 1 Question 13 Page 271 For , the maximum height of 6.1 m occurs at time t = 1.0 s. 24.9 10 1h t t= − + + For h x , the maximum height of 6.1 m occurs at a horizontal distance of x = 17.7 m.
20.0163 0.577 1
16 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 1 Question 14 Page 271
( )( ) ( )( )
( ) ( )( )
( )
2
2
2 22
22
2
2 3
2 3
2 1 1
2 1 1 1
1 4
y x x
x x
x x
x x
x
= − + +
= − − +
= − − + − − − +
= − − + − − − +
= − − +
3
3
The diver's maximum height is 4 m. Chapter 6 Section 1 Question 15 Page 271
( )( )( ) ( )( )
2
2
2 2 2
2 2 2
2
2 84 1025
2 42 1025
2 42 21 21 1025
2 42 21 2 21 1025
2 21 143
C t t
t t
t t
t t
t
= − +
= − +
= − + − +
= − + − +
= − +
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) ( )( )4 0.5 120 20R x x= − + c) ( )( )
( )( ) ( )( )
( ) ( )( )
( )
2
2
2 22
22
2
4 0.5 120 20
10 20 480
10 2 480
10 2 1 1 480
10 2 1 10 1 480
10 1 490
R x x
x x
x x
x x
x x
x
= − +
= − + +
= − − +
= − − + − − − +
= − − + − − − +
= − − +
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 17
Chapter 6 Section 1 Question 17 Page 272 a) b) y =
( )( )( ) (( )
2
2
2 2 2
2 2
2
1.5 6 7
1.5 4 7
1.5 4 2 2 7
1.5 4 4 1.5 2 7
1.5 2 13
y x x
x x
x x
x x
x
= + −
= + −
= + + − −
= + + −
= + −
) −
x x
x x
x x
x x
x
− − +
= − + +
= − + + − +
( )( )( ) ( )( )( )
2
2
2 2 2
2 2 2
2
0.1 2 1
0.1 20 1
0.1 20 10 10 1
0.1 20 10 0.1 10 1
0.1 10 11
= − + + − − +
= − + +
( )2
)
The minimum point is at (–2, –13). The maximum point is at (–10, 11). c) d)
( )( )( )( )
2
2
2 2 2
2 2
2
0.3 3
0.3 10
0.3 10 5 5
0.3 10 5 0.3 5
0.3 5 7.5
y x x
x x
x x
x x
x
= +
= +
= + + −
= + + −
= + −
( )( ) ( )( )( )( ) ( )(
( )
2
2
2 22
2 22
2
1.25 5
1.25 4
1.25 4 2 2
1.25 4 2 1.25 2
1.25 2 5
y x x
x x
x x
x x
x
= − +
= − −
= − − + − − −
= − − + − − − −
= − − + The minimum point is at (–5, –7.5). The maximum point is at (2, 5). e) f)
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
0.5 6 12
0.5 12 12
0.5 12 6 6 12
0.5 12 6 0.5 6 12
0.5 6 6
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − − +
= − −
( )( )( ) ( )( )( )
2
2
2 2 2
2 2 2
2
0.02 0.6 9
0.02 30 9
0.02 30 15 15 9
0.02 30 15 0.02 15 9
0.02 15 4.5
y x x
x x
x x
x x
x
= − − −
= − + −
= − + + − −
= − + + − −
= − + −
−
The minimum point is at (6, –6). The maximum point is at (–15, –4.5). Chapter 6 Section 1 Question 18 Page 272
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
0.2 1.6 4.2
0.2 8 4.2
0.2 8 4 4 4.2
0.2 8 4 0.2 4 4.2
0.2 4 1
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − − +
= − +
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 Page 272 a) c = 6, h = 4; ( )22 8 16 4x x x+ + = + b) b = 12, h = 6; ( 22 12 36 6x x x+ + = + )or b = –12, h = –6; ( )22 12 36 6x x x− + = − c) b = –10, c = 27; ( )2 2
2
5 2 10 25 2
10 27
x x x
x x
− + = − + +
= − + Chapter 6 Section 1 Question 20 Page 272
( )( )( ) ( )( )
2
2
2 2 2
2 2 2
2
0.15 9 195
0.15 60 195
0.15 60 30 30 195
0.15 60 30 0.15 30 195
0.15 30 60
d v v
v v
v v
v v
v
= − +
= − +
= − + − +
= − + − +
= − +
Minimum drag occurs at a speed of 30 km/h. Chapter 6 Section 1 Question 21 Page 272
( )( ) ( )( )
( ) ( )( )
( )
2
2
2 22
22
2
10 21
10 21
10 5 5 21
10 25 1 5 21
5 4
y x x
x x
x x
x x
x
= − + −
= − − −
= − − + − − − −
= − − + − − − −
= − − +
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 19
Chapter 6 Section 1 Question 23 Page 273 Let the length be represented by x. The width will be 10 – x.
( )
( )( ) ( )( )( )( ) ( )( )
( )
2
2
2 22
2 22
2
10
10
10
10 5 5
10 5 1 5
5 25
A x x
x x
x x
x x
x x
x
= −
= − +
= − −
= − − + − − −
= − − + − − − −
= − − +
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.
( )
( )( ) ( )( )( )( ) ( )( )
( )
2
2
2 22
2 22
2
200 2
2 200
2 100
2 100 50 50
2 100 50 2 50
2 50 5000
A x x
x x
x x
x x
x x
x
= −
= − +
= − −
= − − + − − −
= − − + − − − −
= − − +
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 2y ax bx c= + + . Since the y-intercept is 5, c = 5. Substitute the coordinates of the points (2, –3) and (–1, 12) into 2 5y ax bx= + + to form a linear system of equations that can be solved for a and b.
( ) ( )
2
23 = +− 12 1 1− −2
5
54 2 8
2
y ax bx
a ba b
= + +
+
+ = −
57
y ax bx
a ba b
= + +
=
( ) ( )
2
2
5
+ +
= −
2 4a b+ = − 7a b− = 2 4a b+ = − 7a b− =
7a b
bb
− =
= −
5
5
+
Substitute a = 1 into equation .
3 31
aa==
7
6 1− =
( ) ( )( )( )( ) ( )
( )
2
2 22
2 22
2
6 5
6 3 3
6 3 3
3 4
y x x
x x
x x
x
= − +
= − + − − − +
= − + − − − +
= − −
The vertex is at (3, –4). Chapter 6 Section 1 Question 27 Page 273
2
2
2 22
2 22
2 2
2 2
2 2
2 4
y ax bx cba x x ca
b b ba x x ca a a
b b ba x x a ca a a
b ba x ca a
= + +
⎛ ⎞= + +⎜ ⎟⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞= + + − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞= + + − +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠
⎛ ⎞= + − +⎜ ⎟⎝ ⎠
The x-coordinate of the vertex is 2ba
− .
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
2R , and the base is
2S . Apply the Pythagorean
theorem to find S.
2 22
2 2
2 2
2 23
4 43
3
S R R
S R
S R
S R
⎛ ⎞ ⎛ ⎞+ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
=
=
The side length of the triangle is 3R . Chapter 6 Section 1 Question 29 Page 273 There are ways to draw the names. Two of these include Jane and Farhad. The
probability that they will go together is
4 3 12× =2 1
12 6= . Answer D.
Chapter 6 Section 1 Question 30 Page 273
( )
1000 3000
30100
30
1000 10
10
googol
=
=
=
Answer D
22 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 2 Solve Quadratic Equations Chapter 6 Section 2 Question 1 Page 279 a) b) 5 or 2x x= − = − 3 or 4x x= = − c) d) 1 or 7x x= = 0 or 9x x= = −
e) 3 or 52
x x= − = f) 1 4 or 2 3
x x= = −
g) 5 3 or 3 4
x x= =
Chapter 6 Section 2 Question 2 Page 279 a)
( )( )
2 8 12 02 6 0
2 0 or 6 02 or 6
x xx x
x xx x
+ + =
+ + =
+ = + == − = −
Check by substituting both solutions in the original equation. For x = –2: For x = –6:
( ) ( )
2
2
8 12
8 22 10
2
x x= + +
= + +−−
=
L.S. 0 =R.S.
( ) ( )
2
2
8 12
8 26 6 10
x x= + +
= + +
=
L.S. 0
=
L.S. 0
−−
−−
=R.S.
L.S. = R.S. L.S. = R.S. The roots are –2 and –6. b)
( )( )
2 9 18 03 6 0
3 0 or 6 03 or 6
h hh h
h hh h
+ + =
+ + =
+ = + == − = −
Check by substituting both solutions in the original equation. For h = –3: For h = –6:
( ) ( )
2
2
9 18
9 83 10
3
h h= + +
= + +
=R.S.
( ) ( )
2
2
9 18
9 86 6 10
h h= + +
= + +
=
L.S. 0
−−
=R.S.
L.S. = R.S. L.S. = R.S. The roots are –3 and –6.
MHR • Principles of Mathematics 10 Solutions 23
c)
( )
2 3 03 03 0 or 0
3
m mm m
m mm
+ =
+ =
+ = == −
Check by substituting both solutions in the original equation. For m = –3: For m = 0:
( ) ( )2 333− −+
2 3
0
m m= +
=
=
L.S. 0 0 =R.S.
( ) ( )
2
2
3
0030
m m= +
= +
=
L.S. =R.S.
L.S. = R.S. L.S. = R.S. The roots are –3 and –0. d)
( )( )
2 18 56 04 14 0
4 0 or 14 04 or 14
w ww w
w ww w
− + =
− − =
− = − == =
Check by substituting both solutions in the original equation. For w = 4: For w = 14:
( ) ( )
2
2
18 56
18 54 4 60
w w= − +
= − +
=
L.S. 0=R.S.
( )241 ( )
2 18 56
18 51 60
4
w w= − +
= − +
=
L.S. 0
0 0
=R.S.
L.S. = R.S. L.S. = R.S. The roots are 4 and 14. e)
( )
2 2 02 02 0 or 0
2
x xx x
x xx
− =
− =
− = ==
Check by substituting both solutions in the original equation. For x = 2: For x = 0:
( ) ( )
2
2
2
0222
x x= −
= −
=
L.S. =R.S.
( ) ( )
2
2
2
0020
x x= −
= −
=
L.S. =R.S.
L.S. = R.S. L.S. = R.S. The roots are 2 and 0.
24 MHR • Principles of Mathematics 10 Solutions
f)
( )( )
2 17 30 02 15 0
2 0 or 15 02 or 15
c cc c
c cc c
− + =
− − =
− = − == =
Check by substituting both solutions in the original equation. For c = 2: For c = 15:
( ) ( ) +
2
2
17 30
17 32 2 00
c c= − +
= −
=
L.S. 0 =R.S.
( )251 ( )
2 17 30
17 31 00
5
c c= − +
= − +
=
L.S. 0
2
0
=R.S.
L.S. = R.S. L.S. = R.S. The roots are 2 and 15. g)
( )( )
2 9 22 011 2 0
11 0 or 2 011 or 2
n nn n
n nn n
+ − =
+ − =
+ = − == − =
Check by substituting both solutions in the original equation. For n = –11: For n = 2:
( ) ( )
2
211 1
9 22
9 20
1
n n
−
= −
+
=
−
+
= −
L.S. =R.S.
( ) ( )
2
2
9 22
2 9 20
2 2
n n= + −
= + −
=
L.S. 0
0 0
=R.S.
L.S. = R.S. L.S. = R.S. The roots are –11 and 2. h)
( )
2 11 011 011 0 or 0
11
y yy y
y yy
− =
− =
− = ==
Check by substituting both solutions in the original equation. For y = 11: For y = 0:
( ) ( )
2
2
11
11011 11
y y= −
= −
=
L.S. =R.S.
( ) ( )
2
2
11
10 1 00
y y= −
= −
=
L.S. =R.S.
L.S. = R.S. L.S. = R.S. The roots are 11 and 0.
MHR • Principles of Mathematics 10 Solutions 25
Chapter 6 Section 2 Question 3 Page 280 a) b)
( ) ( )( ) ( )( )( )
2
2
2
3 28 93 27 9
3 27 9
3 9 9
9 3 1 0
x xx x x
x x x
x x x
x x
+ + =
+ + + =
+ + + =
+ + + =
+ + =
00
0
0
0
0
0
( ) ( )( ) ( )( )( )
2
2
2
4 19 154 4 15 15 0
4 4 15 15
4 1 15 1
1 4 15 0
k kk k k
k k k
k k k
k k
+ + =
+ + + =
+ + + =
+ + + =
+ + =
9 0 or 3 1 0
19 or 3
x x
x x
+ = + =
= − = −
1 0 or 4 15 0151 or 4
k k
k k
+ = + =
= − = −
00
0
0
c) d)
( ) ( )( ) ( )
( )( )
2
2
2
8 22 158 12 10 15
8 12 10 15
4 2 3 5 2 3 0
2 3 4 5
y yy y y
y y y
y y y
y y
− + =
− − + =
− + − + =
− − − =
− − =
( ) ( )
216 1 04 1 4 1 0
bb b
− =
+ + − =
2 3 0 or 4 5 0
3 5 or 2 4
y y
y y
− = − =
= =
4 1 0 or 4 1 01 1 or 4 4
b b
b b
+ = − =
= − =
e) f)
( )
210 30 010 3 0
m mm m
+ =
+ = ( )
2
2
4 12 9
2 3
x x
x
0
0
− + =
− =
10 0 or 3 0
0 or 3m mm m= += =
=−
2 3 0 32
x
x
− =
=
26 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 2 Question 4 Page 280 a)
( )( )
2
2
5 45 4 0
1 4 01 0 or 4 0
1 or 4
x xx x
x xx x
x x
+ = −
+ + =
+ + =
+ = + == − = −
b)
( )( )
2
2
8 158 15 0
3 5 03 0 or 5 0
3 or 5
c cc c
c cc c
c c
+ = −
+ + =
+ + =
+ = + == − = −
c) d)
( )( )
2
2
13 1213 12 0
12 1 012 0 or 1 0
12 or 1
k kk x
k kk k
k k
= −
− + =
− − =
− = − == =
( )
2
2
2
1 22 1 0
1 01 0
1
b bb b
bb
b
+ = −
+ + =
+ =
+ == −
e) f)
( )( )
2
2
300 2020 300 0
10 30 010 0 or 30 0
10 or 30
m mm mm m
m mm m
= −
+ − =
− + =
− = + == = −
( )
2
2
77 07 07 0 or 0
7
y yy y
y yy y
y
=
− =
− =
− = ==
MHR • Principles of Mathematics 10 Solutions 27
Chapter 6 Section 2 Question 5 Page 280 a) b)
( ) ( )( ) ( )( )( )
2
2
2
2
2 72 7 6 0
2 4 3 6 0
2 4 3 6 0
2 2 3 2 0
2 2 3 0
m mm m
m m m
m m m
m m m
m m
= − −
+ + =
+ + + =
+ + + =
+ + + =
+ + =
6
( ) ( )( ) ( )( )( )
2
2
2
2
9 89 8 0
9 9 8 8 0
9 9 8 8 0
9 1 8 1 0
1 9 8 0
x xx x
x x x
x x x
x x x
x x
= +
− − =
− + − =
− + − =
− + − =
− + =
2 0 or 2 3 0
32 or 2
m m
m m
+ = + =
= − = −
1 0 or 9 8 081 or 9
x x
x x
− = + =
= = −
c) d)
( )
2
2
2
4 124 12 9 0
2 3 0
y yy y
y
− = −
− + =
− =
9
( ) ( )( ) ( )
( )( )
2
2
2
2
5 2 1616 2 5 0
16 8 10 5 0
16 8 10 5 0
8 2 1 5 2 1 0
2 1 8 5 0
p pp p
p p p
p p p
p p p
p p
− = −
− − =
+ − − =
+ + − − =
+ − + =
+ − =
2 3 0
3 2
y
y
− =
=
2 1 0 or 8 5 01 5 or 2 8
p p
p p
+ = − =
= − =
e) f)
( ) ( )( ) ( )
( )( )
2
2
2
2
12 10 3712 37 10 0
12 40 3 10 0
12 40 3 10 0
4 3 10 1 3 10 0
3 10 4 1 0
mm m
m m m
m m m
m m m
m m
= −
+ − =
+ − − =
+ + − − =
+ − + =
+ − =
m 7
0
( ) ( )( ) ( )( )( )
2
2
2
2
3 223 22 7 0
3 21 7 0
3 21 7 0
3 7 7
7 3 1 0
w ww w
w w w
w w w
w w w
w w
+ = −
+ + =
+ + + =
+ + + =
+ + + =
+ + =
3 10 0 or 4 1 0
10 1 or 3 4
m m
m m
+ = − =
= − =
7 0 or 3 1 017 or 3
w w
w w
+ = + =
= − = −
28 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 2 Question 6 Page 280 a)
( )( )
2
2
10 16 0 Divide both sides by 1.10 16 0
8 2 0
x xx xx x
− − − = −
+ + =
+ + =8 0 or 2 0
8 or 2x x
x x+ = + =
= − = −
b)
( )( )
2
2
3 24 45 0 Divide both sides by 3.8 15 0
3 5 0
t tt t
t t
+ + =
+ + =
+ + =3 0 or 5 0
3 or 5t t
t t+ = + =
= − = −
c)
( ) ( )( ) ( )( )( )
2
2
2
2
2
6 15 96 15 9 0 Divide both sides by 3.2 5 3 0
2 2 3 3 0
2 2 3 3 0
2 1 3 1 0
1 2 3 0
d dd dd d
d d d
d d d
d d d
d d
+ = −
+ + =
+ + =
+ + + =
+ + + =
+ + + =
+ + =
1 0 or 2 3 031 or 2
d d
d d
+ = + =
= − = −
d)
( ) ( )( ) ( )( )( )
2
2
2
2
2
10 32 610 32 6 0 Divide both sides by 2.
5 16 3 05 15 3 0
5 15 3 0
5 3 1 3 0
3 5 1 0
g gg gg g
g g g
g g g
g g g
g g
− + =
− + − = −
− + =
− − + =
− + − + =
− − − =
− − =
3 0 or 5 1 013 or 5
g g
g g
− = − =
= =
MHR • Principles of Mathematics 10 Solutions 29
Chapter 6 Section 2 Question 7 Page 280
( )( )
2
2
2 3 02 3 0
1 3 01 0 or 3 0
1 or 3
x xx x
x xx x
x x
− + + =
− − =
+ − =
+ = − == − =
The ball has travelled 3 m horizontally when it hits the ground. Chapter 6 Section 2 Question 8 Page 280
( )( )
( ) ( )( ) ( )
( )( )
2
2
2
2
10 2 3 54
2 3 20 30 542 17 84 0
2 24 7 84 0
2 24 7 84 0
2 12 7 12 0
12 2 7 0
x x
x x xx x
x x x
x x x
x x x
x x
+ − =
− + − =
+ − =
+ − − =
+ + − − =
+ − + =
+ − =
12 0 or 2 7 0
712 or 2
x x
x x
+ = − =
= − =
The negative answer is inadmissible. If x is 3.5 cm, the area is 54 cm2. Chapter 6 Section 2 Question 9 Page 280 a) ( )( )5 4x x− − = 0 b) ( )( )2 3x 0x+ − = Chapter 6 Section 2 Question 10 Page 280 a) ( )( )
2
6 7
42 0
x x
x x
− + =
+ − =
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
( )( )2
2
3 2 5 4 0
15 12 10 8 015 2 8 0
x x
x x xx x
− + =
+ − − =
+ − =
30 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 2 Question 12 Page 280 a) Answers will vary. For example: 2 2 0
( 1)( 2) 0x x
x x− − =
+ − =
b) Answers will vary. For example: 26 1
(2 1)(3 1) 0x x
x x0+ − =
+ − =
Chapter 6 Section 2 Question 13 Page 280 Answers will vary. For example: 2 3 0x x− + = There are no integers whose product is 3 and sum is –1. Chapter 6 Section 2 Question 14 Page 280
( )
( )( )
22 2
2 2
2
2
1 29
2 1 8412 2 840 0 Divide both sides by 2.
420 020 21 0
x x
x x xx xx x
x x
+ − =
+ − + =
− − =
− − =
+ − =
20 0 or 21 020 or 21
x xx x
+ = − == − =
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 Question 16 Page 281 a) For n = 1: For n = 2:
( )( )
1
11 12
S n n= +
= +
=
( )( )
1
12 26
S n n= +
= +
=
The formula works for n = 1 and n = 2. b) ( )
( )5 5
1
130
S n n= +
= +
=
The sum of the first five even numbers is 30. c) ( )
( )( )
2
1 306
306 017 18 0
n n
n nn n
+ =
+ − =
− + =17 0 or 18 0
17 or 18n n
n n− = + =
= = −
The value of n is 17. Chapter 6 Section 2 Question 17 Page 281 ( )( )
( )( )
2
2
2
3.90 0.10 120 20 700
468 78 12 2 7002 66 232 0 Divide both sides by 2.
33 116 04 29 0
n n
n n nn nn nn n
− + =
+ − − =
− + − = −
− + =
− − =
4 0 or 29 04 or 29
n nn n
− = − == =
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.
( ) ( )
( )( )
2 22
2 2 2
2
7 8
14 49 16 642 15 0
5 3 0
x x x
x x x x xx x
x x
+ + = +
+ + + = + +
− − =
− + =
5 0 or 3 05 or 3
x xx x
− = + == = −
−
The negative answer is inadmissible. The width is 5 m and the length is 12 m. Chapter 6 Section 2 Question 20 Page 281
( )( )
( )( )( )
2
2
2
2
2
1125 1 4500
0 1125 2 1 4500
0 1125 2250 1125 45000 1125 2250 3375
0 1125 2 3
0 1125 3 1
P t
t t
t tt t
t t
t t
= − −
= − + −
= − + −
= − −
= − −
= − +
3 0 or 1 03 or 1
t tt t
− = + == =
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 Question 21 Page 281 a)
( )( )
2 3 21 2
y yy y
+ + =
+ + =
00
0
1 0 or 2 01 or 2
y yy y
+ = + == − = −
The coefficients are the same for the two equations, but the solutions for 2 23 2y yy x+ + = are y = –x and y = –2x. b) i)
( )( )
2 23 22 0
y yx xy x y x
+ + =
+ + =
0
x
)
00
0
0
0=
0 or 2 0 or 2
y x y xy x y
+ = + == − = −
ii)
( ) (( ) ( )
( )( )
2 2
2 2
2 2
5 6 85 10 4 8
5 10 4 8
5 2 4 2
2 5 4
y yx xy yx yx x
y yx yx x
y y x x y x
y x y x
− − =
− + − =
− + − =
− + − =
− +
2 0 or 5 4 042 or 5
y x y x
y x y
− = + =
= = x−
iii)
( )
2 2
2 2
2
1 1 1 09 3 44 12 9
2 3
y yx x
y yx x
y x
− + =
− + =
− =
0
0
2 3 032
y x
y x
− =
=
34 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 3 Graph Quadratics Using the x-Intercepts Chapter 6 Section 3 Question 1 Page 288 Factor and solve the corresponding quadratic equation by letting y = 0. a) b)
( )(
2 5 60 2y x x
x x= + +
= + + )3 )4( )(
2 11 280 7y x x
x x= − +
= − −
2 0 or 3 02 or 3
x xx x
+ = + == − = −
7 0 or 4 07 or 4
x xx x
− = − == =
The x-intercepts are –2 and –3. The x-intercepts are 7 and 4. c) d)
( )
2 90 9y x x
x x= +
= + ( )(
2 5 240 8y x x
x x= + −
)3= + −
0 or 3 03
x xx
= += −=
8 0 or 3 0
8 or 3x x
x x+ = − =
= − =
The x-intercepts are 0 and –9. The x-intercepts are –8 and 3. e) f)
( )(
2 2 80 4y x x
x x= − −
= − + )2 )3( )(
2 9 360 12y x x
x x= + −
= + −
4 0 or 2 04 or 2
x xx x
− = + == = −
12 0 or 3 0
12 or 3x x
x x+ = − =
= − =
The x-intercepts are 4 and –2. 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) b)
( ) ( )( ) (
( )( )
2
2
2
4 20 90 4 2 18 9
0 4 2 18 9
0 2 2 1 9 2 1
0 2 1 2 9
y x xx x x
x x x
x x x
x x
= + +
= + + +
= + + +
= + + +
= + +
)
( )
28 60 2 4 3y x x
x x= −
= −
2 1 0 or 2 9 0
1 9 or 2 2
x x
x x
+ = + =
= − = −
2 0 or 4 3 030 or 4
x x
x x
= − =
= =
c) d)
( ) ( )( ) (
( )( )
2
2
2
6 11 70 6 3 14 7
0 6 3 14 7
0 3 2 1 7 2 1
0 2 1 3 7
y x xx x x
x x x
x x x
x x
= − −
= + − −
= + + − −
= + − +
= + −
)( ) ( )
( ) ( )( )( )
2
2
2
5 6 80 5 10 4 8
0 5 10 4 8
0 5 2 4 2
0 2 5 4
y x xx x x
x x x
x x x
x x
= + −
= + − −
= + + − −
= + − +
= + −
2 1 0 or 3 7 0
1 7 or 2 3
x x
x x
+ = − =
= − =
2 0 or 5 4 042 or 5
x x
x x
+ = − =
= − =
e) f)
( ) ( )( ) ( )
( )( )
2
2
2
3 13 40 3 12 4
0 3 12 4
0 3 4 1 4
0 4 3 1
y x xx x x
x x x
x x x
x x
= − +
= − − +
= − + − +
= − − −
= − −
( )
2
2
4 20 2
0 2 5
y x x
x
5= − +
= −
4 0 or 3 1 0
14 or 3
x x
x x
− = − =
= =
2 5 0 52
x
x
− =
=
36 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 3 Question 3 Page 289 a) Factor and solve the corresponding quadratic equation.
( )(
2 9 140 2y x x
x x= + +
= + + )7 The x-intercepts are –2 and –7. Find the x-coordinate of the vertex, and then, the y-coordinate.
( )
2
2
2 72
92
9 14
9 1492 2
254
x
y x x
− + −=
= −
= + +
⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎠
= −
−
9⎝
−
The vertex is 9 25,2 4
⎛ ⎞− −⎜ ⎟⎝ ⎠
.
b) Factor and solve the corresponding quadratic equation.
( )( )
2 6 80 2y x x
x x= − +
= − − 4
The x-intercepts are 2 and 4. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
2
2 42
3
3
6 8
63 81
x
y x x
+=
=
= − +
= − +
= −
The vertex is ( ) . 3, 1−
MHR • Principles of Mathematics 10 Solutions 37
c) Factor and solve the corresponding quadratic equation.
( )( )(
2
2
4 5
0 4
0 5
y x x
x x
x x
= − − +
= − + −
= − + − )5
1
The x-intercepts are –5 and 1. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
22 − +−
5 12
24
4 2
5
59
x
y x x
− +=
= −
= − − +
= −−
=
The vertex is ( )2,9− . d) Factor and solve the corresponding quadratic equation.
( )
2 50 5y x x
x x= − −
= − +
The x-intercepts are 0 and –5. Find the x-coordinate of the vertex, and then, the y-coordinate.
( )
2
2
−−5 52
0 52
52
5
5
4
225
x
y x x
+ −=
= −
= − −
⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎝ ⎠
=
⎠
The vertex is 5 25,2 4
⎛ ⎞⎜ ⎟ . −⎝ ⎠
38 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 3 Question 4 Page 289 a) Factor and solve the corresponding quadratic equation.
( )(
2 90 3y x
x x= −
= + − )3
The x-intercepts are –3 and 3. Find the x-coordinate of the vertex, and then, the y-coordinate.
( )
2
2
3 32
990
09
x
y x
− +=
=
= −
= −
= −
The vertex is ( )0, 9− . b) Factor and solve the corresponding quadratic equation.
( )( )(
2
2
10 9
0 10
0 9
y x x
x x
x x
= − + −
= − − +
= − − − )9
1
The x-intercepts are 9 and 1. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )25 5+ −
2
9 12
510 9
10 916
x
y x x
+=
=
= − + −
= −
=
The vertex is ( ) . 5,16
MHR • Principles of Mathematics 10 Solutions 39
c) Factor and solve the corresponding quadratic equation.
( )
2
2
12 36
0 6
y x x
x
= − +
= −
The x-intercept is 6. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
26 6
6 62
612 36
12 360
x
y x x
+=
=
= − +
= − +
=
The vertex is ( )6,0 . d) Factor and solve the corresponding quadratic equation.
( )(
2160 4 4y x
)x x= −
= + −
The x-intercepts are –4 and 4. Find the x-coordinate of the vertex, and then, the y-coordinate.
( )
2
2
4 42
016
11
066
x
y x
− +=
=
= −
= −
=
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.
( ) ( )( ) ( )
( )( )
2
2
2
2 15 70 2 14 7
0 2 14 7
0 2 7 7
0 7 2 1
y x xx x x
x x x
x x x
x x
= + +
= + + +
= + + +
= + + +
= + +
7 0 or 2 1 0
17 or 2
x x
x x
+ = + =
= − = −
The zeros are –7 and 12
− .
Find the x-coordinate of the vertex, and then, the y-coordinate.
2
2
172
2154
2 15 7
2 15 715 15⎞ ⎛ ⎞+ +⎟ ⎜ ⎟⎝ ⎠
− −
1698
4 4
x
y x x
⎛ ⎞− + −⎜ ⎟⎝ ⎠=
= −
= + +
⎛= ⎜⎝ ⎠
= −
The vertex is 15 169,4 8
⎛ ⎞− −⎜ ⎟⎝ ⎠
.
MHR • Principles of Mathematics 10 Solutions 41
b) Factor and solve the corresponding quadratic equation.
( ) ( )( ) ( )
( )( )
2
2
2
12 16 50 12 6 10 5
0 12 6 10 5
0 6 2 1 5 2 1
0 2 1 6 5
y x xx x x
x x x
x x x
x x
= − +
= − − +
= − + − +
= − − −
= − −
2 1 0 or 6 5 0
1 5 or 2 6
x x
x x
− = − =
= =
The zeros are 12
and 56
.
Find the x-coordinate of the vertex, and then, the y-coordinate.
2
22 23
1 52 6
22312 16 5
12 16 5
13
3
x
y x x
+=
=
= − +
⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
= −
+
The vertex is 2 1,3 3
⎛ ⎞−⎜ ⎟⎝ ⎠
.
42 MHR • Principles of Mathematics 10 Solutions
c) Factor and solve the corresponding quadratic equation.
( )( )( ) ( )
( ) ( )( )( )
2
2
2
2
8 13 6
0 8 13 6
0 8 16 3 6
0 8 16 3 6
0 8 2 3 2
0 2 8 3
y x x
x x
x x x
x x x
x x x
x x
= − − +
= − + −
= − + − −
⎡ ⎤= − + + − −⎣ ⎦⎡ ⎤= − + − +⎣ ⎦
= − + −
8 3 0 or 2 0
3 or 28
x x
x x
− = + =
= = −
The zeros are 38
and . 2−
Find the x-coordinate of the vertex, and then, the y-coordinate.
( )
2
213 1316
3 28
213168 13 6
8 136
36132
1
x
y x x
+ −=
= −
= − − +
⎛ ⎞ ⎛ ⎞= − − +⎜ ⎟ ⎜ ⎟⎝ ⎝
=
6⎠⎠
−−
The vertex is 13 361,16 32
⎛ ⎞⎜ ⎟ . − −⎝ ⎠
MHR • Principles of Mathematics 10 Solutions 43
d) Factor and solve the corresponding quadratic equation.
( ) ( )( ) ( )
( )( )
2
2
2
8 17 90 8 8 9 9
0 8 8 9 9
0 8 1 9 1
0 1 8 9
y x xx x x
x x x
x x x
x x
= + +
= + + +
= + + +
= + + +
= + +
1 0 or 8 9 0
91 or 8
x x
x x
+ = + =
= − = −
The zeros are –1 and 98
− .
Find the x-coordinate of the vertex, and then, the y-coordinate.
2
2
918
21716
8 17 9
8 1 17 97 1716 16
⎞ ⎛ ⎞+ +⎟ ⎜ ⎟− −
132
x
y x x
⎛ ⎞− + −⎜ ⎟⎝ ⎠=
= −
= + +
⎛= ⎜⎝ ⎠ ⎝ ⎠
= −
The vertex is 17 1,16 32
⎛ ⎞− −⎜ ⎟⎝ ⎠
.
44 MHR • Principles of Mathematics 10 Solutions
e) Factor and solve the corresponding quadratic equation.
( )
2
2
16 24 9
0 4 3
y x x
x
= − +
= −
4 3 0
34
x
x
− =
=
The zero is 34
.
Find the x-coordinate of the vertex, and then, the y-coordinate.
2
2
3 34 4
23416 24 9
16 24 9
0
3 34 4
x
y x x
+=
=
= − +
⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
+
The vertex is 3 ,04
⎛ ⎞⎜ ⎟⎝ ⎠
.
MHR • Principles of Mathematics 10 Solutions 45
f) Factor and solve the corresponding quadratic equation.
( )( )( ) ( )
( ) ( )( )( )
2
2
2
2
4 18 8
0 2 2 9 4
0 2 2 8 4
0 2 2 8 4
0 2 2 4 4
0 2 4 2 1
y x x
x x
x x x
x x x
x x x
x x
= − − −
= − + +
= − + + +
⎡ ⎤= − + + +⎣ ⎦⎡ ⎤= − + + +⎣ ⎦
= − + +
4 0 or 2 1 0
14 or 2
x x
x x
+ = + =
= − = −
The zeros are –4 and 12
− .
Find the x-coordinate of the vertex, and then, the y-coordinate.
2
2
142
2944 18 8
4 18
49
94
4
94
x
y x x
⎛ ⎞− + −⎜ ⎟⎝ ⎠=
= −
= − − −
⎛ ⎞ ⎛ ⎞= − − −⎜ ⎟ ⎜ ⎟
8⎝ ⎠ ⎠
=
⎝−−
The vertex is 9 49,4 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 ( )6y ax x= − . To find a, substitute (x, y) = (5, 5).
( ) ( )2
2
65 25 305 51
1 ( 6)6
5 = −5 5a aa aa
ay x x
x x
= −= −
− == − −
= − +
b) Since the x-intercepts are –4 and 6, the equation has the form ( )( )4 6y a x x= + − . To find a, substitute (x, y) = (1, –5).
( )( )
( )( )
( )2
2
4 65 2515
1 4 651 2 2451 2 245 5 5
5 = + −− 1 1aa
a
y x x
x x
x x
− = −
=
= + −
= − −
= − −
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 ( )( )8 2y a x x= + + . To find a, substitute (x, y ) = (–7, –10).
( )( )
( )( )( )2
2
10 52
2 8 2
2 10 16
2 20 32
aa
ay x x
x x
x x
=
− = −=
= + +
= + +
= + +
8 210 7 7+ +− − −
b) Since the x-intercepts are –5 and –1, the equation has the form ( )( )5 1y a x x= + + . To find a, substitute (x, y) = (–3, 3).
( )( )
( )( )
( )2
2
3 434
3 5 143 6 543 9 154 2 4
aa
a
y x x
x x
x x
= −
− =
= − + +
= − + +
= − − −
5 133 3= + +− −
c) Since the x-intercepts are 0 and 8, the equation has the form ( )8y ax x= − . To find a, substitute (x, y) = (6, 6).
( )( )
( )
( )2
2
86 1212
1 821 82
42
6
1
6 6aa
a
y x x
x x
x x
= −
= −
− =
= − −
= − −
= − +
48 MHR • Principles of Mathematics 10 Solutions
d) Since the x-intercepts are –4 and 5, the equation has the form ( )( )4 5y a x x= + − . To find a, substitute (x, y) = (1, –6).
( )( )
( )(( )2
2
4 56 20
0.30.3 4 5
0.3 20
)
0.3 0
1
.
1
3 6
aa
ay x x
x x
x x
= +
− = −=
= + −
= − −
= − −
)2
6 −−
Chapter 6 Section 3 Question 8 Page 290 a)
( )( )(
2
2
4
0 4
0 2
h d
d
d d
= − +
= − −
= − + − The d-intercepts are at –2 and 2. The width of the garage is 4 m. Find the vertex.
( )
2
2
2 22
40 +
40
4
d
h d
− +=
=
−
= −
= +
=
2
The height of the garage is 4 m. b) c) The relation is valid for . h must be positive. 2 d− ≤ ≤
MHR • Principles of Mathematics 10 Solutions 49
Chapter 6 Section 3 Question 9 Page 290 a)
( )( )( ) ( )
( ) ( )( )( )
2
2
2
2
3 11 4
0 3 11 4
0 3 12 4
0 3 12 4
0 3 4 4
0 4 3 1
y x x
x x
x x x
x x x
x x x
x x
= − + +
= − − −
= − − + −
⎡ ⎤= − − + −⎣ ⎦⎡ ⎤= − − + −⎣ ⎦
= − − +
4 0 or 3 1 014 or 3
x x
x x
− = + =
= = −
The zeros are 4 and 13
− .
b) The relation is valid for . 0 4x≤ ≤ c) The rocket has travelled 4 m horizontally when is lands on the ground. d) Find the vertex.
2
211
1 432
1163 11 4
3 11
14.06
8
116
x
y x x
− +=
=
= − + +
⎛ ⎞ ⎛ ⎞= − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
4
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 ( )( )100 100y a x x= + − . To find a, substitute (x, y) = (0, 4).
( )( )
( )( )
( )2
2
100 1004 10 000
12500
1 100 1002500
1 10 0002500
1 42
4 0
50
0
0
aa
a
y x x
x
x
= + −
= −
− =
= − + −
= − −
= − +
( )2801 42500
1.44
y = − +
=
A point on the arch that is 20 m horizontally from one end is 1.44 m high.
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 x-intercept will be another 8 units to the right, or 5. Since the x-intercepts are –11 and 5, the equation has the form ( )( )11 5y a x x= + − . To find a, substitute (x, y) = (–3, 7).
( )( )
( )( )
( )2
2
11 57 64
764
7 11 5647 6 55
647 21 38
64 32 45
6
aa
a
y x x
x x
x x
+
= −
− =
= − + −
= − + −
= +− −
7 33= −−−
The y-intercept is 38564
.
52 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 3 Question 12 Page 290 a)
( )
( )
( ) ( )
( ) ( )
( )( )
2
2
2
2
1 16 3635 35 3510 16 363510 2 18 363510 2 18 363510 2 183510 2 1835
h d d
d d
d d d
d d d
d d d
d d
= − + +
= − − −
= − + − −
⎡ ⎤= − + − +⎣ ⎦
⎡ ⎤= − + − +⎣ ⎦
= − + −
2
18
The d-intercepts are –2 and 18. b) c) The relation is valid for . The height must be positive. 0 d≤ ≤
d) The ball was kicked at a height of 3635
m.
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)
b)
( )
21 2125 25
10 1125
d w
w w
= − +
= − − 0
w
0
The zeros are 0 and 10. The relation is valid for 0 1w≤ ≤ , since d must be positive. c) The road is 10 m wide. d) Find the vertex.
( ) ( )2 25 5+
2
0 102
51 2
125 251
125 250.2
w
d w w
+=
=
= − +
= −
=
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; or 2 10 25y x x= + + 2 10 25y x x= − +
b) n = 6 or –6; or 2 6y x x= − + − 9 2 6 9y x x= − − −
c) n = 1; 216 8 1y x x= − +
MHR • Principles of Mathematics 10 Solutions 55
Chapter 6 Section 3 Question 16 Page 291 a) The Marble Heads:
( )( )(
2
2
2 3
0 2
0 3
y x x
x x
x x
= − + +
= − − −
= − − + )3
1
−
3 0 or 1 0
3 or 1x x
x x− = + =
= =
The XY Team:
( )( )( ) ( )
( ) ( )( )( )
2
2
2
2
2 3
0 2 3
0 2 2 2 3 3
0 2 2 2 3 3
0 2 2 1 3 1
0 2 1 2 3
y x x
x x
x x x
x x x
x x x
x x
= − + +
= − − −
= − + − −
⎡ ⎤= − + + − −⎣ ⎦⎡ ⎤= − + − +⎣ ⎦
= − + −
1 0 or 2 3 0
31 or 2
x x
x x
+ = − =
= − =
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:
( ) ( )
2
2
1 32
12 3
12 34
1
x
y x x
− +=
=
= − + +
= − + +
=
The XY Team:
2
21 14
312
2142 3
2 3
3.1254
x
y x x
− +=
=
= − + +
⎛ ⎞ ⎛ ⎞= − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
The Marble Heads' marble flies higher by 0.875 m.
MHR • Principles of Mathematics 10 Solutions 57
Chapter 6 Section 3 Question 17 Page 291
a)
( )
( )( )
2
2
1 50100
10 5000100
10 5000 5000100
y x
x
x x
= − +
= − −
= − + −
The width of the hangar is 2 5000 , or about 141.42 m. 141.42 6.0823.24
Six planes can fit side by side inside the hangar. b) The width of the airplane hangar decreases as the height increases.
( )
( )( )
2
2
2
1 50100
1 50100
10 48001
2
001 4800 4800
100
y x
x
x
x x
= − +
= − +
= − −
= − − +
The width of the hangar is 2 4800 , or about 138.56 m. 139.56 5.9623.24
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: 2 8 18y x x= − + Parabola B: 2 12 37y x x= − − − 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
( )( )( )( )( )
2
2 2
2 2
4 3 4
4 3 3 4
4 4 3 12 4
x x a x x c
x a x a x x c
x a x a x x c
+ + = + +
+ + + = + +
+ + + = + +
( )4 3 1
12 4 1114
1211124
33
aa
a
c a
+ =
+ =
= −
=
⎛ ⎞= −⎜ ⎟⎝ ⎠
= −
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 Question 21 Page 291 a) ( )
( ) ( )( )( )
2
2
3 12 3 27 0
3 12 3 27 0
3 9 3 3 0
3 9 or 32 or 1
x x
x x
x x
x x 3x x
− + =
− + =
− − =
= == =
b) ( )
( )( )( ) ( )
( )( )
2 3
2 3
2
2 3 2 128 0
2 3 2 2 128 0
2 24 2 128 0
2 8 2 16 0
2 8 or 2 163 or 4
x x
x x
x x
x x
x x
x x
+− + =
− + =
− + =
− − =
= == =
MHR • Principles of Mathematics 10 Solutions 59
Chapter 6 Section 3 Question 22 Page 291 a) There are possible outcomes. Three of these, BBG, BGB, and GBB, are
favourable. The probability of exactly 2 boys is
2 2 2 8× × =38
.
b) The possible outcomes are JBB BJB BBJ JBG BJG BGJ JGB GJB GBJ JGG GJG GGJ
Half of these have 2 boys. The probability of 2 boys is 12
.
c) John is a boy. This eliminates some of the possible families.
60 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 4 The Quadratic Formula Chapter 6 Section 4 Question 1 Page 300 a) For 7x2 + 24x + 9 = 0, a = 7, b = 24, b) For 2x2 + 4x – 7 = 0, a = 2, b = 4, and c = –7. and c = 9.
( )( )( )
2
224 2
(4 7 97
42
42
24 32414
24 1814
b b acxa
− ± −=
− ± −=
− ±=
− ±=
)( )( )
2 42
2
4 724
4
b b aca−
− ±
− ±=
x − ±=
2 44 2 72
−−=
The roots are –3 and 37
− . The roots are 4 724
− + and 4 724
− − .
c) For 4x2 – 12x + 9 = 0, a = 4, b = –12, d) For 2x2 – 7x + 4 = 0, a = 2, b = –7, and c = 4. and c = 9.
( ) ( )( )( )
2
2
42
42
12 08
32
b b acxa
− ± −=
−
±=
±=
=
12 124
4 9−
( ) ( )( )( )
2
2
42
47 72
± −− 2 42
7 174
b b aca
− ± −
±=
=
x =
The root is 32
. The roots are 7 174
+ and 7 174
− .
MHR • Principles of Mathematics 10 Solutions 61
e) For 3x2 + 5x – 1 = 0, a = 3, b = 5, f) For 16x2 + 24x + 9 = 0, a = 16, b = 24, and and c = –1. c = 9.
( )( )( )
2 42
2
5 376
5
b b acxa
− ± −=
− ±
− ±=
2 45 3 13
−−
( )( )( )
2
224 2=
4 96116
42
42
24 032
34
b b aca−
− ± −=
− ±=
= −
x − ±=
The roots are 5 376
− + and 5 376
− − . The root is 34
− .
Chapter 6 Section 4 Question 2 Page 300 a) For 3x2 + 14x + 5 = 0, a = 3, b = 14, and c = 5.
( )( )( )
2
214 14 3 53
42
42
14 1366
14 2 346
7 343
b b acxa
− ± −=
− ± −=
− ±=
− ±=
− ±=
The exact roots are 7 343
− + and 7 343
− − . The approximate roots are –0.39 and –4.28.
62 MHR • Principles of Mathematics 10 Solutions
b) For 8x2 + 12x + 1 = 0, a = 8, b = 12, and c = 1.
( )( )( )
2
212 12 8 18
42
42
12 11216
12 4 716
3 74
b b acxa
− ± −=
− ± −=
− ±=
− ±=
− ±=
The exact roots are 3 74
− + and 3 74
− − . The approximate roots are –0.09 and –1.41.
c) For 4x2 – 7x – 1 = 0, a = 4, b = –7, and c = –1.
( ) ( )( )( )
2
2
4
7 4
2
42
7 58
6
b b acxa
−
− ± −
=
=
−
=
±
7 14
± −
The exact roots are 7 658
+ and 7 658
− . The approximate roots are 1.88 and –0.13.
d) For 10x2 – 45x – 7 = 0, a = 10, b = –45, and c = –7.
( ) ( )( )( )
2 42
2
45 2302
50
b b acxa
− ± −=
±
±=
=
245 45 4 10 710−− −
The exact roots are 45 230520
+ and 45 230520
− . The approximate roots are 4.65 and –0.15.
MHR • Principles of Mathematics 10 Solutions 63
e) For –5x2 + 16x – 2 = 0, a = –5, b = 16, and c = –2.
( )( )( )
2
216
42
42
16
16 5 25
−
−
−
1610
2
b b acxa
− ± −=
− ± −
− ±=
−
=
The exact roots are 16 21610
− +−
and 16 21610
− −−
. The approximate roots are 0.13 and 3.07.
f) For –6x2 + 17x + 5 = 0, a = –6, b = 17, and c = 5.
( )( )( )
2
217 17 6 56
−
−
42
42
17 40912
17 40912
b b acxa
− ± −=
− ± −=
− ±=
−±
=
The exact roots are 17 40912+ and 17 409
12− . The approximate roots are 3.10 and –0.27.
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.
( )( )( )
2
214 14 5 35
−
42
42
14 25610
14 1610
b b acxa
− ± −=
− ± −=
− ±=
− ±=
The x-intercepts are 3 and –0.2.
Find the vertex. For the x-coordinate, use2bxa
= − .
( )
( ) ( )
2
2
21.45 14 3
5 1412.
31.84 1.4
x
y x x
= −
=
= − −
= − −
= −
145
−
The vertex is (1.4, –12.8). The axis of symmetry is . 1.4x =
MHR • Principles of Mathematics 10 Solutions 65
b) For 2x2 – 5x – 12 = 0, a = 2, b = –5, and c = –12.
( ) ( )( )( )
2
2
42
42
5 12110
14 1
5 2 1
110
b b acxa
− −
− ± −=
=
±=
− ±=
5 22
± −
The x-intercepts are 4 and –1.5.
Find the vertex. For the x-coordinate, use2bxa
= − .
( )
( ) ( )
2
2
21.252 5 12
2 51
121.255.125
1.25
x
y x x
= −
=
= − −
= − −
= −
52
−
The vertex is (1.25, –15.125). The axis of symmetry is . 1.25x =
66 MHR • Principles of Mathematics 10 Solutions
c)
( )
2
2
10 25
0 5
y x x
x
= + +
= +
5 0 5
xx
+ == −
The x-intercept is –5. Find the vertex.
( ) ( ) +
2
2
5 52
510 25
10 250
5 5
x
y x x
− −
− −=
= −
= + +
= +
=
The vertex is ( )5,0− . The axis of symmetry is . 5x = −
MHR • Principles of Mathematics 10 Solutions 67
d)
( )
2
2
9 24 1
0 3 4
y x x
x
= − +
= −
6
3 4 0
4 3
x
x
− =
=
The x-intercept is 43
.
Find the vertex.
2
24 43
4 43 3
2439 24 16
9 243
0
x
y x x
+=
=
= − +
⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
16
The vertex is 4 ,03
⎛ ⎞⎜ ⎟ . ⎝ ⎠
The axis of symmetry is 43
x = .
68 MHR • Principles of Mathematics 10 Solutions
e) For x2 – 2x + 3 = 0, a = 1, b = –2, and c = 3.
( ) ( )( )( )
2
2
42
42
2
2 21
± −− 1
8
3
2
b b acxa
− ± −=
± −=
=
There are no x-intercepts.
Find the vertex. For the x-coordinate, use2bxa
= − .
( )
( )
2
2
21
2 32 1
21 3
21
−x
y x x
= −
=
= − +
= − +
=
The vertex is ( ) . 1,2 The axis of symmetry is 1x = .
MHR • Principles of Mathematics 10 Solutions 69
f) For –x2 – 3x – 3 = 0, a = –1, b = –3, and c = –3.
( ) ( )( )( )
2
2
42
42
3 32
b b acxa
− ± −=
−=
±=
−−
3 3 1 31
± −
−
− −
There are no x-intercepts.
Find the vertex. For the x-coordinate, use2bxa
= − .
( )
( ) ( )2
31
1.5 3 31.5− −
−−
− −
2
21.5
3 3
0.75
x
y x x
= −
= −
= − − −
= −
= −
The vertex is ( ) . 1.5, 0.75− − The axis of symmetry is . 1.5x = − 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) ( )2
2
3 2
6 1
y x
x x
= − +
= − + 1Let y = 0 and use the quadratic formula with a = 1, b = –6, and c = 11.
( ) ( )( )( )
2
2
4
61
± −6 1
2
4 112
6 82
b b acxa
−
− ± −=
=
± −=
There are no x-intercepts. b) ( )2
2
2 4
4
y x
x x
= − − +
= − +Factor and solve the corresponding quadratic equation.
( )
20 40 4
x xx x
= − +
= − −
The x-intercepts are 0 and 4. c) ( )2
2
2 4 5
2 16 2
y x
x x
= + −
= + + 7Let y = 0 and use the quadratic formula with a = 2, b = 16, and c = 27.
( ) ( )( )( )
2
16
42
42
16 404
b b acxa
− ± −=
− ±=
− ±=
216 2 272−
The x-intercepts are 16 404
− ± .
MHR • Principles of Mathematics 10 Solutions 71
d) ( )2
2
2 3 1
2 12 1
y x
x x
= − + −
= − − − 9Let y = 0 and use the quadratic formula with a = –2, b = –12, and c = –19.
( ) ( )( )( )
2
22
42
4
12 84
2
b b acxa
− ± −=
± −=
±=
−−
12 1 2 912
−
−
−−
There are no x-intercepts. e) ( )25y x= −Factor and solve the corresponding quadratic equation.
( )20 55 0
5
xx
x
= −
− ==
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.
( ) ( )( )( )
2
1
42
2
1 1.20.2
1 1.20.2
b b acda
− ± −=
− ±=
− ±=
−±
=
1
21 0.1 0.50.14−− −
−
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
2
2.6 0.1 0.50 0.1 2.
d dd d
= − + +
= − + −Use the quadratic formula with a = –0.1, b = 1, and c = –2.1.
( ) ( )( )( )
2
1
42
2
1 0.160.2
1 0.40.2
b b acda
− ± −=
− ±=
− ±=
−− ±
=−
21 0.1 2.10.14−− − −
−
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.
( ) ( )( )( )
2
2
42
42
8.4 8.4 4.9 1.54.9
−
−
8.4 99.960.2
8.4 99.969 8.
b b acta
− ± −=
− ± −=
− ±=
−±
=
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 2bta
= − .
( ) ( )
2
2
2( )0.86
4.9 8.4 1.5
4.9
8.44.9
0.86 0.86
−
8.4 1.55
t
h t t
= −
= − + +
= − + +
The ball reaches a height of 5 m. Chapter 6 Section 4 Question 8 Page 301 Answers will vary.
74 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 4 Question 9 Page 301 a) 2
2
7 127 12 9
x xx x
− =
− − =
90
Use the quadratic formula with a = 7, b = –12, and c = –9.
( ) ( )( )( )
2 42
2
12 39641
b b acxa
− ± −=
±
±=
=
212 1 4 97
2 7−− −
x 2.28 or x –0.56. b) 2
2
4 12 134 13 12 0
x xx x
= −
+ − =
Use the quadratic formula with a = 4, b = 13, and c = –12.
( ) ( )( )( )
2
213 13 4 124
−
42
42
13 3618
13 198
b b acxa
− ± −=
− ± −=
− ±=
− ±=
8
x = 0.75 or x = –4. c) 2
2
4 2.8 4.4 2.8 4.8 0
x xx x
= +
− − =Use the quadratic formula with a = 4, b = –2.8, and c = –4.8.
( ) ( )( )( )
2
22.
42
42
2.8 84.648
2
8 4 4.
.8 9.28
b b acxa
− −
− ± −=
± −=
±=
±=
2.8 84
x = 1.5 or x –0.8.
MHR • Principles of Mathematics 10 Solutions 75
d) ( )2
3 8
3 8 1 0
x x
x x
− = −
− + =
1
Use the quadratic formula with a = 3, b = –8, and c = 1.
( ) ( )( )( )
2
2
42
48 83
± −− 3 12
8 526
b b acxa
− ± −=
±=
=
)3
x 2.54 or x 0.13 e) ( ) (2
2
2
3 2
6 9 2 64 15 0
x x
x x xx x
− = − +
− + = − −
− + =Use the quadratic formula with a = 1, b = –4, and c = 15.
( ) ( )( )( )
2
2
42
4 1 1542
4 442
b b acxa
−
− ± −=
=
± −=
)5
41
± −
There are no real roots. f) ( ) ( )(2
2 2
2
3 2 5 2
6 9 4 253 6 34 0
x x x
x x xx x
+ = + −
+ + = −
− + + =Use the quadratic formula with a = –3, b = 6, and c = 34.
( ) ( )( )( )
2
6
42
42
6 4446
b b acxa
− ± −=
− ±
− ±=
=
−
26 3 343
− −
−
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.
( ) ( )( )( )
2
20 0.0044 21.30.0044−
−
0
42
42
0.374880.0088
b b acxa
− ± −=
± −=
±=
−
33
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) 2
2
6 4.9 8.10 4.9 8.1
t tt t
= − + +
= − + −Use the quadratic formula with a = –4.9, b = 8.1, and c = –3.
( ) ( )( )( )
2
28.1 8.1 4.9 34.9
− −
−
42
42
8.1 6.819.8
b b acta
− ± −=
− ± −=
±=
−−
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 1.092+ , or about 0.83 s.
MHR • Principles of Mathematics 10 Solutions 77
Chapter 6 Section 4 Question 12 Page 301 a)
( )
2
2
4.2 0.2 1.6 4.20 0.2 1.60 0.2 8
x xx xx x
= − +
= −
= −So, x = 0 or x = 8. The half-pipe is 8 m wide. b) 2
2
2.2 0.2 1.6 4.20 0.2 1.6 2
x xx x
= − +
= − +Use the quadratic formula with a = 0.2, b = –1.6, and c = 2.
( ) ( )( )( )
2
21.6 10.2
− .6
42
4 0.2
1.6 0
2 2
.960.4
b b acxa
− ± −=
± −
±=
=
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.
( ) ( )( )( )
2
2.96 0.0048
42
42
0.96 0.30720.0096
b b acva
− ± −=
± −=
± −=
0.96 0 640.0048
−
There are no v-intercepts.
b) For the v-coordinate of the vertex, use 2bva
= − .
0.960.0048)−
2(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
a) Compare 14 1404
x ±= with
2 42
b b acxa
− ± −= .
( ) ( )2
2
142 4
2
14 4 2 1407
2 14 7 0
baa
cc
x x
= −==
− − =
=
− + =
b) Compare 11 1456
x − ±= with
2 42
b b acxa
− ± −= .
( ) ( )2
2
112 6
3
11 4 3 1452
3 11 2 0
baa
cc
x x
===
− =
= −
+ − =
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 ( )1
2n n −
segments.
MHR • Principles of Mathematics 10 Solutions 79
d) ( )
2
11000
22 2000 0
n n
n n
−=
− − =
Use the quadratic formula with a = 1, b = –2, and c = –2000.
( ) ( )( )( )
2 42
2
2 80042
b b acna
− ± −=
±=
=
2 42 2 1 002 01
± −− −
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 2 6 9 0x x− + = . 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 Question 19 Page 302
2 2 16x y+ = 2 9y x= −
Substitute x2 – 9 for y in equation .
( )
2 2
22
4 2
2
16
16
17 0
9
65
x y
x x
xx
+ =
+ =
− + =
−
Use the quadratic formula with a = 1, b = –17, and c = 65.
( ) ( )( )( )
22
2
42
17 ± −
5. 18
42
17 2
17 1 651
92
b b acxa
−
− ± −=
=
±=
19.
xxy x
=±
= −
So, x 11.19 or x 5.81. 2
2
11.193.
11
359
92
..19
19
xxy x
=±
= −= −=
2
2
5.812.41
99
3
== −
−
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 Question 21 Page 302
2
11
1 0
xx
x x
= +
− − =
Use the quadratic formula with a = 1, b = –1, and c = –1.
( ) ( )( )( )
2 42
2
1 52
b b acxa
− ± −=
±=
=
2 411 1 11
± −−−
The roots are 1 52+ and 1 5
2− .
1 5 1 5 1 52 2 4
1
⎛ ⎞⎛ ⎞− +=⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟
⎝ ⎠⎝ ⎠= −
−
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: 15768
.
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 Question 24 Page 303 a – b + c = 0 a – c + d = 0 a + 2b + 2d = 0 b – d = 3 Add equations and to obtain 2 1a b d− + = . Rearrange equation : . 3b d− + = −Substitute into equation . 2 1a b d− + = 2 1
23−
2
aa==
0b d+ + =
Substitute a = 2 into equation . a + 2b + 2d = 0
2 2 1b d+ = −Add equations and . b – d = 3
1b d+ = − 2 2
1bb==
+
Substitute b = 1 into equation to determine that d = –2. Substitute a = 2 and b = 1 into equation . a – b + c = 0 2 − + 0
11 c
c== −
21
12
abcd
=== −= −
MHR • Principles of Mathematics 10 Solutions 83
Chapter 6 Section 5 Solve Problems Using Quadratic Equations Chapter 6 Section 5 Question 1 Page 311 a) 24.9 45 2h t t= − + + b) Let h = 0 and use the quadratic formula with a = –4.9, b = 45, and c = 2.
( ) ( )( )( )
2
245 45 4.9 24.9
−
−
42
42
45 2064.29.8
b b acta
− ± −=
− ± −=
− ±=
−
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 Page 311
a) For the t-coordinate of the vertex, use 2bta
= − .
2494.−
5 5
9( )5
t = −
=.5
124h = − + +
=
( ) ( )24.9 49 1
The maximum height of the firework is 124 m above the ground. b) 2
2
100 4.9 49 1.50 4.9 49 98.5
t tt t
= − + +
= − + −Use the quadratic formula with a = –4.9, b = 49, and c = –98.5.
( ) ( )( )( )
2
249
42
42
49 470.49.8
b b acta
− ± −=
− ± −=
=±−
−
49 4.9 98.54.9−
− −
So, t 2.79 or t 7.21. The firework is more than 100 m above the ground over the time interval . 2.79 7.21t≤ ≤
84 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 5 Question 3 Page 312
( )2
16 35
16 35 0
x x
x x
+ =
+ − =
Use the quadratic formula with a = 1, b = 16, and c = –35.
( ) ( ) )(( )
2
216
42
42
16 3962
b b acxa
− ± −=
− ± −=
− ±=
16 1 351
−
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
( )2
1 3306
3306 0
x x
x x
+ =
+ − =
Use the quadratic formula with a = 1, b = 1, and c = –3306.
( ) ( )( )( )
2
1 1
42
42
1 13 2252
1 1152
b b acxa
− ± −=
− ±=
− ±=
− ±=
2 1 33061
− −
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
( )2
2 323
2 323 0
x x
x x
+ =
+ − =
Use the quadratic formula with a = 1, b = 2, and c = –323.
( ) ( )( )( )
2
2 31 231
−−
42
42
2 1296
2 2
22 36
2
b b acxa
− ± −=
− ±=
− ±=
− ±=
So, x = 17 or x = –19. The numbers are 17 and 19, or –17 and –19. Chapter 6 Section 5 Question 6 Page 312
( ) ( )
( )( )
2 22
2 2 2
2
2
2 3 7
4 12 9 14 492 2 40 0
20 05 4 0
x x x
x x x x xx x
x xx x
+ = + +
+ + = + + +
− − =
− − =
− + =
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) 20.05 0.95 0.5y x x= − + +
86 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 5 Question 8 Page 312
2
2
2
( )600 1250
4.0
V r hr
r
r
π
π
π
=
=
=
=
The radius of the cylinder is 40 mm. Chapter 6 Section 5 Question 9 Page 312
( )
( )2
1 4 202
4 40
4 40 0
b b
b b
b b
+ =
+ =
+ − =
Use the quadratic formula with a = 1, b = 4, and c = –40.
( ) ( )( )( )
2
4 4
42
42
4 1762
b b acxa
− ± −=
− ±
− ±=
=
2 1 401− −
So, x 4.6 or x –8.6. The length of the base is 46 mm. Chapter 6 Section 5 Question 10 Page 312
( )22
2 2
2
2
1 365
2 1 3652 2 364 0
182 0
x x
x x xx x
x x
+ + =
+ + + =
+ − =
+ − =
Use the quadratic formula with a = 1, b = 1, and c = –182.
( ) ( )( )( )
2
1 1
42
42
1 7292
1 272
b b acxa
− ± −=
− ±=
− ±=
− ±=
2 1 1821
− −
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 2 2x y+ = 3
xyx
Area: 33xy =Solve for y in the perimeter equation and substitute into the area equation.
( )2
3333
11.5 33 0
11.5
x x
=
=
− + − =
x−
Use the quadratic formula with a = –1, b = 11.5, and c = –33.
( ) ( )( )( )
2
211.5 11.5 1 3
42
42
11.5 0.252
1
3
1.5 0.52
1−
b b acxa
− ± −=
− ± −=
− ±=
−− ±
=
−
−
−
So, x = 5.5 or x = 6.
One dimension is 6 cm. The other dimension is 336
, or 5.5 cm.
Chapter 6 Section 5 Question 12 Page 312
( )2
2 12001200 2180 000
1200 2 180 000
2 1200 180 000 0
y xy x
xyx x
x x
+ == −=
− =
− + − =
Use the quadratic formula with a = –2, b = 1200, and c = –180 000.
( ) ( ) )(( )
2
21200 12
42
42
1200 04
b b acxa
− ± −=
− ± −
−=
±
=
−
00 2 180 0002
−
−
−
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 Question 13 Page 313
( ) ( )
( )( )
2 22
2 2 2
2
2 4
4 4 8 14 12 0
6 4 0
x x x
x x x x xx x
x x
+ + = +
+ + + = + +
− − =
− + =
6
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
( )22 2
2
2
10 6
101 3636
1010.597
x x
x
x
x
+ =
=
=
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) 24.9 36.85 0.61h t t= − + + 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.
( ) ( )( )( )
2
236.85 36.85 4
42
42
.9 0.614.9
36.85 1369.889.8
b b acta
− ± −=
− ± −=
− ±=
−
−
−
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 2bta
= − .
2( )3
36. 584.9−
3.76 63.7
. 67
t = −
4.9 36.85 0.61
4.9 36.85 0.6169.89
h t t= − + +
= − + +
c)
( ) ( )
2
2
The maximum height reached was 69.89 m the time is 3.76 s.
90 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Section 5 Question 16 Page 313 a) 24.9 15 1h t t= − + + b) 2
( ) ( )2 15 11 1+ +
4.9 15 1
4.911.1
h t t= − + +
= −
= 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.
( ) ( )( )( )
2
215 15 4.9 14.9
−
−
42
42
15 2448
.69.
b b acta
− ± −=
− ± −=
− ±−
=
So, t –0.1 or t 3.1. The ball lands after 3.1 s.
d) For the t-coordinate of the vertex, use 2bta
= − .
154.9)−
1.5 1.5
2(1.5
t = −
15 11 52.
h t t= − + +
= − + +
( ) ( )
2
2
4.9 15 1
4.9
The ball reached a maximum height of 12.5 m after 1.5 s.
MHR • Principles of Mathematics 10 Solutions 91
e)
( ) ( )2 15 11 1+ +
20.81 15 1
0.8115.2
h t t= − + +
= −
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.
( ) ( )( )( )
2
215 15 0.81 10.81
−
−
42
42
15 228.241.62
b b acta
− ± −=
− ± −=
− ±=
−
So, t –0.1 or t 3.1. The ball lands after 18.6 s.
For the t-coordinate of the vertex, use 2bta
= − .
150.81)−
9.3
2(9.3
t = −
1.4
h t t= − + +
= − + +
( ) ( )
2
2
0.81 15 1
0.8 9.31 1570
The ball reached a maximum height of 70.4 m after 9.3 s.
92 MHR • Principles of Mathematics 10 Solutions
f) h t
( ) ( )1 1
2
2
11.55 15 1
11.55 15 14.5
t= − + +
= − + +
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.
( ) ( )( )( )
2
215 15 11.55 111.55
−
−
42
42
15 271.223.1
b b acta
− ± −=
− ± −=
− ±=
−
So, t –0.1 or t 1.4. The ball lands after 1.4 s.
For the t-coordinate of the vertex, use 2bta
= − .
1511.55)−
0.65 0. 56
2(0.65
t = −
15 15.9
h t t= − + +
= − + +
=
( ) ( )
2
2
11.55 15 1
11.55
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) ( )( )10 0.5 30 2R x= − + x , where x represents the number of price reductions. b) ( )( )
( )( )
2
2
10 0.5 30 2
300 50
150150 5
0 51 10
R x x
x xx x
x 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) ( )( )
( )( )
2
10 0.5 30 2
0 300 50 20 15
R x x
x xx x
= − +
= + −
= − +
So, x = 20 or x = –15.
The x-intercepts are –15 and 20. The x-coordinate of the vertex is 15 202
− + , or 2.5.
The maximum revenue occurs after 2.5 price reductions. The price is ( )10 , or $8.75. 2.5 0.5− 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 ( )( )
( )( )
2
2
2
30 2 20 2 1064
600 100 4 10644 100 464 0
25 116 04 29 0
x x
x xx x
x xx x
+ + =
+ + =
+ − =
+ − =
− + =
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 ( )( ) ( )(
2
2
2
24 2 15 2 1.5 15 24
360 78 4 5404 78 180 02 39 90 0
x x
x xx xx x
+ + =
+ + =
+ − =
+ − =
)
Use the quadratic formula with a = 2, b = 39, and c = –90.
( ) ( )( )( )
2
239 3
42
42
39 22414
b b acxa
− ± −=
− ± −=
− ±=
9 2 902
−
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
( )2
3 2 1496
3 2 1496 0
w w
w w
+ =
+ − =
Use the quadratic formula with a = 3, b = 2, and c = –1496.
( ) )( )(( )
2
2 2
42
42
2 17 9566
2 1346
b b acwa
− ± −=
− ±=
− ±=
− ±=
2 3 14963
− −
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) ( )( )
2
2
40 2 50 2 875
2000 180 4 8754 180 1125 0
x x
x xx x
− − =
− + =
− + =
Use the quadratic formula with a = 4, b = –180, and c = 1125.
( ) ( )( )( )
2
2180 1804
− 4 1125
42
42
180 14 4008
180 1208
b b acxa
− ± −=
± −=
±=
±=
= ×
=
=
So, x = 37.5 or x = 7.5. The side length of the squares being removed is 7.5 cm. c) V x
( )875875 7.56562.5
The volume of the box is 6562.5 cm2.
MHR • Principles of Mathematics 10 Solutions 97
Chapter 6 Section 5 Question 22 Page 314 ( )( )
2
2
21 2 15 2 216
315 72 4 2164 72 99 0
x x
x xx x
− − =
− + =
− + =
Use the quadratic formula with a = 4, b = –72, and c = 99.
( ) ( )( )( )
2
2
42
472 724
− 4 992
72 36008
72 608
b b acxa
− ± −=
± −=
±=
±=
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 ( )( ) ( )( )
( )( )
2
2
20 16 0.6 20 16
320 36 19236 128 0
4 32 0
x x
x xx xx x
− − =
− + =
− + =
− − =
So, x = 4 or x = 32. The width of the cut is 4 cm. Chapter 6 Section 5 Question 24 Page 314
( )( ) ( )( )
( )( )
2
2
120 15 20 152
300 35 15035 150 0
5 30 0
x x
x xx xx x
− − =
− + =
− + =
− − =
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.
( )
2
2
2
23
0 300 2323
90 00023 23
90 000
y ax
a
a
y x
= −
= −
=
= −
Chapter 6 Section 5 Question 26 Page 314 a) b) 20.008 0.383 8.726y x x= − + 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) 2 22 7 4
6 71x x x x
x+ + = − −
= −
There is one point of intersection because the resulting equation is linear. b) 2 2
2
3 12 16 2 45 8 13 0
3x x x xx x− + = − − +
− + =
Use the quadratic formula.
( ) ( )( )( )
2
28 85
± −−
42
42
8 1961
5
0
13
b b acxa
− ± −=
± −=
=
There are no points of intersection because the resulting quadratic equation has no real roots. c) 2 2
2
6 10 5 30 464 24 36 0
x x x xx x
− + = − +
− + − =Use the quadratic formula.
( ) ( )( )( )
2
224
42
42
24 08
b b acxa
− ± −=
− ± −=
− ±=
−
+
24 4 364
−
−
−
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) , where WC represents the wind chill temperature and w represents the wind speed.
20.0032 0.425 6WC w w= −
b) The QuadReg operation results in a linear relation, since the coefficient of the x2-term is 0.
13WC t= − , 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 Page 315 Solve two equations with two unknowns. Using the maximum height reached, the equation will be of the form ( )2 14y a x h= − + . Use the two points (0, 4) and (24, 0).
( )( )
2
2
2
14
1
40
10
0
0 0
a h
a h
ah
= − +
− = −
− =
a h
a h
( )( )
2
2
14
14 4
24
2
= − +
− = −
2
10 ah−
= ( )214 576 48a h− = − + h
Substitute the expression for a from equation into equation .
( )
( )
2
22
2 2
2
2
14 576 48
1014 576 48
14 5760 480 104 480 5760 0
120 1440 0
a h h
h hh
h hh hh h
− = − +
−⎛ ⎞− = − +⎜ ⎟⎝ ⎠
− = − + −
− − + =
+ − =
h
Use the quadratic formula.
( ) ( )( )( )
2
120
42
2
120 201602
b b acha
− ± −=
− ±
− ±=
=
2120 4 1 14401−− −
So, h –131 or h 11. Substitute h = 11 into equation .
21110
10121
a
a
−=
= −
MHR • Principles of Mathematics 10 Solutions 101
Now , use the equation ( )210 11 14121
y x= − − + with h = 10.
( )2
2
2
10 11 14121
1210 10 220 1210 169410 220 72 0
01
6
x
x xx
= − − +
= − + − +
− + =
xUse the quadratic formula.
( ) ( )( )( )
2
220
42
4 102
220 1
726
9 36020
b b acxa
− ± −=
± −
±=
=
220 210
−
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.
1
2
3
4
5
11235
fffff
===
==
102 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Review Chapter 6 Review Question 1 Page 316 a)
( )( )( )( )
2
2
2 2 2
2 2 2
2
8 7
8 7
8 4 4
8 4 4
4 23
y x x
x x
x x
x x
x
= + −
= + −
= + + − −
= + + − −
= + −
7
7
MHR • Principles of Mathematics 10 Solutions 103
104 MHR • Principles of Mathematics 10 Solutions
7
7
b)
( )( )( )( )
2
2
2 2 2
2 2 2
2
2 7
2 7
2 1 1
2 1 1
1 6
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + +
104 MHR • Principles of Mathematics 10 Solutions
MHR • Principles of Mathematics 10 Solutions 105
6
6
c)
( )( )( )( )
2
2
2 2 2
2 2 2
2
4 6
4 6
4 2 2
4 2 2
2 2
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + +
MHR • Principles of Mathematics 10 Solutions 105
106 MHR • Principles of Mathematics 10 Solutions
3
3
d)
( )( )( )( )
2
2
2 2 2
2 2 2
2
6 3
6 3
6 3 3
6 3 3
3 12
y x x
x x
x x
x x
x
= + −
= + −
= + + − −
= + + − −
= + −
106 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Review Question 2 Page 316 a)
( )( )( )( )
2
2
2 2 2
2 2 2
2
12 30
12 30
12 6 6 30
12 6 6 30
6 6
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + − The vertex is (–6, –6). b)
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
14 50
14 50
14 7 7 50
14 7 7 50
7 1
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − − +
= − + The vertex is (7, 1). c)
( )( ) ( )( )( )( ) ( )( )
( )
2
2
2 22
2 22
2
6 7
6 7
6 3 3 7
6 3 1 3
3 2
y x x
x x
x x
x x
x
= − + −
= − − −
= − − + − − − −
= − − + − − − − −
= − − +
7
The vertex is (3, 2).
MHR • Principles of Mathematics 10 Solutions 107
d)
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
5 40 76
5 8 76
5 8 4 4 76
5 8 4 5 4 7
5 4 4
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − − +
= − −
6
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)
( )( )
2 10 21 03 7 0
3 0 or 7 03 or 7
x xx x
x xx x
+ + =
+ + =
+ = + == − = −
Check by substituting both solutions in the original equation. For x = –3: For x = –7:
( ) ( )
2
2
10 21
13 130 20
x x= + +
= + +−−
=
L.S. 0 =R.S.
( ) ( )
2
2
10 21
17 7 10 20
x x= + +
= + +
=
L.S. 0
−−
=R.S.
L.S. = R.S. L.S. = R.S. The roots are –3 and –7.
108 MHR • Principles of Mathematics 10 Solutions
b)
( )( )
2 8 20 02 10 0
2 0 or 10 02 or 10
m mm m
m mm m
+ − =
− + =
− = + == = −
0 8 20
8 200
m m= −
+
=
+
Check by substituting both solutions in the original equation. For m = 2: For m = –10:
( ) ( )
2
2
8 20
2 8 20
2 0
m m= + −
= + −
=
L.S. =R.S.
( ) ( )
2
210 10− −= −
L.S. 0 =R.S.
L.S. = R.S. L.S. = R.S. The roots are 2 and –10. c)
( ) ( )( ) ( )( )( )
2
2
2
2
6 21 9 0 Divide both sides by 3.2 7 3 0
2 6 3 0
2 6 3 0
2 3 1 3 0
3 2 1 03 0 or 2 1 0
13 or 2
y yy y
y y y
y y y
y y y
y yy y
y y
+ + =
+ + =
+ + + =
+ + + =
+ + + =
+ + =
+ = + =
= − = −
Check by substituting both solutions in the original equation.
For y = –3: For y = 12
− :
( ) ( )2 1 93 3− +−+
26 21 9
6 20
y y= + +
=
=
L.S. 0 =R.S. 2
2
6 21 9
6 21 9
0
1 12 2
y y= + +
⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
L.S. 0=R.S.
− −
L.S. = R.S. L.S. = R.S.
The roots are –3 and – 12
.
MHR • Principles of Mathematics 10 Solutions 109
d)
( ) ( )( ) ( )( )( )
2
2
2
5 13 65 15 2 6
5 15 2 6
5 3 2 3
3 5 2 0
n nn n n
n n n
n n n
n n
+ − =
+ − − =
+ + − − =
+ − + =
+ − =
00
0
0
3 0 or 5 2 0
23 or 5
n n
n n
+ = − =
= − =
Check by substituting both solutions in the original equation.
For n = –3: For n = 25
:
( ) ( )2 3 63 3− −−+
25 13 6
5 10
n n= + −
=
=
L.S. 0 =R.S. 2
2
5 13 6
2 25
5 13
05
n n= + −
⎛ ⎞ ⎛ ⎞ 6= + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
L.S. 0=R.S.
L.S. = R.S. L.S. = R.S.
The roots are –3 and 25
.
110 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Review Question 5 Page 316 a)
( )( )
2
2
8 98 9 0
1 9 0
y yy y
y y
= +
− − =
+ − = 1 0 or 9 0
1 or 9y y
y y+ = − =
= − = b)
( )( )
2
2
8 78 7 0
1 7 0
x xx x
x x
− = −
− + =
− − = 1 0 or 7 0
1 or 7x x
x x− = − =
= = c)
( ) ( )( ) ( )( )( )
2
2
2
2
3 103 10 7 0
3 3 7 7 0
3 3 7 7 0
3 1 7 1 0
1 3 7 0
m mm m
m m m
m m m
m m m
m m
= − −
+ + =
+ + + =
+ + + =
+ + + =
+ + =
7
1 0 or 3 7 0
71 or 3
m m
m m
+ = + =
= − = −
d)
( )
2
2
2
30 25 925 30 9 0
5 3
x xx x
x
− =
− + =
− = 0 5 3 0
3 5
x
x
− =
=
MHR • Principles of Mathematics 10 Solutions 111
e)
( ) ( )( ) ( )
( )( )
2
2
2
2
8 5 18 14 5 0
8 4 10 5 0
8 4 10 5 0
4 2 1 5 2 1 0
2 1 4 5 0
k kk k
k k k
k k k
k k k
k k
= − +
− + =
− − + =
− + − + =
− − − =
− − =
4
2 1 0 or 4 5 0
1 5 or 2 4
k k
k k
− = − =
= =
f)
( ) ( )( ) ( )( )( )
2
2
2
2
3 2 53 5 2 0
3 3 2 2 0
3 3 2 2 0
3 1 2 1 0
1 3 2 0
x xx x
x x x
x x x
x x x
x x
+ = −
− + =
+ + + =
+ + + =
+ + + =
+ + =
1 0 or 3 2 0
21 or 3
x x
x x
+ = + =
= − = −
Chapter 6 Review Question 6 Page 316
( ) ( )
( )
2 22
2 2 2
2
2
3 1 3 1
9 6 1 9 612 012 012 0
x x x
1x x x x xx xx x
x x
+ = + −
+ + = + − +
− + =
− =
− =
0 or 12 0 or 12x x
x= −
==
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.
( )(
2 8 120 2y x x
x x= + +
= + + )6
The x-intercepts are –2 and –6. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
24 + +− 4
2 62
48 12
8 124
x
y x x
− −=
= −
= + +
−=
= −
The vertex is ( )4, 4− − . b) Factor and solve the corresponding quadratic equation.
( )(
2 4 50 1y x x
x x= − −
= + − )5
The x-intercepts are –1 and 5. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
2
1 52
24
52 2
5
49
x
y x x
− +=
=
= − −
= −
−
= − The vertex is ( )2, 9− .
MHR • Principles of Mathematics 10 Solutions 113
c) Factor and solve the corresponding quadratic equation.
( )( )(
2
2
6 27
0 6
0 9
y x x
x x
x x
= − − +
= − + −
= − + − )27
3
The x-intercepts are –9 and 3. Find the x-coordinate of the vertex, and then, the y- coordinate.
( ) ( )
2
23 − +− 3
9 32
36 27
6 2736
x
y x x
− +=
= −
= − − +
−= −
=
The vertex is ( )3,36− .
114 MHR • Principles of Mathematics 10 Solutions
d) Factor and solve the corresponding quadratic equation.
( ) ( )( ) ( )
( )( )
2
2
2
3 10 80 3 6 4 8
0 3 6 4 8
0 3 2 4 2
0 2 3 4
y x xx x x
x x x
x x x
x x
= + +
= + + +
= + + +
= + + +
= + +
2 0 or 3 4 042 or 3
x x
x x
+ = + =
= − = −
The x-intercepts are –2 and – 43
.
Find the x-coordinate of the vertex, and then, the y-coordinate.
2
25 10 853 3⎞ ⎛ ⎞+ +⎟ ⎜ ⎟
⎝ ⎠− −
423
253
3 10 8
3
13
x
y x x
− −=
= −
= + +
⎛= ⎜⎝ ⎠
= −
The vertex is 5 1,3 3
⎛ ⎞− −⎜ ⎟⎝ ⎠
.
MHR • Principles of Mathematics 10 Solutions 115
e) Factor and solve the corresponding quadratic equation.
( )
2 30 3y x x
x x= − −
= − +
The x-intercepts are 0 and –3. Find the x-coordinate of the vertex, and then, the y-coordinate.
2
2
0 3232
3 12
3
9
32
4
32
x
y x x
−
−=
= −
= − − +
⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
−
The vertex is 3 9,2 4
⎛ ⎞−⎜ ⎟⎝ ⎠
.
f) Factor and solve the corresponding quadratic equation.
( )(
2 40 2y x
x x= −
= + − )2
The x-intercepts are –2 and 2. Find the x-coordinate of the vertex, and then, the y-coordinate.
( )
2
2
2 22
440
04
x
y x
− +=
=
= −
= −
= −
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; or 2 20 100y x x= + + 2 20 100y x x= − + b) k = –49; 24 28 4y x x= − + − 9 Chapter 6 Review Question 10 Page 316 a) For –3x2 – 2x + 5 = 0, a = –3, b = –2, and c = 5.
( ) ( )( )( )
2
2
42
42
2 646
5
2 86
b b acxa
− ± −=
=
±−
=
±=
−
2 2 33
± −−
−
−
The roots are 1 and 53
− .
b) For 9x2 – 8x – 3 = 0, a = 9, b = –8, and c = –3.
( ) ( )( )( )
2
2
42
42
8
8 8 39
± −− −
17218
8 2 4318
9
9
34 4
b b acxa
− ± −=
=
=
=
±=
±
±
The roots are 4 439
+ and 4 439
− .
MHR • Principles of Mathematics 10 Solutions 117
c) For 5x2 + 7x + 1 = 0, a = 5, b = 7, and c = 1.
( ) )( )(( )
2 42
42
7 2910
7 7
b b acxa
− ± −=
− ±=
− ±=
2 5 15−
The roots are 7 2910
− + and 7 2910
− − .
d) For 25x2 + 90x + 81 = 0, a = 25, b = 90, and c = 81.
( ) ( )( )( )
2
2
42
42
90 0
90 90 25 15
82
0595
b b acxa
− ± −=
− ± −=
− ±=
= −
The root is 95
− .
Chapter 6 Review Question 11 Page 317
2
2
2
0.0034 0.004 0.30.0034 0.004 0.3
0 0.0034 0.004 125.3125
d s ss ss s
= + −
= + −
= + −
Use the quadratic formula with a = 0.0034, b = 0.004, and c = –125.3.
( ) ( )( )( )
2
20.004 0.0 0.0
42
42
0.004 1.7040.0068
b b acsa
− ± −=
− ± −=
− ±=
04 034 125.30.0034
−
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) 24.9 4 3h t t= − + + b) 2
2
4.9 4 30 4.9 4h t t
t t= − + +
= − + + 3Use the quadratic formula with a = –4.9, b = 4, and c = 3.
( ) )( )(( )
2 42
2
4 74.89.8
4
b b acta
− ± −=
− ±=
− ±=
−
2 44 4.9 34.9
−
−
−
So, t –0.47 or t 1.29. The diver enters the water after 1.29 s. c) 2
2
2
4.9 4 33.5 4.9 4 3
0 4.9 4 0.5
h t tt tt t
= − + +
= − + +
= − + −Use the quadratic formula with a = –4.9, b = 4, and c = –0.5.
( ) )( )(( )
2
4 4
42
2
4 6.29.8
b b acta
− ± −=
− ±=
− ±=
−
2 44 .94.9
0.5− −
−
−
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 0.66t≤ ≤
MHR • Principles of Mathematics 10 Solutions 119
Chapter 6 Review Question 13 Page 317 ( )( ) ( )(
2
2
5 2 7 2 2 5 7
35 24 4 704 24 35 0
x x
x xx x
+ + =
+ + =
+ − =
)
Use the quadratic formula with a = 4, b = 24, and c = –35.
( ) ( )( )( )
2
224
42
42
24 11368
1.2
b b acxa
− ± −=
− ± −
− ±=
=
=
24 4 354
−
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) ( )( )6 4 0.25R x= + − x
)
, where R is the revenue in dollars, and x is the number of price reductions. b) ( )(
2
2
6 4 0.25
30 24 2.5 0.250 6 2.5 0.25
R x x
x xx x
= + −
= + −
= − + −
Use the quadratic formula with a = –0.25, b = 2.5, and c = –6.
( ) ( )( )( )
2
2
42
42
2.5
2.5 0.250.5
2.5 0.5.0 5
b b acxa
− ± −=
− ± −=
− ±=
−− ±
=−
2.5 0.25 60.25
− −
−
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 2bxa
= − .
22.50.2− 5( )
5
x = −
=
)
( )( )( )(5
6 4 0.25
6 4 0.2530.
525
R x x= + −
= + − ×
=
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)
( )( )( )( )
2
2
2 2 2
2 2 2
2
6 4
6 4
6 3 3 4
6 3 3 4
3 5
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + − The vertex is (–3, –5). b)
( )( ) ( )( )( )( ) ( )( )
( )
2
2
2 22
2 22
2
8 3
8 3
8 4 4 3
8 16 4 1 4 3
4 13
y x x
x x
x x
x x
x
= − + −
= − − −
= − − + − − − −
= − − + − − − − −
= − − + The vertex is (4, 13). c)
( )( )( ) (( )
2
2
2 2 2
2 2 2
2
3 24 10
3 8 10
3 8 4 4 10
3 8 4 3 4 1
3 4 38
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + −
) 0
The vertex is (–4, –38).
122 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Chapter Test Question 2 Page 318 a)
( )( )
2 5 4 01 4 0
1 0 or 4 01 or 4
x xx x
x xx x
− + =
− − =
− = − == =
b)
( )( )
29 1 03 1 3 1 0
3 1 0 or 3 1 01 1 or 3 3
yy y
y y
y y
− =
+ − =
+ = − =
= − =
c)
( )( )
2
2
3 103 10 0
5 2 05 0 or 2 0
5 or 2
x xx x
x xx x
x x
= +
− − =
− + =
− = + == = −
d)
( )
2
2
9 12 4 0
3 2 03 2 0
2 3
b b
bb
b
− + =
− =
− =
=
e)
( ) ( )( ) ( )( )( )
2
2
2
2
3 13 103 13 10 0
3 15 2 10 0
3 15 2 10 0
3 5 2 5 0
5 3 2 05 0 or 3 2 0
25 or 3
x xx x
x x x
x x x
x x x
x xx x
x x
+ =
+ − =
+ − − =
+ + − − =
+ − + =
+ − =
+ = − =
= − =
MHR • Principles of Mathematics 10 Solutions 123
f)
( )
26 30 06 5 0
5 0 or 6 05 or 0
m mm m
m mm m
+ =
+ =
+ = == − =
g)
( )
2
2
5 105 10 05 2 0
2 0 or 5 02 or 0
x xx xx x
x xx x
=
− =
− =
− = == =
h)
( )
2
2
2
4 1 44 4 1 0
2 1 02 1 0
1 2
d dd d
dd
d
+ = −
+ + =
+ =
+ =
= −
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)
( )(
2 2 350 7y x x
x x= + −
= + − )5The x-intercepts are –7 and 5. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
21 12 3−+−
7 52
12 35
536
x
y x x
− +=
= −
= + −
−
= −
=
)0
4
The axis of symmetry is x = –1. The vertex is (–1, –36). b)
( )( )(
2
2
20
0 2
0 5
y x x
x x
x x
= − − +
= − + −
= − + −The x-intercepts are –5 and 4. Find the x-coordinate of the vertex, and then, the y-coordinate.
( )0.5 50.− −( )
2
2
5 42
0.520
202 .250
x
y x x
− +=
= −
= − − +
= − − +
=
The axis of symmetry is x = –0.5. The vertex is (–0.5, 20.25).
MHR • Principles of Mathematics 10 Solutions 125
c)
( )
23 60 3 2y x x
x x= − −
= − +The x-intercepts are 0 and –2. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )21 6 1−− −
2
2 02
13 6
33
x
y x x
− +=
= −
= − −
= −
=
The axis of symmetry is x = –1. The vertex is (–1, 3). d)
( )
2
2
10 25
0 5
y x x
x
= − +
= −The x-intercept is 5. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
25 5
5 52
510 25
10 250
x
y x x
+=
=
= − +
= − +
=
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.
( ) ( )( )( )
2 42
2
11 1698
11 138
b b acxa
− ± −=
±=
±=
=
2 411 11 4 34
± −− −
The roots are 3 and 14
− .
b) 2
2
5 75 7 0
x xx x
+ =
+ − =For x2 + 5x – 7 = 0, a = 1, b = 5, and c = –7.
( ) )( )(( )
2 42
2
5 532
5
b b acxa
− ± −=
− ±
− ±=
=
2 45 1 71
−−
The roots are 5 532
− + and 5 532
− − .
c) 2
2
9 30 259 30 25
x xx x
= −
− + = 0For 9x2 – 30x + 25 = 0, a = 9, b = –30, and c = 25.
( ) ( )( )( )
2
230 39
− 0
42
4 9 22
30 018
5
5
3
b b acxa
− ± −=
± −=
±
=
=
The root is 53
.
MHR • Principles of Mathematics 10 Solutions 127
d) For 7k2 – 9k + 3 = 0, a = 7, b = –9, and c = 3.
( ) ( )( )( )
2
2
4
97
± −9 7
2
42
914
3
3
b b acka
−
− ± −=
=
± −=
3
There are no real roots. e) 2
2
4 94 9 3 0
s ss s
− = −
− + =For 4s2 – 9s + 3 = 0, a = 4, b = –9, and c = 3.
( ) ( )( )( )
2
2
42
49 94
± −− 4 32
9 338
b b acsa
− ± −=
±=
=
The roots are 9 338
+ 3 and 9 338
− .
f) 2
2
3 73 7
t tt t
− =
− − = 0For 3t2 – t – 7 = 0, a = 3, b = –1, and c = –7.
( ) ( )( )( )
2
2
4
1 3
2
42
1 56
8
b b acta
−
− ± −
=
=
−
=
±
1 73
± −
The roots are 1 856
+ and 1 856
− .
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.
( ) ( )( )( )
2
212 12 3 63
42
42
12 726
12 3 86
4 82
b b acxa
− ± −=
− ± −=
− ±=
− ±=
− ±=
The roots are 4 82
− + and 4 82
− − .
b) Use the quadratic formula with a = 1, b = –8, and c = 3.
( ) ( )( )( )
2
2
42
48 81
± −− 1 32
8 522
b b acxa
− ± −=
±=
=
The roots are 8 522
+ and 8 522
− .
c) Use the quadratic formula with a = 4, b = 0, and c = –10.
( ) ( )( )( )
2
20
42
42
1608
4
b b acma
− ± −=
−
±=
=
0 104
± −
The roots are 1608
+ and 1608
− .
MHR • Principles of Mathematics 10 Solutions 129
d)
( )( )
2
2
2
2
5 10 55 10 5 0
5 2 1
5 1 01 0
1
x xx x
x x
xx
x
− + =
− + − =
− − + =
− − =
− ==
0
6
The root is 1. e) ( )
( )( )
2
2
2
5 1
10 25 1610 9 0
1 9 0
k
k kk kk k
− =
− + =
− + =
− − = The roots are 1 and 9.
f)
( )
2
2
2
1 02 2
2 1
1 0
x x
x x
x
+ + =
+ + =
+ =
0
)1
The root is –1. g) ( ) ( )(
( )
2
2 2
2
2 1 2
2 4 2 3 27 07 0
m m m
m m m mm m
m m
− = + +
− + = + +
− =
− =The roots are 0 and 7. h) ( )( ) 2
2 2
2
5 2 3 1 4 5
15 2 4 511 7 0
x x x
x x xx x
+ − = +
+ − = +
+ − =Use the quadratic formula with a = 11, b = 1, and c = –7.
( ) ( )( )( )
2 42
42
1 309
1
22
b b acxa
− ± −=
− ±
− ±=
=
2 11 1 711− −
The roots are 1 30922
− + and 1 30922
− − .
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.
( )20
5−
2 2
2
2
2
bda
= −
= −
=
2 121
0
h d d= − + +
= − + +
=
0
( ) ( )
2
2
5 20 1
5
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 x-axis. 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) ( )( )
2
5 3
2 15 0
x x
x x
− + =
− − =
b)
( )( )2
1 3 02 5
2 1 5 3 0
10 11 3 0
x x
x x
x x
⎛ ⎞⎛ ⎞− − =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
− − =
− + =
MHR • Principles of Mathematics 10 Solutions 131
Chapter 6 Chapter Test Question 11 Page 318 a)
( )
2 40 4h d d
d d= − +
= − − The d-intercepts are 0 and 4. The shed is 4 m wide. The d-coordinate of the vertex is 2.
( ) ( )22 2+
2 4
44
h d d= − +
= −
=
4
The height of the shed is 4 m. b) c) The relation is valid for values 0 d≤ ≤ . h must be positive. Chapter 6 Chapter Test Question 12 Page 318 Find t-coordinate of the vertex. Then, find the C-coordinate.
( )
2
216
963
−
bta
= −
= −
=
( ) ( )
2
21
3 96 1014
3 96 10142 6
6 64
1
C t t= − +
= − +
=
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
( )( )2
2
2
1 2 1 6 3 2402
12 3 48012 483
48312
6.34
x x
xx
x
x
+ − =
− =
=
=
For an area of 240 m2, the value of x is 6.34 m. Chapter 6 Chapter Test Question 14 Page 319
2
2
2
0.0052 0.1320 0.0052 0.130 0.0052 0.13 20
d s ss ss s
= +
= +
= + −
Use the quadratic formula with a = 0.0052, b = 0.13, and c = –20.
( ) ( )( )( )
2
20.13 0 0
42
42
0.13 0.43290.0104
b b acsa
− ± −=
− ± −=
− ±=
.13 .0052 200.0052
−
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.
( ) )( )(( )
2 42
42
3 252
3 52
3 3
b b acda
− ± −=
− ±=
− ±=
−− ±
=−
4
2 411
−−
−
The zeros are –1 and 4. b) c) The relation is valid for . h must be positive. 0 d≤ ≤ d) Van hits the water at 4 m.
e) The d-coordinate of the vertex is 1 42
− + , or 1.5.
( ) ( )1.5 1.5
2
2
3 4
3 46 25.
h d d= − + +
= − + +
=
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.
( )14
4.9−
1.4 4.1
2
21.4
bta
= −
= −
2.51 .52
h t t= − + +
= − + +
( ) ( )
2
2
4.9 14 2.5
4.9 14
The maximum height of 12.5 m occurs at 1.4 s. b) 2
2
2
4.9 14 2.50.5 4.9 14 2.5
0 4.9 14 2
h t tt tt t
= − + +
= − + +
= − + +Use the quadratic formula with a = –4.9, b = 14, and c = 2.
( ) ( )( )( )
2
214 14 4.9 34.9
−
−
42
42
14 2358
.29.
b b acda
− ± −=
− ± −=
− ±−
=
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 Page 319 Let x and y represent the legs of the triangle.
( )
( )( )
2 2 2
22
2 2
2
2
2323
17
289
529 46 2892 46 240 0
23 120 0
2
0
3
8 15
x yy x
x y
x
x x xx xx xx x
x
+ == −
+ =
+ =
+ − + =
− + =
− + =
− − =
−
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.
2
2
8 512648
xxx
=
==
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 First Differences
–2 9 –1 7 –2 0 5 –2 1 3 –2 2 1 –2
The first differences are constant. The relation is linear. b)
x y First Differences
Second Differences
–2 –3 –1 3 6 0 5 2 –4 1 3 -2 –4 2 -3 -6 –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 is the graph of 2 2y x= + 2y x= translated 2 units upward. b) The graph of is the graph of ( 23y x= + ) 2y x= translated 3 units to the left.
c) The graph of is the graph of 2 2y x= + 214
y = − x vertically
compressed by a factor of 14
and reflected in the x-axis.
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
( )( )3y a x x= + − 5 . Substitute (x, y) = (1, –4) and solve for a.
( )( )4 = +− 1 13 54 1614
aa
a
−
− = −
=
An equation for the parabola is ( )(1 3 54
y x x= + − ) .
Chapter 6 Chapters 4 to 6 Review Question 5 Page 320
a) 44
1221
16
− =
=
b) ( )( )
22
133
19
−− =
−
=
c) d) 025 1= 1 188
− =
e) ( ) f) 121− =13
33 14 3
4127646427
−⎛ ⎞⎜ ⎟ =⎝ ⎠ ⎛ ⎞
⎜ ⎟⎝ ⎠
=
=
Chapter 6 Chapters 4 to 6 Review Question 6 Page 320 a) 20.8 years = 4 × 5.2 years
4120 1.252
⎛ ⎞ =⎜ ⎟⎝ ⎠
After 20.8 years, 1.25 g of Cobalt-60 remains. b) 36.4 years = 7 × 5.2 years
7120 0.156 252
⎛ ⎞ =⎜ ⎟⎝ ⎠
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
( ) 2
2 2
2
3 2 1
6 35 3
A x x x
x x xx x
= + −
= + −
= +
Chapter 6 Chapters 4 to 6 Review Question 8 Page 320 a) b) ( )( )( ) 2
2
3 3 3 3
9
n n n n n
n
+ − = − + −
= −
9 252 25 10h h h+ = + +
c) ( )( ) 2
2
4 2 2 4
6 8
d d d d d
d d
− − = − − +
= − +
8 d) ( )( ) 2
2
3 7 7 3
10 21
m m m m m
m m
21+ + = + + +
= + +
e) ( )( ) 23 5 3 5 9 25t t t− + = − f) ( )2 27 14x x x 49− = − + Chapter 6 Chapters 4 to 6 Review Question 9 Page 320 a) ( )( ) ( )
( )
2
2
3 2
3 1 2 5 6 15 2 5
6 13 5
6 13 5
x x x x x x x
x x x
x x x
+ − = − + −
= − −
= − −
b) ( ) ( )( )2 2 2
2
2 3 2 2 4 12 9 4
3 12 13
k k k k k k
k k
+ − + − = + + − +
= + + c) 5 ( )( ) ( ) ( )2 2 2
2 2
2
4 3 1 3 4 5 3 11 4 9 24 16
15 55 20 9 24 1624 79 4
y y y y y y y
y y y yy y
− + + − = − − + − +
= − − + − +
= − − d) ( )( ) ( )
( )
2 2
2 2
2 2
3 2 3 3 2 3 6 4 9 6
3 6 5 6
18 15 18
a b a b a ab ab b
a ab b
a ab b
+ − = − + −
= + −
= + −
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) ( )( )2 12 27 3 9y y y y+ + = + + b) ( )( )2 2 3 3 1x x x x+ − = + − c) d) ( )(2 22 21 21 1n n n n+ + = + + ) ( )( )2 8 15 5 3p p p p− + = − − e) ( )( )2 2 15 5 3x x x x+ − = + − f) 2 5 24 cannot be factoredk k+ + Chapter 6 Chapters 4 to 6 Review Question 12 Page 321 a) b) ( 22 12 36 6p p p+ + = + ) ( )229 6 1 3 1d d d+ + = − c) ( )( )2 49 7 7x x x− = + − d) ( )224 20 25 2 5a a a− + = − e) ( )
( )(
2 28 18 2 4 9
2 2 3 2 3
t t
t t
− = −
= + − ) f) ( )( )2 24 2a b a b a b− = + − 2
Chapter 6 Chapters 4 to 6 Review Question 13 Page 321 a) ( )( )
( )(( )( )( )( )
2
2
2
2
11; 11 10 1 10
11; 11 10 1 10
7; 7 10 2 5
7; 7 10 2 5
k m m m m
k m m m m
k m m m m
k m m m m
= + + = + +
= − − + = − −
= + + = + +
= − − + = − −
)
b) ( )
( )
22
22
12; 9 12 4 3 2
12; 9 12 4 3 2
k a a a
k a a a
= − + = −
= − + + = +
MHR • Principles of Mathematics 10 Solutions 141
Chapter 6 Chapters 4 to 6 Review Question 14 Page 321
( )( )
2
2
4 36 81
2 9
A x x
x
π
π
= + +
= +
The radius of the circle is 2 . The diameter is 4x + 18. 9x +
142 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Chapters 4 to 6 Review Question 15 Page 321 a)
( )( )( )( )
2
2
2 2 2
2 2 2
2
6 16
6 16
6 3 3 16
6 3 3 16
3 25
y x x
x x
x x
x x
x
= + −
= + −
= + + − −
= + + − −
= + − b)
( )( ) ( )( )( )( ) ( )
( )
2
2
2 22
2 22
2
8 7
8 7
8 4 4
8 4 4
4 9
y x x
x x
x x
x x
x
= − +
= − +
= − + − − − +
= − + − − − +
= − −
7
7
0
0
c)
( )( )( )( )
2
2
2 2 2
2 2 2
2
4 10
4 10
4 2 2 1
4 2 2 1
2 6
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − +
= + +
MHR • Principles of Mathematics 10 Solutions 143
144 MHR • Principles of Mathematics 10 Solutions
8
) 5
7
d)
( )( ) ( )( )( )( ) ( )( )
( )
2
2
2 22
2 22
2
6 8
6 8
6 3 3 8
6 3 1 3
3 1
y x x
x x
x x
x x
x
= − + −
= − − −
= − − + − − − −
= − − + − − − − −
= − − + e)
( )( )( ) (( )
2
2
2 2 2
2 2 2
2
2 8 5
2 4 5
2 4 2 2 5
2 4 2 2 2
2 2 3
y x x
x x
x x
x x
x
= + +
= + +
= + + − +
= + + − − +
= + − f)
( )( )( ) ( )( )( )
2
2
2 2 2
2 2 2 2
2
2 12 7
2 6 7
2 6 3 3 7
2 6 3 3 2 3
2 3 11
y x x
x x
x x
x x
x
= − − −
= − + −
= − + + − −
= − + + − − − −
= − + +
144 MHR • Principles of Mathematics 10 Solutions
Chapter 6 Chapters 4 to 6 Review Question 16 Page 321 a)
( )( )
2 14 24 02 12 0
2 0 or 12 02 or 12
x xx x
x xx x
− + =
− − =
− = − == =
Check by substituting both solutions in the original equation. For x = 2: For x = 12:
( ) ( ) +
2
2
14 24
14 22 2 40
x x= − +
= −
=
L.S. 0 =R.S.
( )221 ( )
2 14 24
14 21 40
2
x x= − +
= − +
−−
=
L.S. 0
=
L.S. 0
=R.S.
L.S. = R.S. L.S. = R.S. The roots are 2 and 12. b)
( )( )
2 4 21 07 3 0
7 0 or 3 07 or 3
n nn n
n nn n
+ − =
+ − =
+ = − == − =
Check by substituting both solutions in the original equation. For n = –7: For n = 3:
( ) ( )
2
2
4 21
4 17 20
7
n n= + −
= + −
=R.S.
( ) ( )
2
2
4 21
3 4 20
3 1
n n= + −
= + −
=
L.S. 0 =R.S.
L.S. = R.S. L.S. = R.S. The roots are –7 and 3.
MHR • Principles of Mathematics 10 Solutions 145
c)
( )( )
2 16 04 4 0
4 0 or 4 04 or 4
mm m
m mm m
− =
+ − =
+ = − == − =
Check by substituting both solutions in the original equation. For m = –4: For m = 4:
( )24 16− )24 16
2 16
0
m= −
=
=
L.S. 0 2 16
0
m =R.S.
(− −
= −
=
=
L.S. 0 =R.S.
L.S. = R.S. L.S. = R.S. The roots are 4 and –4. d)
( ) ( )( ) ( )( )( )
2
2
2
2 5 22 4 2
2 4 2
2 2 2
2 2 1 0
y yy y y
y y y
y y y
y y
+ + =
+ + + =
+ + + =
+ + + =
+ + =
00
0
0
2 0 or 2 1 0
12 or 2
y y
y y
+ = + =
= − = −
Check by substituting both solutions in the original equation.
For y = –2: For y = 12
− :
( ) ( )2 22 5 2+ +− −
22 5 2
20
y y= + +
=
=
L.S. 0 =R.S. 2
2
2 5 2
1 12 5 2
02 2
y y= + +
⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
=
L.S. 0=R.S.
− −
L.S. = R.S. L.S. = R.S.
The roots are –2 and 12
− .
146 MHR • Principles of Mathematics 10 Solutions
e)
( )( )
2
2
2
2
7 70 1757 70 175 0
7 10 25 0
7 5 05 0
5
t tt t
t t
tt
t
= −
− + =
− + =
− =
− ==
Check by substituting the solution in the original equation.
( )
2
2
7
71
575
t=
=
=
L.S.
)1
( )70 17570 17
55 5
17
t= −
= −
=
R.S.
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)
( )(
2 3 20 2y x x
x x= − +
= − − The x-intercepts are 2 and 1. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
2
1 22
1.53 2
1.5 1.53 20.25
x
y x x
+=
=
= − +
= − +
= −
The vertex is ( ) . 1.5, 0.25−
MHR • Principles of Mathematics 10 Solutions 147
b)
( )(
2 160 4y x
x x= −
= + − )4 The x-intercepts are –4 and 4. Find the x-coordinate of the vertex, and then, the y-coordinate.
( )20 16−
2
4 42
016
16
x
y x
− +=
=
= −
= −
=
The vertex is ( ) . 0, 16− c)
( ) ( )( ) ( )
( )( )
2
2
2
2 5 120 2 8 3 12
0 2 8 3 12
0 2 4 3 4
0 4 2 3
y x xx x x
x x x
x x x
x x
= − −
= − + −
= − + −
= − + −
= − + The x-intercepts are 4 and . 1.5− Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
2
1.5 42
1.252 5 12
1.252 515.125
1.25
x
y x x
− +=
=
= − −
= −
= −
12−
The vertex is ( ) . 1.25, 15.125−
148 MHR • Principles of Mathematics 10 Solutions
d)
( )( )(
2
2
7 12
0 7
0 3
y x x
x x
x x
= − − −
= − + +
= − + + )12
4 The x-intercepts are –3 and –4. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
2
3 42
3.57 12
3.5− 3.57 120.25
x
y x x
− −=
= −
= − − −
= −
=
−− −
The vertex is ( )3.5,0.25− . e)
( )
23 60 3 2y x x
x x= − +
= − − The x-intercepts are 0 and 2. Find the x-coordinate of the vertex, and then, the y-coordinate.
( ) ( )
2
2
0 2x +=
21
3 6
33
1 6 1
y x x=
= − +
= − +
=
he vertex is ( )1,3 .T
MHR • Principles of Mathematics 10 Solutions 149
Chapter 6 Chapters 4 to 6 Review Question 18 Page 321 The equation is of the form ( )6y ax x= + . Substitute (x, y) = –3, 4) and solve for a.
( )( )
( )
2
64 949
4 694 89
3
3
4 3aa
a
y x x
x x
= −
− =
= − +
= − −
= +− −
Chapter 6 Chapters 4 to 6 Review Question 19 Page 321 a) Use the quadratic formula with a = 1, b = –6, and c = 1.
( ) ( )( )( )
2
2
42
46 61
± −− 1 12
6 322
b b acxa
− ± −=
±=
=
The roots are 6 322
+ and 6 322
− .
b) Use the quadratic formula with a = 3, b = –5, and c = 1.
( ) ( )( )( )
2
2
42
45 53
± −− 3 12
5 136
b b acxa
− ± −=
±=
=
The roots are 5 136
+ and 5 136
− .
150 MHR • Principles of Mathematics 10 Solutions
c) Use the quadratic formula with a = 3.2, b = –5.6, and c = –7.1.
( ) ( )( )( )
2
25.6 3.2
42
42
5.6 122.246.4
b b acxa
− −
− ± −=
± −=
±=
5.6 7.13.2
The roots are 5.6 122.246.4
+ and 5.6 122.246.4
− .
d) Use the quadratic formula with a = 2, b = 5, and c = –9.
( ) )( )(( )
2 42
2
5 974
5
b b acxa
− ± −=
− ±
− ±=
=
2 45 2 92
−−
The roots are 5 974
− + and 5 974
− − .
e) Use the quadratic formula with a = 3, b = –8, and c = 3.
( ) ( )( )( )
2
2
42
48 83
± −− 3 32
8 286
b b acxa
− ± −=
±=
=
The roots are 8 286
+ and 8 286
− .
MHR • Principles of Mathematics 10 Solutions 151
Chapter 6 Chapters 4 to 6 Review Question 20 Page 321 Perimeter: 2 2 8x y+ =Area: 2xy =Rearrange the perimeter equation to obtain an expression for y. Then, substitute the expression into the area equation.
( )2
2
2 2 844
22
4 2 04 2 0
4
x yx y
y x
xyx
x x
x
x
x
−
+ =+ =
= −
=
=
− + − =
− + =Use the quadratic formula with a = 1, b = –4, and c = 2.
( ) ( )( )( )
2
2
42
42
4
4 41
± −− 1
8
2
2
b b acxa
− ± −=
±=
=
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
( )( )2
20 500 20
10 000 100 20
R x x
x x
= + −
= + −
Calculate the x-coordinate of the vertex.
( )100
20−
2
22.5
bxa
= −
= −
=
The fare that will maximize the revenue is $22.50.
152 MHR • Principles of Mathematics 10 Solutions
