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Active Maths 2 (Strands 1–5): Ch 17 Solutions
Chapter 17 Exercise 17.1
1
Q. 1. x2 + 10x + 21 = 0
(x + 3)(x + 7) = 0
x + 3 = 0 OR x + 7 = 0
x = −3 OR x = −7
Checking solutions
x2 + 10x + 21 = 0
(−3)2 + 10(−3) + 21 = 0
9 − 30 + 21 = 0
30 − 30 = 0
0 = 0 True
∴ x = −3 is a solution.
x2 + 10x + 21 = 0
(−7)2 + 10(−7) + 21 = 0
49 − 70 + 21 = 0
70 − 70 = 0
0 = 0 True
∴ x = −7 is a solution.
Q. 2. x2 − 3x + 2 = 0
(x − 1)(x − 2) = 0
x − 1 = 0 OR x − 2 = 0
x = 1 OR x = 2
Checking solutions
x2 − 3x + 2 = 0
(1)2 − 3(1) + 2 = 0
3 − 3 = 0
0 = 0 True
∴ x = 1 is a solution.
x2 − 3x + 2 = 0 (2)2 − 3(2) + 2 = 0 4 − 6 + 2 = 0 6 − 6 = 0 0 = 0 True
∴ x = 2 is a solution.
Q. 3. x2 − 5x = 0 x(x − 5) = 0 x = 0 OR x − 5 = 0 x = 5
Checking solutions
x2 − 5x = 0 x2 − 5x = 0(0)2 − 5(0) = 0 (5)2 − 5(5) = 0
0 − 0 = 0 25 − 25 = 0True 0 = 0
∴ x = 0 is a solution True∴ x = 5 is a solution
Q. 4. x2 − x − 20 = 0
(x + 4)(x − 5) = 0
x + 4 = 0 OR x − 5 = 0
x = −4 OR x = 5
Checking solutions
x2 − x − 20 = 0 x2 − x − 20 = 0(−4)2 − (−4) − 20 = 0 (5)2 − (5) − 20 = 0 16 + 4 − 20 = 0 25 − 5 − 20 = 0 20 − 20 = 0 25 − 25 = 0 0 = 0 0 = 0
True True∴ x = −4 is a solution ∴ x = 5 is a solution
2 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 5. x2 − 15x + 56 = 0
(x − 7)(x − 8) = 0
x − 7 = 0 OR x − 8 = 0
x = 7 OR x = 8
Q. 6. x2 − 64 = 0
x2 = 64
x = ± √___
64
x = ±8
Q. 7. 3x2 + 8x + 5 = 0
(3x + 5)(x + 1) = 0
3x + 5 = 0 OR x + 1 = 0
3x = −5 OR x = −1
x = − 5 __ 3
Checking solutions3x2 + 8x + 5 = 0 3x2 + 8x + 5 = 0
3(−1)2 + 8(−1) + 5 = 0
3 − 8 + 5 = 0
8 − 8 = 0
0 = 0
True
∴ x = –1 is a solution
3 ( − 5 __ 3 ) 2 + 8 ( − 5 __ 3 ) + 5 = 0
3 × 25 ___ 9
– 40 ___ 3 + 5 = 0
25 ___ 3 − 40 ___ 3 + 15 ___ 3 = 0
40 ___ 3 − 40 ___ 3 = 0
0 = 0True
∴ x = − 5 __ 3 is a solution
3 93
(Keeping fractions as an improper fraction instead of a mixed fraction is a useful non-calculator technique for checking solutions and for use in further calculations.)
Q. 8. 4p2 − 16 = 0 (÷ 4)
p2 − 4 = 0
p2 = 4
p = ± √__
4
p = ±2
Q. 9. 7q2 + 8q + 1 = 0
(7q + 1)(q + 1) = 0
7q + 1 = 0 OR q + 1 = 0
7q = − 1 OR q = −1
q = − 1 __ 7
3Active Maths 2 (Strands 1–5): Ch 17 Solutions
Checking solutions
7q2 + 8q + 1 = 0 7q2 + 8q + 1 = 0
7 ( − 1 __ 7 ) 2 + 8 ( − 1 __ 7 ) + 1 = 0 7(−1)2 + 8(−1) + 1 = 0
7 ( 1 ___ 49
) − 8 __ 7
+ 1 = 0
1 __ 7
− 8 __ 7
+ 7 __ 7
= 0
8 __ 7
− 8 __ 7
= 0
0 = 0
True
∴ q = − 1 __ 7
is a solution
7 − 8 + 1 = 0
8 − 8 = 0
0 = 0
True
∴ q = –1 is a solution
Q. 10. 4x2 − 16x = 0 (÷ 4) x2 − 4x = 0 x(x − 4) = 0 x = 0 OR x − 4 = 0 x = 4
Q. 11. 3y2 − 2y − 1 = 0 (3y + 1)(y − 1) = 0 3y + 1 = 0 OR y − 1 = 0 3y = −1 OR y = 1
y = − 1 __ 3
Q. 12. 3x2 + 20x − 7 = 0 (3x − 1)(x + 7) = 0 3x − 1 = 0 OR x + 7 = 0 3x = 1 OR x = −7
x = 1 __ 3
Q. 13. 3x2 + 19x + 6 = 0 (3x + 1)(x + 6) = 0 3x + 1 = 0 OR x + 6 = 0 3x = −1 OR x = −6
x = − 1 __ 3
Q. 14. −2x2 + 11x = 0 (× −1) 2x2 − 11x = 0 x (2x − 11) = 0 x = 0 OR 2x − 11 = 0 2x = 11
x = 11 ___ 2
Q. 15. 81 − 4x2 = 0 (9)2 − (2x)2 = 0 (9 + 2x)(9 − 2x) = 0 9 + 2x = 0 OR 9 − 2x = 0 2x = −9 OR −2x = −9
x = − 9 __ 2 OR x = 9 __ 2
x = ± 9 __ 2
Q. 16. 5a2 = 2a
5a2 − 2a = 0 a(5a − 2) = 0 a = 0 OR 5a − 2 = 0 5a = 2 a = 2 __ 5
Q. 17. 100x2 − 25 = 0 (÷ 25)
4x2 − 1 = 0
4x2 = 1
x2 = 1 __ 4
x = ± √__
1 __ 4
x = ± 1 __ 2
Q. 18. 2x2 + 2x − 12 = 0 (÷ 2)
x2 + x − 6 = 0
(x − 2)(x + 3) = 0
x − 2 = 0 OR x + 3 = 0
x = 2 OR x = −3
4 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 19. 25x2 − 25 = 0 (÷25) x2 − 1 = 0 x2 = 1 x = ± √
__ 1
x = ±1
Q. 20. 2b2 − b = 10
2b2 − b − 10 = 0
(2b − 5)(b + 2) = 0
2b − 5 = 0 OR b + 2 = 0
2b = 5 OR b = −2
b = 5 __ 2
Q. 21. 2x2 − 3x = 0
x(2x − 3) = 0
x = 0 OR 2x − 3 = 0
2x = 3
x = 3 __ 2
Q. 22. (3x + 4)(x − 3) = 10
3x2 − 9x + 4x − 12 − 10 = 0
3x2 − 5x − 22 = 0
(3x − 11)(x + 2) = 0
3x − 11 = 0 OR x + 2 = 0
3x = 11 OR x = −2
x = 11 ___ 3
Q. 23. 625x2 − 196 = 0
(25x)2 − (14)2 = 0
(25x + 14)(25x − 14) = 0
25x + 14 = 0 OR 25x − 14 = 0
25x = −14 OR 25x = 14
x = − 14 ___ 25 OR x = 14 ___ 25
Q. 24. 5y2 = 7y
5y2 − 7y = 0
y(5y − 7) = 0
y = 0 OR 5y − 7 = 0
5y = 7
y = 7 __ 5
Q. 25. 3x2 − 24x + 36 = 0 (÷ 3)
x2 − 8x + 12 = 0
(x − 2)(x − 6) = 0
x − 2 = 0 OR x − 6 = 0
x = 2 OR x = 6
Q. 26. 5y2 − 125 = 0
5y2 = 125
y2 = 25
y = ±5
Q. 27. (2x + 1)(x − 3) − 4 = 0
2x2 − 6x + x − 3 − 4 = 0
2x2 − 5x − 7 = 0
(2x − 7)(x + 1) = 0
2x − 7 = 0 OR x + 1 = 0
2x = 7
x = 7 __ 2
Q. 28. 4x2 + 8x + 3 = 0 (2x + 1)(2x + 3) = 0 2x + 7 = 0 OR 2x + 3 = 0 2x = −1 OR 2x = −3
x = − 1 __ 2 OR x = − 3 __ 2
Q. 29. 6x2 + 7x + 2 = 0
(2x + 1)(3x + 2) = 0
2x + 1 = 0 OR 3x + 2 = 0
2x = −1 OR 3x = −2
x = − 1 __ 2 OR x = − 2 __ 3
Q. 30. 9x2 − 7x − 2 = 0 (9x + 2)(x − 1) = 0 9x + 2 = 0 OR x − 1 = 0 9x = −2
x = − 2 __ 9
Q. 31. −6x2 + 5x + 1 = 0 (× −1) 6x2 − 5x − 1 = 0 (6x + 1)(x − 1) = 0 6x + 1 = 0 OR x − 1 = 0 6x = −1 OR x = 1
x = − 1 __ 6
5Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 32. 4x2 − 13x + 3 = 0
(4x − 1)(x − 3) = 0
4x − 1 = 0 OR x − 3 = 0
4x = 1 OR x = 3
x = 1 __ 4
Q. 33. 10x2 − 37x + 7 = 0 (5x − 1)(2x − 7) = 0 5x − 1 = 0 OR 2x − 7 = 0 5x = 1 OR 2x = 7
x = 1 __ 5 OR x = 7 __ 2
Q. 34. 80a2 − 245 = 0 (÷5) 16a2 − 49 = 0 16a2 = 49
a2 = 49 ___ 16
a = ± √___
49 ___ 16
a = ± 7 __ 4
Q. 35. 4b2 + 16b + 15 = 0
(2b + 3)(2b + 5) = 0
2b + 3 = 0 OR 2b + 5 = 0
2b = −3 OR 2b = −5
b = − 3 __ 2 OR b = − 5 __ 2
Q. 36. 8c2 − 6 + 13c = 0
8c2 + 13c − 6 = 0
(8c − 3)(c + 2) = 0
8c − 3 = 0 OR c + 2 = 0
8c = 3 OR c = −2
c = 3 __ 8
Q. 37. 21 = 10m2 + 29m
10m2 + 29m − 21 = 0
(5m − 3)(2m + 7) = 0
5m − 3 = 0 OR 2m + 7 = 0
5m = 3 OR 2m = –7
m = 3 __ 5 OR m = − 7 __ 2
Checking solutions (non calculator working shown)
10m2 + 29m − 21 = 0
10 ( 3 __ 5 ) 2 + 29 ( 3 __ 5 ) − 21 = 0
10 × 9 ____ 2 5 5 + 87 ___ 5 − 21 = 0
18 ___ 5 + 87 ___ 5 − 105 ____ 5 = 0
105 ____ 5 − 105 ____ 5 = 0
0 = 0 True
∴ m = 3 __ 5 is a solution
10m2 + 29m − 21 = 0
10 ( − 7 __ 2 ) 2 + 29 ( − 7 __ 2 ) − 21 = 0
1 0 5 × 49 ___ 4 2 − 203 ____ 2 − 21 = 0
245 ____ 2 − 203 ____ 2 − 42 ___ 2 = 0
245 ____ 2 − 245 ____ 2 = 0
0 = 0 True
∴ m = − 7 __ 2 is a solution
Q. 38. 1 __ 4 x2 + 3x + 5 = 0
x2 + 12x + 20 = 0 (× 4)
(x + 10)(x + 2) = 0
x = −2 OR x = −10
Q. 39. x2 − 2.4x − 6.4 = 0 (× 10) 10x2 − 24x − 64 = 0 (÷ 2) 5x − 12x − 32 = 0 (5x + 8)(x − 4) = 0 5x + 8 = 0 OR x − 4 = 0 5x = −8 OR x = 4
x = − 8 __ 5
Q. 40. x2 __ 3 + x __ 2 − 5 __ 6 = 0 (× 6)
2x2 + 3x − 5 = 0 (2x + 5)(x − 1) = 0 2x + 5 = 0 OR x − 1 = 0 2x = −5 OR x = 1
x = − 5 __ 2
2
6 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 41. 3x2 ___ 5 − 11x ____ 10 − 1 = 0 (× 10)
6x2 − 11x − 10 = 0
(3x + 2)(2x − 5) = 0
3x + 2 = 0 OR 2x − 5 = 0
3x = −2 OR 2x = 5
x = − 2 __ 3 OR x = 5 __ 2
Q. 42. 1 __ 2 x2 − 17x ____ 8 + 1 __ 2 = 0 (× 8)
4x2 − 17x + 4 = 0
(4x − 1)(x − 4) = 0 4x − 1 = 0 OR x − 4 = 0 4x = 1 OR x = 4
x = 1 __ 4
Exercise 17.2
Q. 1. x2 + 8x + 15 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 1, b = 8, c = 15
b2 − 4ac = (8)2 − 4(1)(15)
= 64 − 60 = 4
x = −8 ± √__
4 _________ 2 × 1
x = −8 + 2 _______ 2 OR x = −8 − 2 _______ 2
x = −6 ___ 2 OR x = −10 ____ 2
∴ x = −3 OR x = −5
Q. 2. x2 + 7x + 10 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 1, b = 7, c = 10 b2 − 4ac = (7)2 − 4(1)(10) = 9
x = −7 ± √__
9 _________ 2 × 1
x = −7 + 3 _______ 2 OR x = −7 − 3 _______ 2
∴ x = −2 OR x = −5
Q. 3. x2 + x − 6 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 1, b = 1, c = −6 b2 − 4ac = (1)2 − 4(1)(−6) = 1 + 24 = 25
x = −1 ± √___
25 __________ 2 × 1
x = −1 + 5 _______ 2 OR x = −1 − 5 _______ 2
x = 4 __ 2 OR x = −6 ___ 2
∴ x = 2 OR x = −3
Q. 4. 2x2 − 13x + 20 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 2, b = −13, c = 20 b2 − 4ac = (−13)2 − 4(2)(20) = 169 − 160 = 9
x = −(−13) ± √
__ 9 _____________ 2 × 2
x = 13 + 3 _______ 4 OR x = 13 − 3 _______ 4
x = 16 ___ 4 OR x = 10 ___ 4
x = 4 OR x = 5 __ 2
Checking solutions
2x2 − 13x + 20 = 0 2(4)2 − 13(4) + 20 = 0 32 − 52 + 20 = 0 52 − 52 = 0 0 = 0 True ∴ x = 4 is a solution
2x2 − 13x + 20 = 0
2 ( 5 __ 2 ) 2 − 13 ( 5 __ 2 ) + 20 = 0
12 1 _ 2 − 32 1 _ 2 + 20 = 0
32 1 _ 2 − 32 1 _ 2 = 0
0 = 0 True
∴ x = 5 __ 2 is a solution.
7Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 5. 2x2 − 7x + 5 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 2, b = −7, c = 5 b2 − 4ac = (−7)2 − 4(2)(5) = 9
x = –(–7) ± √
__ 9 __________ 2(2) = 7 ± 3 ______ 4
x = 7 + 3 ______ 4 OR x = 7 − 3 ______ 4
x = 5 __ 2 OR x = 1
Q. 6. 4x2 − 17x + 13 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 4, b = −17, c = 13 b2 − 4ac = (−17)2 − 4(4)(13) = 289 − 208 = 81
x = −(−17) ± √
___ 81 ______________ 2 × 4
x = 17 + 9 _______ 8 OR x = 17 − 9 _______ 8
x = 26 ___ 8 OR x = 8 __ 8
x = 13 ___ 4 OR x = 1
Q. 7. 5x2 − 18x + 9 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 5, b = −18, c = 9 b2 − 4ac = (−18)2 − 4(5)(9) = 324 − 180 = 144
x = −(−18) ± √
____ 144 _______________ 2 × 5
x = 18 + 12 ________ 10 OR x = 18 − 12 ________ 10
x = 30 ___ 10 OR x = 6 ___ 10
x = 3 OR x = 3 __ 5
Q. 8. 7x2 + 25x + 12 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 7, b = 25, c = 12 b2 − 4ac = (25)2 − 4(7)(12) = 625 − 336 = 289
x = −25 ± √____
289 ____________ 2 × 7
x = −25 + 17 _________ 14 OR x = −25 − 17 _________ 14
x = −8 ___ 14 OR x = −42 ____ 14
x = − 4 __ 7 OR x = −3
Q. 9. 3x2 = 4x + 7 3x2 − 4x − 7 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 3, b = −4, c = −7 b2 − 4ac = (−4)2 − 4(3)(−7) = 16 + 84 = 100
x = −(−4) ± √
____ 100 ______________ 2 × 3
x = 4 + 10 _______ 6 OR x = 4 − 10 _______ 6
x = 14 ___ 6 OR x = −6 ___ 6
x = 7 __ 3 OR x = −1
Q. 10. 4x2 = 4x + 8
∴ 4x2 − 4x − 8 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 4, b = −4, c = −8
b2 − 4ac = (−4)2 − 4(4)(−8)
= 16 + 128
= 144
x = −(−4) ± √
____ 144 ______________ 2 × 4
x = 4 + 12 _______ 8 OR x = 4 − 12 _______ 8
x = 16 ___ 8 OR x = −8 ___ 8
x = 2 OR x = −1
8 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 11. 23x + 20 = −6x2
6x2 + 23x + 20 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 6, b = 23, c = 20
b2 − 4ac = (23)2 − 4(6)(20)
= 529 − 480
= 49
x = −23 ± √___
49 ___________ 2 × 6
x = −23 + 7 ________ 12 OR x = −23 − 7 ________ 12
x = −16 ____ 12 OR x = −30 ____ 12
x = −4 ___ 3 OR x = −5 ___ 2
Q. 12. x2 + 7 __ 2 x = 2 (×2)
2x2 + 7x = 4 (–4)
2x2 + 7x − 4 = 0
x = −b ± √________
b2 − 4ac _______________ 2a
a = 2, b = 7, c = −4
b2 − 4ac = (7)2 − 4(2)(−4)
= 49 + 32
= 81
x = −7 ± √___
81 __________ 2 × 2
x = −7 + 9 _______ 4 OR x = −7 − 9 _______ 4
x = 2 __ 4 OR x = −16 ____ 4
x = 1 __ 2 OR x = −4
Exercise 17.3
Q. 1. x2 + 6x + 1 = 0
a = 1, b = 6, c = 1
b2 − 4ac = (6)2 − 4(1)(1)
= 36 − 4 = 32
∴ x = −6 ± √___
32 __________ 2 × 1
x = −6 + √___
32 __________ 2 OR x = −6 − √___
32 __________ 2
Note √___
32 = √_______
16 × 2
= √___
16 × √__
2
= 4 √__
2
∴ x = −6 + 4 √__
2 __________ 2 OR x = −6 − 4 √__
2 __________ 2
Dividing by 2 gives
x = −3 + 2 √__
2 OR x = −3 − 2 √__
2
Q. 2. x2 + 8x + 5 = 0
a = 1, b = 8, c = 5
b2 − 4ac = (8)2 − 4(1)(5)
= 64 − 20
= 44
∴ x = −8 ± √___
44 __________ 2 × 1
x = −8 + √___
44 __________ 2 OR x = −8 − √___
44 _________ 2
Give answer to 3 decimal places
x = −0.683|3... OR x = −7.316|6...
∴ x = −0.683 OR x = −7.317
Q. 3. y2 − 2y − 29 = 0
a = 1, b = −2, c = −29
b2 − 4ac = (−2)2 − 4(1)(−29)
= 4 + 116
= 120
∴ y = −(−2) ± √
____ 120 ______________ 2 × 1
y = 2 ± √____
120 _________ 2 OR = 2 – √____
120 _________ 2
Give answer to 2 d.p.
y = 6.47|7... OR y = −4.47|7
y = 6.48 OR y = −4.48
Q. 4. x2 − 5x − 28 = 0 a = 1, b = −5, c = −28 b2 − 4ac = (−5)2 − 4(1)(−28) = 25 + 112 = 137
∴ x = −(−5) ± √
____ 137 ______________ 2 × 1
9Active Maths 2 (Strands 1–5): Ch 17 Solutions
x = 5 + √____
137 _________ 2 OR x = 5 − √____
137 _________ 2
Give answer to 1 d.p.
x = 8.3|5... OR x = −3.3|5...
x = 8.4 OR x = −3.4
Q. 5. 2x2 − 11x − 9 = 0
a = 2, b = −11, c = −9
b2 − 4ac = (−11)2 − 4(2)(−9)
= 121 + 72
= 193
∴ x = −(−11) ± √
____ 193 _______________ 2 × 2
x = 11 + √____
193 __________ 4 OR x = 11 − √____
193 __________ 4
Give answer to 3 d.p.
x = 6.223|1... OR x = −0.723|1...
x = 6.223 OR x = −0.723
Q. 6. 3x2 − 10x + 4 = 0
a = 3, b = −10, c = 4
b2 − 4ac = (−10)2 − 4(3)(4)
= 100 − 48
= 52
∴ x = −(−10) ± √
___ 52 ______________ 2 × 3
x = 10 + √___
52 _________ 6 OR x = 10 − √___
52 _________ 6
Give answer in surd form
√___
52 = √_______
4 × 13
= √__
4 × √___
13
= 2 √___
13
∴ x = 10 + 2 √___
13 __________ 6 OR x = 10 − 2 √___
13 __________ 6
Dividing by 2 gives
x = 5 + √___
13 ________ 3 OR x = 5 − √___
13 ________ 3
Q. 7. 2x2 − 5x − 21 = 0
a = 2, b = −5, c = −21
b2 − 4ac = (−5)2 − 4(2)(−21)
= 25 + 168= 193
∴ x = −(−5) ± √
____ 193 ______________ 2 × 2
x = 5 + √____
193 _________ 4 OR x = 5 − √____
193 _________ 4
Give answer to 3 d.p.
x = 4.723|1... OR x = −2.223|1...
x = 4.723 OR x = −2.223
Q. 8. 4q2 − q − 13 = 0
a = 4, b = −1, c = −13
b2 − 4ac = (−1)2 − 4(4)(−13)
= 1 + 208
= 209
∴ q = −(−1) ± √
____ 209 ______________ 2 × 4
q = 1 + √____
209 _________ 8 OR q = 1 − √____
209 _________ 8
Give answer to 1 d.p.
∴ q = 1.9|3... OR q = −1.6|8...
q = 1.9 OR q = −1.7
Q. 9. 5x2 + 4x − 5 = 0
a = 5, b = 4, c = −5
b2 − 4ac = (4)2 − 4(5)(−5)
= 16 + 100
= 116
∴ x = −4 ± √____
116 ___________ 2 × 5
x = −4 + √____
116 ___________ 10 OR x = −4 − √____
116 ___________ 10
Give answer to 2 d.p.
x = 0.67|7... OR x = −1.47|7...
∴ x = 0.68 OR x = −1.48
Q. 10. 8x2 − 5x − 11 = 0
a = 8, b = −5, c = −11
b2 − 4ac = (−5)2 − 4(8)(−11)
= 25 + 352 = 377
∴ x = −(−5) ± √
____ 377 ______________ 2 × 8
x = 5 + √____
377 _________ 16 OR x = 5 − √____
377 _________ 16
Give answer to 3 d.p.
x = 1.526|0... OR x = −0.901|0...
x = 1.526 OR x = −0.901
10 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 11. 9a2 − 8a − 24 = 0
a = 9, b = −8, c = −24
b2 − 4ac = (−8)2 − 4(9)(−24)
= 64 + 864
= 928
∴ a = −(−8) ± √
____ 928 ______________ 2 × 9
a = 8 + √____
928 _________ 18 OR a = 8 − √____
928 _________ 18
Give answer to 2 d.p.
a = 2.13|6... OR a = −1.24|7...
a = 2.14 OR a = −1.25
Q. 12. 3x2 − 9x + 5 = 0
a = 3, b = −9, c = 5
b2 − 4ac = (−9)2 − 4(3)(5)
= 81 − 60
= 21
∴ x = −(−9) ± √
___ 21 _____________ 2 × 3
x = 9 + √___
21 ________ 6 OR x = 9 − √___
21 ________ 6
Give answer to 1 d.p.
x = 2.2|6... OR x = 0.7|3...
x = 2.3 OR x = 0.7
Q. 13. 11b2 − 3b − 7 = 0
a = 11, b = −3, c = −7
b2 − 4ac = (−3)2 − 4(11)(−7)
= 9 + 308
= 317
∴ b = −(−3) ± √
____ 317 ______________ 2 × 11
b = 3 + √____
317 _________ 22 OR b = 3 − √____
317 _________ 22
Give answer to 3 d.p.
b = 0.945|6... OR b = −0.672|9...
b = 0.946 OR b = −0.673
Q. 14. 2x2 + 4x + 1 = 0
a = 2, b = 4, c = 1
b2 − 4ac = (4)2 − 4(2)(1)
= 16 − 8 = 8
∴ x = −4 ± √__
8 _________ 2 × 2
x = −4 + √__
8 _________ 4 OR x = −4 − √__
8 _________ 4
Give answer in surd form
Note: √__
8 = √______
4 × 2
= √__
4 × √__
2
= 2 √__
2
∴ x = −4 + 2 √__
2 __________ 4 OR x = −4 − 2 √__
2 __________ 4
Dividing by 2 gives
x = −2 + √__
2 _________ 2 OR x = −2 − √__
2 _________ 2
Q. 15. 4x2 + 10x + 5 = 0 a = 4, b = 10, c = 5 b2 − 4ac = (10)2 − 4(4)(5) = 100 − 80 = 20
∴ x = −10 ± √___
20 ___________ 2 × 4
x = 10 + √___
20 _________ 8 OR x = −10 − √___
20 ___________ 8
Give answer in surd form
Note: √___
20 = √______
4 × 5
= √__
4 × √__
5
= 2 √__
5
∴ x = −10 + 2 √__
5 ___________ 8 OR x = −10 −2 √__
5 __________ 8
Dividing by 2 gives
x = −5 + √__
5 _________ 4 OR x = −5 − √__
5 _________ 4
Q. 16. x2 − 12x + 14 = 0 a = 1, b = −12, c = 14 b2 − 4ac = (−12)2 − 4(1)(14) = 144 − 56 = 88
∴ x = −(−12) ± √
___ 88 ______________ 2 × 1
x = 12 + √___
88 _________ 2 OR x = 12 − √___
88 _________ 2
Give answer in surd form
11Active Maths 2 (Strands 1–5): Ch 17 Solutions
Note: √___
88 = √_______
4 × 22
= √__
4 × √___
22
= 2 √___
22
x = 12 + 2 √___
22 __________ 2 OR x = 12 − 2 √___
22 __________ 2
Dividing by 2 gives
∴ x = 6 + √___
22 OR x = 6 − √___
22
Q. 17. 5x2 − 16x + 9 = 0
a = 5, b = −16, c = 9
b2 − 4ac = (−16)2 − 4(5)(9)
= 256 − 180
= 76
∴ x = −(−16) ± √
___ 76 ______________ 2 × 5
x = 16 + √___
76 _________ 10 OR x = 16 − √___
76 _________ 10
Give answer to 1 d.p.
∴ x = 2.4|7... OR x = 0.7|2...
x = 2.5 OR x = 0.7
Q. 18. 15x2 − 21x + 1 = 0
a = 15, b = −21, c = 1
b2 − 4ac = (−21)2 − 4(15)(1)
= 441 − 60
= 381
∴ x = −(−21) ± √
____ 381 _______________ 2 × 15
x = 21 + √____
381 __________ 30 OR x = 21 − √____
381 __________ 30
Give answer to 2 d.p.
x = 1.35|0... OR x = 0.04|9...
∴ x = 1.35 OR x = 0.05
Q. 19. 4x2 = 1 − 2x
4x2 + 2x − 1 = 0
a = 4, b = 2, c = −1
b2 − 4ac = (2)2 − 4(4)(−1)
= 4 + 16
= 20
∴ x = −2 ± √___
20 __________ 2 × 4
x = −2 + √___
20 __________ 8 OR x = −2 − √___
20 __________ 8
Give answer in surd form
Note √___
20 = √__
4 × √__
5
= 2 √__
5
∴ x = −2 + 2 √__
5 __________ 8 OR x = −2 − 2 √__
5 __________ 8
Dividing by 2 gives
x = −1 + √__
5 _________ 4 OR x = −1 − √__
5 _________ 4
Q. 20. 2 − 10x + 9x2 = 0
9x2 − 10x + 2 = 0
a = 9, b = −10, c = 2
b2 − 4ac = (−10)2 − 4(9)(2)
= 100 − 72
= 28
x = −(−10) ± √
___ 28 ______________ 2 × 9
x = 10 + √___
28 _________ 18 OR x = 10 − √___
28 _________ 18
Give answer in surd form
Note: √___
28 = √______
4 × 7
√___
28 = √__
4 × √__
7
= 2 √__
7
∴ x = 10 + 2 √__
7 _________ 18 OR x = 10 − 2 √__
7 _________ 18
Dividing by 2 gives
x = 5 + √__
7 ___________ 9 OR x = 5 − √__
7 _______ 9
Q. 21. 3x = 20 − 8x2
8x2 + 3x − 20 = 0
a = 8, b = 3, c = −20
b2 − 4ac = (3)2 − 4(8)(−20)
= 9 + 640
= 649
x = −3 ± √____
649 ___________ 2 × 8
x = −3 + √____
649 ___________ 16 OR x = −3 − √____
649 ___________ 16
Give answer to 3 d.p.
∴ x = 1.404|7... OR x = −1.779|7...
∴ x = 1.405 OR x = −1.780
12 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 22. 15x2 − 14x = 9
15x2 − 14x − 9 = 0
a = 15, b = −14, c = −9
b2 − 4ac = (−14)2 − 4(15)(−9)
= 196 + 540
= 736
∴ x = −(−14) ± √
____ 736 _______________ 2 × 15
x = 14 + √____
736 __________ 30 OR x = 14 − √____
736 __________ 30
Give answer in surd form
Note: √____
736 = √________
16 × 46
= √___
16 × √___
46
= 4 √___
46
∴ x = 14 + 4 √___
46 __________ 30 OR x = 14 − 4 √___
46 __________ 30
x = 7 + 2 √___
46 _________ 15 OR x = 7 − 2 √___
46 _________ 15
Q. 23. 8x2 = 3 − 8x
8x2 + 8x − 3 = 0
a = 8, b = 8, c = −3
b2 − 4ac = (8)2 − 4(8)(−3)
= 64 + 96
= 160
x = −8 ± √____
160 ___________ 2 × 8
x = −8 + √____
160 ___________ 16 OR x = −8 − √____
160 ___________ 16
Give answer as a surd
Note: √____
160 = √________
16 × 10 = √
___ 16 × √
___ 10
= 4 √___
10
∴ x = −8 + 4 √___
10 ___________ 16 OR x = −8 −4 √___
10 __________ 16
Dividing by 4 gives
x = −2 + √___
10 __________ 4 OR x = −2 − √___
10 __________ 4
Q. 24. 5x2 = 2x + 11 5x2 − 2x − 11 = 0 a = 5, b = −2, c = −11
b2 − 4ac = (−2)2 − 4(5)(−11) = 4 + 220 = 224
∴ x = −(−2) ± √
____ 224 ______________ 2 × 5
x = 2 + √____
224 _________ 10 OR x = 2 − √____
224 _________ 10
Give answer to 2 d.p.
x = 1.69|6... OR x = −1.29|6... ∴ x = 1.70 OR x = −1.30
Q. 25. 2x2 + 5x + 3 __ 8 = 0 (× 8)
16x2 + 40x + 3 = 0 a = 16, b = 40, c = 3 b2 − 4ac = (40)2 − 4(16)(3) = 1,600 − 192
= 1,408
x = –40 ± √
______ 1,408 _____________ 2 × 16
x = −40 + √
______ 1,408 ______________ 32 OR x =
−40 − √______
1,408 ______________ 32
Give answer to 2 d.p.
x = −0.07|7... OR x = −2.42|2...
∴ x = −0.08 OR x = −2.42
Q. 26. x2 __ 3 + 2x ___ 5 − 4 = 0 (× 15)
5x2 + 6x − 60 = 0
a = 5, b = 6, c = −60
b2 − 4ac = (6)2 − 4(5)(−60)
= 36 + 1,200
= 1,236
∴ x = −6 ± √
______ 1,236 _____________ 2 × 5
x = −6 + √
______ 1,236 _____________ 10 OR x =
−6 − √______
1,236 _____________ 10
Give answer to 3 d.p.
x = 2.915|6... OR x = −4.115|6...
x = 2.916 OR x = −4.116
13Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 27. 4 __ 9 x2 = 3 __ 2 x + 1 __ 5 (× 90)
40x2 = 135x + 18
40x2 − 135x − 18 = 0
a = 40, b = −135, c = −18
b2 − 4ac = (−135)2 − 4(40)(−18)
= 21,105
x = −(−135) ± √
_______ 21,105 ___________________ 2 × 40
x = 135 + √
_______ 21,105 ______________ 80 OR x =
135 − √_______
21,105 ______________ 80
Give answer to 2 d.p.
∴ x = 3.50|3... OR x = −0.12|8...
x = 3.50 OR x = −0.13
Q. 28. (x + 2)2 − 2(2x − 3)2 = 10
(x + 2)(x + 2) − 2(2x − 3)(2x − 3) = 10
x2 + 4x + 4 − 2[4x2 − 12x + 9] = 10
x2 + 4x + 4 − 8x2 + 24x − 18 − 10 = 0
−7x2 + 28x − 24 = 0 (× −1)
7x2 − 28x + 24 = 0
a = 7, b = −28, c = 24
b2 − 4ac = (−28)2 − 4(7)(24)
= 784 − 672
= 112
∴ x = −(–28) ± √
____ 112 ______________ 2 × 7
x = 28 + √____
112 __________ 14 OR x = 28 − √____
112 __________ 14
Give answer as surd.
Note: √____
112 = √_______
16 × 7
= √___
16 × √__
7
= 4 √__
7
∴ x = 28 + 4 √__
7 _________ 14 OR x = 28 − 4 √__
7 _________ 14
Dividing by 2 gives
x = 14 + 2 √__
7 _________ 7 OR x = 14 − 2 √__
7 _________ 7
Exercise 17.4
Q. 1. Solve x2 + 9x + 20 = 0
(x + 4)(x + 5) = 0
x + 4 = 0 OR x + 5 = 0
x = −4 OR x = −5
Use this to solve
(y + 1)2 + 9(y + 1) + 20 = 0
∴ y + 1 = −4 OR y + 1 = −5
y = −5 OR y = −6
Q. 2. Solve x2 − 4x − 32 = 0
(x + 4)(x − 8) = 0
x + 4 = 0 OR x − 8 = 0
x = −4 OR x = 8
Use this to solve
(4t)2 − 4(4t) − 32 = 0
∴ 4t = −4 OR 4t = 8
t = −1 OR t = 2
Q. 3. Solve x2 − 2x − 3 = 0
(x + 1)(x − 3) = 0
x + 1 = 0 OR x − 3 = 0
x = −1 OR x = 3
Use this to solve
(t + 1)2 − 2(t + 1) − 3 = 0
∴ t + 1 = −1 OR t + 1 = 3
t = −2 OR t = 2
Q. 4. (i) Solve x2 − 2x − 15 = 0
(x + 3)(x − 5) = 0
x + 3 = 0 OR x − 5 = 0
x = −3 OR x = 5
(ii) Hence, solve
(3a − 1)2 − 2(3a − 1) − 15 = 0
3a − 1 = −3 OR 3a − 1 = 5
3a = −2 OR 3a = 6
a = − 2 __ 3 OR a = 2
14 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 5. Solve 2x2 + 7x − 4 = 0 (2x − 1)(x + 4) = 0 2x − 1 = 0 OR x + 4 = 0
2x = 1 OR x = −4
x = 1 __ 2
Hence, solve
2 ( 2p − 1 __ 2 ) 2 + 7 ( 2p − 1 __ 2 ) − 4 = 0
∴ 2p − 1 __ 2 = 1 __ 2 OR 2p − 1 __ 2 = −4
2p = 1 OR 2p = −3 1 __ 2
p = 1 __ 2 OR p = − 7 __ 4
Q. 6. Solve x2 + 2x − 24 = 0
(x − 4)(x + 6) = 0
x − 4 = 0 OR x + 6 = 0
∴ x = 4 OR x = −6
Hence, solve
24 − 2(2y + 2) − (2y + 2)2 = 0
i.e. × −1 and rearrange order
(2y + 2)2 + 2(2y + 2) − 24 = 0
∴ 2y + 2 = 4 OR 2y + 2 = −6
2y = 2 OR 2y = −8
y = 1 OR y = −4
Q. 7. (i) Solve 2x2 − 9x − 18 = 0
(2x + 3)(x − 6) = 0
2x + 3 = 0 OR x − 6 = 0
2x = −3 OR x = 6
x = − 3 __ 2
(ii) Hence, solve
2(2a2 − 2)2 − 9(2a2 − 2) − 18 = 0
∴ 2a2 – 2 = −3 ___ 2 OR 2a2 − 2 = 6
2a2 = 1 __ 2 OR 2a2 = 8
a2 = 1 __ 4 OR a2 = 4
∴ a = ± √__
1 __ 4 OR a = ± √__
4
a = ± 1 __ 2 OR a = ±2
∴ a = 1 __ 2 , − 1 __ 2 , 2 OR −2
Q. 8. Solve x2 − 6x − 16 = 0
(x + 2)(x − 8) = 0
x + 2 = 0 OR x − 8 = 0
x = −2 OR x = 8
Hence, solve
(3q2 − 5q)2 − 6(3q2 − 5q) − 16 = 0 ∴ 3q2 − 5q = −2 OR 3q2 − 5q = 8 3q2 − 5q + 2 = 0 OR 3q2 − 5q − 8 = 0 (3q − 2)(q − 1) = 0 OR (3q − 8)(q + 1) = 0
15Active Maths 2 (Strands 1–5): Ch 17 Solutions
3q – 2 = 0 OR q − 1 = 0 OR 3q − 8 = 0 OR q + 1 = 0 3q = 2 OR q = 1 OR 3q = 8 OR q = −1
q = 2 __ 3 OR q = 8 __ 3
∴ q = 2 __ 3 , 1, 8 __ 3 , −1
Q. 9. Solve to 2 d.p. x2 − 2x − 7 = 0
a = 1, b = −2, c = −7
b2 − 4ac = (−2)2 − 4(1)(−7)
= 4 + 28
= 32
Using the quadratic formula:
x = −b ± √________
b2 − 4ac _______________ 2a
gives x = −(−2) ± √
___ 32 _____________ 2 × 1
∴ x = 2 + √___
32 ________ 2 OR x = 2 − √___
32 ________ 2
x = 3.82|8... OR x = −1.82|8...
x = 3.83 OR x = −1.83
Hence, solve
(2p − 5)2 − 2(2p − 5) − 7 = 0 ∴ 2p − 5 = 3.83 OR 2p − 5 = −1.83 2p = 8.83 OR 2p = 3.17 p = 4.415 OR p = 1.585
Q. 10. (i) Solve to 1 decimal place: 5x2 − 2x − 15 = 0
a = 5, b = −2, c = −15
b2 − 4ac = (−2)2 − 4(5)(−15)
= 4 + 300
= 304
∴ x = −(−2) ± √
____ 304 ______________ 2 × 5
x = 2 + √____
304 _________ 10 OR x = 2 − √____
304 _________ 10
x = 1.9|4... OR x = −1.5|4...
x = 1.9 OR x = −1.5
(ii) Hence, solve
5 ( a __ 2 + 1 ) 2 − 2 ( a __ 2 + 1 ) − 15 = 0
∴ a __ 2 + 1 = 1.9 OR a __ 2 + 1 = −1.5
a __ 2 = 0.9 OR a __ 2 = −2.5
a = 1.8 OR a = −5
Q. 11. Solve 12x2 − 11x + 2 = 0
Factors of 24 that add to give –11:
−3 × − 8 = 24
−3 −8 = −11
∴ 12x2 − 11x + 2 = 0
12x2 − 3x − 8x + 2 = 0
3x(4x − 1) − 2(4x − 1) = 0
(3x − 2)(4x − 1) = 0
3x − 2 = 0 OR 4x − 1 = 0
3x = 2 OR 4x = 1
x = 2 __ 3 OR x = 1 __ 4
Hence, solve
12(3y2 + y)2 − 11(3y2 + y) + 2 = 0
∴ 3y2 + y = 2 __ 3 OR 3y2 + y = 1 __ 4
Solving 3y2 + y = 2 __ 3 (× 3)
9y2 + 3y = 2
9y2 + 3y − 2 = 0
16 Active Maths 2 (Strands 1–5): Ch 17 Solutions
(3y − 1)(3y + 2) = 0
3y − 1 = 0 OR 3y + 2 = 0
3y = 1 OR 3y = −2
y = 1 __ 3 OR y = − 2 __ 3
Solving 3y2 + y = 1 __ 4 (× 4)
12y2 + 4y = 1
12y2 + 4y − 1 = 0
(6y − 1)(2y + 1) = 0
6y − 1 = 0 OR 2y + 1 = 0
6y = 1 OR 2y = −1
y = 1 __ 6 OR y = − 1 __ 2
∴ y = 1 __ 3 , − 2 __ 3 , 1 __ 6 , − 1 __ 2
Exercise 17.5
Q. 1. Pair of roots 1, 2
x2 − ( Sum of roots ) x + ( Product of roots
) = 0
∴ x2 − (1 + 2)x + (1 × 2) = 0
x2 − 3x + 2 = 0
Q. 2. Pair of roots 5, 6
x2 − ( Sum of roots ) x + ( Product of roots
) = 0
∴ x2 − (5 + 6)x + (5 × 6) = 0
x2 − 11x + 30 = 0
Q. 3. Pair of roots 4, −2
x2 − ( Sum of roots ) x + ( Product of roots
) = 0
∴ x2 − (4 − 2)x + (4 × −2) = 0
x2 − 2x − 8 = 0
Q. 4. Pair of roots 7, −5
x2 − (7 − 5)x + (7 × − 5) = 0
∴ x2 − 2x − 35 = 0
Q. 5. Pair of roots −11, 11
x2 − (−11 + 11)x + (−11 × 11) = 0
∴ x2 − 121 = 0
Q. 6. Pair of roots 8, 0
x2 − (8 + 0)x + (8 × 0) = 0
∴ x2 − 8x = 0
Q. 7. Pair of roots −5, −4
x2 − (−5 − 4)x + (−5 × −4) = 0
x2 + 9x + 20 = 0
Q. 8. Pair of roots 0, −6
x2 − (0 − 6)x + (0 × −6) = 0
x2 + 6x = 0
Q. 9. Pair of roots ±5
x = ±5
x2 = 25
x2 − 25 = 0
Q. 10. Pair of roots − 1 __ 3 , 2 __ 7
x = − 1 __ 3 OR x = 2 __ 7
3x = −1 OR 7x = 2
3x + 1 = 0 OR 7x − 2 = 0
∴ (3x + 1)(7x − 2) = 0
3x(7x − 2) + 1(7x − 2) = 0
21x2 − 6x + 7x − 2 = 0
21x2 + x − 2 = 0
Q. 11. Roots of x2 + px + q = 0 are 3 and −2
x2 − (3 − 2)x + (3 × − 2) = 0
x2 − x − 6 = 0
∴ p = −1 and q = −6
Q. 12. Roots of x2 + bx + c = 0 are 0 and −4
x = 0 and x = −4
x = 0 and x + 4 = 0
∴ x(x + 4) = 0
x2 + 4x = 0
∴ b = 4 and c = 0
17Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 13. Given roots the same:
x2 − (a + a)x + (a × a) = 0
i.e. x2 − 2ax + a2 = 0
but x2 − 8x + c = 0
∴ 2a = 8 hence a = 4
∴ c = 4 × 4
c = 16
Exercise 17.6
Q. 1. x = positive number
x2 + 3x = 88
x2 + 3x – 88 = 0
(x + 11)(x − 8) = 0
∴ x + 11 = 0 OR x − 8 = 0
x = −11 OR x = 8
Since x is a positive number, the number is 8.
Q. 2. x = 1st number
x + 2 = 2nd number
x(x + 2) = 399
x2 + 2x − 399 = 0
(x − 19)(x + 21) = 0
x − 19 = 0 OR x + 21 = 0
x = 19 OR x = −21
If x = 19 then x + 2 = 21
19 × 21 = 399 as required
x = −21 rejected as −21 ∉ N
∴ The numbers are 19 and 21
Q. 3. Let x = the number
x2 + 15 = 8x
x2 − 8x + 15 = 0
(x − 3)(x − 5) = 0
x − 3 = 0 OR x − 5 = 0
x = 3 OR x = 5
The number could be 3 OR 5
Q. 4. Area of rectangle = length × width
∴ x(x + 2) + (2x + 3)(x + 1) + x(x) = 189
x2 + 2x + 2x2 + 2x + 3x + 3 + x2 = 189
4x2 + 7x − 186 = 0
4x2 − 24x + 31x − 186 = 0
4x(x − 6) + 31(x − 6) = 0
(4x + 31) (x − 6) = 0
4x + 31 = 0 OR x − 6 = 0
4x = −31 OR x = 6
x = − 31 ___ 4
x = − 31 ___ 4 rejected as length is positive
∴ x = 6
Q. 5. Perimeter of rectangle = 160 m
(i) P = 2 × length + 2 × width
If x = length
then 160 = 2x + 2 × width
160 − 2x = 2 × width
160 − 2x _________ 2 = width
∴ width = 80 − x
(ii) Area of rectangle = 1536 m2
Area of rectangle = l × w
∴ x(80 − x) = 1,536
80x − x2 = 1,536 (× −1)
x2 − 80x = −1,536
i.e. x2 − 80x + 1,536 = 0
(x − 32)(x − 48) = 0
∴ x − 32 = 0 OR x − 48 = 0
x = 32 OR x = 48
If x = 32 then length = 32
and width = 80 − 32
= 48
If x = 48 then length = 48
and width = 80 − 48
= 32
∴ The dimensions of the rectangle are 32 m and 48 m.
18 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 6. (i) Area of rectangle = l × w
∴ x(x + 5) = 14
x2 + 5x − 14 = 0
(x − 2)(x + 7) = 0
x − 2 = 0 OR x + 7 = 0
x = 2 OR x = −7
x = −7 rejected as length is positive
∴ x = 2 cm
(ii) Area of parallelogram = b × h
∴ (2x − 5)(x + 3) = 21
2x2 + 6x − 5x − 15 − 21 = 0
2x2 + x − 36 = 0
(2x + 9)(x − 4) = 0
2x + 9 = 0 OR x − 4 = 0
x = − 9 __ 2 OR x = 4
x = − 9 __ 2 rejected as length is positive
∴ x = 4 cm
(iii) Area of triangle = 1 __ 2 bh
∴ 1 __ 2 (4x + 3)(2x − 7) = 34.5 (×2)
(4x + 3)(2x − 7) = 69
8x2 − 28x + 6x − 21 − 69 = 0
8x2 − 22x − 90 = 0 (÷2)
4x2 − 11x − 45 = 0
4x2 − 20x + 9x − 45 = 0
4x (x − 5) + 9 (x − 5)=0
(4x + 9)(x − 5) = 0
4x + 9 = 0 OR x – 5 = 0
x = − 9 __ 4 OR x = 5
x = − 9 __ 4 rejected as length is positive
∴ x = 5 cm
Q. 7. Garden
(i) x + 5
16 – 2x
Area = (x + 5)(16 − 2x)
= 16x − 2x2 + 80 − 10x
= −2x2 + 6x + 80
(ii) x + 5 + 2xx
x
xx16 – 2x + 2x
Garden + Path (total area covered)
Area = (3x + 5)(16)
= 48x + 80
Area of path only
= 48x + 80 − (−2x2 + 6x + 80)
= 48x + 80 + 2x2 − 6x − 80
= 2x2 + 42x
(iii) Given area of path = 67.5 m
2x2 + 42x = 67.5
2x2 + 42x − 67.5 = 0
a = 2, b = 42, c = −67.5
b2 − 4ac = (42)2 − 4(2)(−67.5) = 2,304
x = −42 ± √
______ 2,304 ______________ 4 = −42 ± 48 _________ 4
x = −22.5 OR x = 1.5
x = −22.5 rejected, as length is positive
∴ x = 1.5 m
Q. 8. (i) g = –2t2 + 20t + 7
In 1997 t = 7
∴ g = −2(7)2 + 20(7) + 7
g = 49
49,000 games sold in 1997
(ii) 1990 t = 0
g = −2(0)2 + 20(0) + 7 = 7
1991 t = 1
g = −2(1)2 + 20(1) + 7 = 25
19Active Maths 2 (Strands 1–5): Ch 17 Solutions
1992 t = 2
g = −2(2)2 + 20(2) + 7 = 39
1993 t = 3
g = −2(3)2 + 20(3) + 7 = 49
1994 t = 4
g = −2(4)2 + 20(4) + 7 = 55
1995 t = 5
g = −2(5)2 + 20(5) + 7 = 57
1996 t = 6
g = −2(6)2 + 20(6) + 7 = 55
1997 g = 49
1998 g = 39
1999 g = 25
1990 to 1994 = 175,000 games
1995 to 1999 = 225,000 games
∴ 1995 to 1999 more games were sold.
(iii) 25,000 games sold
∴ −2t2 + 20t + 7 = 25
−2t2 + 20t − 18 = 0
2t2 − 20t + 18 = 0 (÷ 2)
t2 − 10t + 9 = 0
(t − 1)(t − 9) = 0
∴ t − 1 = 0 OR t − 9 = 0
t = 1 OR t = 9
25,000 games were sold in 1991 and 1999
(iv) After 2000 the number of games sold gives a negative answer.
e.g. for 2001 t = 11
∴ g = –2(11)2 + 20(11) + 7
g = −15 i.e. −15,000 games and thus the formula is no longer valid
Q. 9. Sold n bars for (n + 1) cents. Solve for money taken each day:
n(n + 1) = 2(n + 1)(n − 2)
n2 + n = 2[n2 − 2n + n − 2]
n2 + n = 2n2 − 2n − 4
0 = n2 − 3n − 4
0 = (n + 1)(n − 4)
n + 1 = 0 OR n − 4 = 0
n = −1 OR n = 4
Since n ≥ 0, n = −1 rejected.
∴ n = 4
Q. 10. (i) Perimeter = 46 m
Length = x
Perimeter = 2l + 2w
∴ 2x + 2w = 46
2w = 46 − 2x
w = 23 − x
width = 23 − x
(ii) x – 8
x – 8
23
– x
x – 8x – 8x
Total area of path and pool:
length = x + x − 8 + x − 8
= 3x − 16
width = 23 − x + x − 8 + x − 8
= x + 7
Total area = (3x − 16)(x + 7)
= 3x2 + 5x − 112
Area of path = 140 m2
∴ 3x2 + 5x − 112 − x(23 − x) = 140
3x2 + 5x − 112 − 23x + x2 = 140
4x2 − 18x − 252 = 0
2x2 − 9x − 126 = 0
(2x − 21)(x + 6) = 0
2x − 21 = 0 OR x + 6 = 0
2x = 21 OR x = −6 ← reject
x = 10.5
Dimensions of pool = 10.5 and 23 − 10.5 = 12.5
Length = 10.5 m, width = 12.5 m
20 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 11. C(x) = 9 + 16x − 0.5x2
C = cost in euros
x = no. of chairs produced
(i) If x = 10
C = 9 + 16(10) − 0.5(10)2
C = 119
costs €119
(ii) 9 + 16x −0.5x2 = 135
−0.5x2 + 16x −126 = 0
x2 − 32x + 252 = 0
(x − 14)(x − 18)
x = 14 OR x = 18
Answer: 14 chairs
(iii) If x = 15
C = 9 + 16(15) − 0.5(15)2
C = €136.50
Chairs sold at €30 each
∴ Profit = €30 × 15 − €136.50
= €313.50
Q. 12. h(t) = −5t2 + 18t + 8
(i) When t = 0
h = −5(0)2 + 18(0) + 8
h = 8
Object fired from height of 8 m
(ii) Average speed in 1st second
when t = 1, h = −5(1)2 + 18(1) + 8
h = 21
Average Speed: Distance ________ time = 21 − 8 _______ 1
∴ Average speed = 13 m/s
(iii) When h = 0
−5t2 + 18t + 8 = 0
5t2 − 18t − 8 = 0
5t2 − 20t + 2t − 8 = 0
5t(t − 4) + 2(t − 4) = 0
(5t + 2)(t − 4) = 0
5t + 2 = 0 OR t − 4 = 0
t = − 2 __ 5 OR t = 4
Since t ≥ 0 ∴ t = 4 seconds.
It will take 4 seconds to hit the ground.
Q. 13. N(t) = 16t2 − 200t + 625
(i) When t = 0
N = 16(0)2 − 200(0) + 625
= 625
625 spores present
(ii) At t = 3
N = 16(3)2 − 200(3) + 625
= 169
169 spores after 3 minutes
(iii) 625 spores present initially
84% of 625 = 525 spores killed
∴ 100 spores left
∴ 16t2 − 200t + 625 = 100
16t2 − 200t + 525 = 0
a = 16, b = −200, c = 525
b2 − 4ac = (−200)2 − 4(16)(525)
= 6,400
Using the quadratic formula:
t = −(−200) ± √
______ 6,400 _________________ 2 × 16
t = 200 + 80 _________ 32 OR t = 200 − 80 _________ 32
t = 8 3 _ 4 OR t = 3 3 _ 4
∴ The shortest time is 3 3 _ 4 minutes
(iv) The laboratory technician may have recommended not releasing the fungicide for commercial use as 3 3 _ 4 minutes may be too long to wait to kill 84% of the spores initially present.
Q. 14. (i) Area of field
= (8x + 6)(13x + 11)
= 104x2 + 88x + 78x + 66
= 104x2 + 166x + 66
21Active Maths 2 (Strands 1–5): Ch 17 Solutions
(ii) Area covered by barn
= x(x + 5)
= x2 + 5x
(iii) Area unbuilt = 1,476 m2
∴ 104x2 + 166x + 66 − (x2 + 5x) = 1,476
103x2 + 161x − 1,410 = 0
a = 103, b = 161, c = −1,410
b2 − 4ac = (161)2 − 4(103)(−1,410)
= 606,841
Using the quadratic formula:
x = −161 ± √
________ 606,841 _________________ 2 × 103
x = −161 + 779 ____________ 206 OR x = −161 − 779 ____________ 206
x = 3 OR x = −4.56...
as x > 0, x = −4.56... rejected ∴ x = 3 Dimensions of the field for x = 3
are: 8x + 6 = 8(3) + 6 = 30 m and 13x + 11 = 13(3) + 11 = 50 m Dimensions of field: 50 m by 30 m
Q. 15. y = −2x2 + 9x + 1
y = height above ground (m)
x = horizontal distance (m)
(i) When x = 0
y = −2(0)2 + 9(0) + 1
y = 1
Height of ball: 1 m
(ii) When x = 2
y = −2(2)2 + 9(2) + 1
y = 11
Height of ball: 11 m
(iii) When y = 0 (i.e. ball landed)
⇒ −2x2 + 9x + 1 = 0
2x2 − 9x − 1 = 0
a = 2, b = −9, c = −1
b2 − 4ac = (−9)2 − 4(2)(−1)
= 81 + 8 = 89
Using the quadratic formula:
x = −(−9) ± √
___ 89 _____________ 2 × 2
x = 9 + √___
89 ________ 4 OR x = 9 − √___
89 ________ 4
x = 4.608... OR x = −0.108...
Since x ≥ 0 reject x = −0.108
x = 4.608...m (× 100)
∴ x = 460.8... cm
The ball landed 461 cm from the batter (to the nearest cm).
(iv) The batter wouldn’t have been pleased to hit the ball such a short distance of only 461 cm.
Q. 16. Using Pythagoras’ Theorem:
(i) (6x)2 + (3x − 7)2 = (8x − 21)2
36x2 + 9x2 − 42x + 49 = 64x2 − 336x + 441 45x2 − 42x + 49 = 64x2 − 336x + 441 0 = 19x2 − 294x + 392 a = 19, b = −294, c = 392 b2 − 4ac = (−294)2 − 4(19)(392) = 56,644
Using the quadratic formula:
x = −(−294) ± √
_______ 56,644 ___________________ 2 × 19
x = 294 + √
_______ 56,644 ______________ 38
OR
x = 294 − √
_______ 56,644 ______________ 38
x = 14 OR x = 28 ___ 19
If x = 28 ___ 19 then
3x − 7 = ( 3 × 28 ___ 19 − 7 ) = −2 11 __ 19
which is invalid, as length cannot be negative ∴ x = 28 __ 19 is rejected
∴ x = 14
22 Active Maths 2 (Strands 1–5): Ch 17 Solutions
(ii) Two shorter sides
(7x − 2)2 + (2x − 8)2
= 49x2 − 28x + 4 + 4x2 − 32x + 64
= 53x2 − 60x + 68
Longest side (hypotenuse)
(6x + 2)2 = 36x2 + 24x + 4
If it is a right-angled triangle
53x2 − 60x + 68 = 36x2 + 24x + 4
17x2 − 84x + 64 = 0
a = 17, b = −84, c = 64
b2 − 4ac = (−84)2 − 4(17)(64)
= 2,704
Using the quadratic formula
x = −(−84) ± √
______ 2,704 ________________ 2 × 17
x = 84 + √
______ 2,704 ____________ 34 OR x =
84 − √______
2,704 ____________ 34
x = 4 OR x = 16 ___ 17
If x = 4, then one of the sides
2x − 8 = 2(4) − 8 = 0
If x = 16 ___ 17 then 2 ( 16 ___ 17 ) − 8 = −6 2 __ 17
We cannot have a negative length side or a side of length zero.
∴ Abdul has indeed made a mistake.
Q. 17. (i) x = 4 ± √___
76 ________ 6
x = −b ± √________
b2 − 4ac _______________ 2a
∴ 2a = 6 ⇒ a = 3
−b = 4 ⇒ b = −4
b2 − 4ac = 76
∴ (−4)2 − 4(3)(c) = 76
16 − 12c = 76
−12c = 60
c = −5
The quadratic equation of the form
ax2 + bx + c = 0
where a = 3, b = −4, c = 5
is 3x2 − 4x − 5 = 0
(ii) 2x2 − 5x + 9 = 0
a = 2, b = −5, c = 9
b2 − 4ac = (−5)2 − 4(2)(9)
= 25 − 72
= −47
Since the discriminant b2 − 4ac is negative, the equation has no real solutions, i.e. we cannot evaluate √
_____ −47 .
(iii) A possible correction to the equation is to change +9 to −9,
so that b2 − 4ac = 25 + 72
= 97 which is positive.
Q. 18.
27x
36x
60 km
N
E
Where x = time in hours
(27x)2 + (36x)2 = 602
729x2 + 1,296x2 = 3,600
2,025x2 = 3,600
x2 = 3,600
______ 2,025
x = √______
3,600
______ 2,025
x = 60 ___ 45
x = 1 1 _ 3 hours
x = 1 hour 20 mins
Started at 08:50 then 60 km apart after 1 hour 20 mins
∴ time = 10:10
Q. 19. y = − 9x2 ___ 20 + 5x ___ 2 + 11
y = height above sea level (m)
x = horizontal distance from cliff face (m)
(i) When x = 0
y = −9 ___ 20 (0)2 + 5 __ 2 (0) + 11
Pebble is 11 m high
23Active Maths 2 (Strands 1–5): Ch 17 Solutions
(ii) When x = 3
y = −9 ___ 20 (3)2 + 5 __ 2 (3) + 11
y = 14 9 __ 20
Height of pebble is 14 9 __ 20 m (or 14.45 m)
(iii) When y = 0
−9 ___ 20 x2 + 5 __ 2 x + 11 = 0 (× −20)
9x2 − 50x − 220 = 0
a = 9, b = −50, c = −220
b2 − 4ac = (−50)2 − 4(9)(−220) = 10,420
∴ x = −(−50) ± √
_______ 10,420 _________________ 2 × 9
x = 50 + √
_______ 10,420 _____________ 18 OR x =
50 − √_______
10,420 _____________ 18
x = 8.44|88... OR x = −2.8932...
as x > 0 reject x = −2.8932...
∴ x = 8.45 to 2 d.p.
The pebble is 8.45 m (to two decimal places) from the cliff face.
Q. 20. d = 0.0058v2 + 0.20v
d = stopping distance (m)
v = speed (km/hr)
(i) v = 70 km/hr
d = 0.0058(70)2 + 0.20(70)
d = 42.42
Stopping distance = 42.42 m
(ii) If d = 100 m
0.0058v2 + 0.20v = 100 (× 10,000)
58v2 + 2,000v − 1,000,000 = 0 (÷ 2)
29v2 + 1,000v − 500,000 = 0
a = 29, b = 1,000, c = −500,000
b2 − 4ac = (1,000)2 − 4(29)(−500,000) = 59,000,000
Using the quadratic formula:
v = −1,000 ± √
___________ 59,000,000 _____________________ 2 × 29
v = 115.192... OR v = −149.674...
Since v ≥ 0, v = −149.67... rejected
The speed of the car was 115 km/hr (to nearest km/hr).
24 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 21. (i) y
y
xxx
y + y + x + x + x = 200
2y + 3x = 200
2y = 200 − 3x
y = 200 − 3x _________ 2
(ii) Area of one of the fields
x × y __ 2 = x × 1 __ 2 × ( 200 − 3x _________ 2 )
= 200x − 3x2 __________ 4
(iii) Area of field = 833 1 _ 3 m2
∴ 200x − 3x2 __________ 4 = 833 1 _ 3 (× 4)
200x − 3x2 = 10,000
_______ 3 (× 3)
600x − 9x2 = 10,000
9x2 − 600x + 10,000 = 0
9x2 − 300x − 300x + 10,000 = 0
3x(3x − 100) − 100(3x − 100) = 0
(3x − 100)(3x − 100) = 0
∴ 3x − 100 = 0
3x = 100
x = 33 1 _ 3
If x = 33 1 _ 3 and y = 200 − 3x _________ 2
then y = 200 − 3 ( 33 1 _ 3 )
____________ 2
∴ y = 50
The value of x is 33 1 __ 3 m and the value of y is 55 m
Q. 22. R(x) = 80x − 1 __ 2 x2
x = no. of furniture units
(i) When x = 50
R = 80(50) − 1 __ 2 (50)2
R = 4,000 − 1,250
R = 2,750
Revenue of €2,750
(ii) If R = €3,000
80x − 1 __ 2 x2 = 3,000 (× −2)
−160x + x2 = −6,000
x2 − 160x + 6,000 = 0
(x − 60)(x − 100) = 0
∴ x − 60 = 0 OR x − 100 = 0
x = 60 OR x = 100
60 units must be sold
(iii) c(x) = 10x + 50
where c = cost
If x = 25 then c = 10 × 25 + 50
c = 300
Cost is €300 to produce 25 units
(iv) If c = €750
then 10x + 50 = 750
10x = 700
x = 70
∴ 70 units manufactured
(v) Profit = sales revenue − cost
If profit is €1,950 then
80x − 1 __ 2 x2 − (10x + 50) = 1,950
− 1 __ 2 x2 + 70x − 50 − 1,950 = 0
− 1 __ 2 x2 + 70x − 2,000 = 0 (×−2)
x2 − 140x + 4,000 = 0
(x − 40)(x − 100) = 0
∴ x = 40 OR x = 100
40 units must be made
Q. 23. h = 22 + 60t − 10t2
t = time in seconds
h = height in metres
(i) When h = 0
22 + 60t − 10t2 = 0 (×−1)
10t2 − 60t − 22 = 0 (÷ 2)
5t2 − 30t − 11 = 0
a = 5, b = −30, c = −11
25Active Maths 2 (Strands 1–5): Ch 17 Solutions
b2 − 4ac = (−30)2 − 4(5)(−11)
= 900 + 220
= 1,120
Using the quadratic formula
t = −(−30) ± √
______ 1,120 ________________ 2 × 5
t = 30 + √
______ 1,120 ____________ 10 OR t =
30 − √______
1,120 ____________ 10
t = 6.346... OR t = −0.346...
t = 6.3 to 1 d.p.
The cannonball hits the ground after 6.3 seconds to 1 d.p.
(ii) When t = 0
h = 22 + 60(0) − 10(0)2
h = 22
The cannonball was fired from 22 m
(iii)
height
time
36.346
–0.346
The maximum height occursat t = 3 seconds
i.e. h = 22 + 60(3) − 10(3)2
= 22 + 180 − 90
= 112
The maximum height the cannonball reaches is 112 m
∴ 120 m is not possible
Alternative solution:
Using a table
t 0 1 2 3 4 5 6
h 22 72 102 112 102 72 22
From the table of values one can see that after 3 seconds the cannonball’s height starts to reduce from its maximum of 112 m
Algebraically, 22 + 60t − 10t2 = 120 does not have a real solution
Revision Exercises
Q. 1. (a) (i) Solve
x2 + x − 20 = 0
(x − 4)(x + 5) = 0
x = 4 OR x = −5
(ii) x2 + 3x − 70 = 0
(x − 7)(x + 10) = 0
x = 7 OR x = −10
(iii) x2 − 25 = 0
x2 = 25
x = ± √___
25
x = ±5
(iv) x2 − 7x = 0
x(x − 7) = 0
∴ x = 0 OR x = 7
26 Active Maths 2 (Strands 1–5): Ch 17 Solutions
(b) Using quadratic formula
x = −b ± √________
b2 − 4ac _______________ 2a
(i) x2 + 5x − 3 = 0
a = 1, b = 5, c = −3
b2 − 4ac = (5)2 − 4(1)(−3) = 37
∴ x = −5 ± √___
37 __________ 2 × 1
x = −5 + √___
37 __________ 2 OR x = −5 − √___
37 __________ 2
x = 0.54|1... OR x = −5.54|1...
x = 0.54 OR x = −5.54 to 2 d.p.
(ii) x2 − 7x − 1 = 0 a = 1, b = −7, c = −1 b2 − 4ac = (−7)2 − 4(1)(−1) = 53
∴ x = −(−7) ± √
___ 53 _____________ 2 × 1
x = 7 + √___
53 ________ 2 OR x = 7 − √___
53 ________ 2
x = 7.14|0... OR x = −0.14|0...
x = 7.14 OR x = −0.14 to 2 d.p.
(iii) x2 − 4x − 2 = 0
a = 1, b = −4, c = −2
b2 − 4ac = (−4)2 − 4(1)(−2) = 24
∴ x = −(−4) ± √
___ 24 _____________ 2 × 1
x = 4 + √___
24 ________ 2 OR x = 4 − √___
24 ________ 2
x = 4.44|9... OR x = −0.44|9...
x = 4.45 OR x = −0.45 to 2 d.p.
(c) Solve x2 − 10x + 21 = 0
(x − 3)(x − 7) = 0
x = 3 OR x = 7
Hence solve
(2t + 1)2 − 10(2t + 1) + 21 = 0
2t + 1 = 3 ∴ 2t = 2 OR 2t + 1 = 7 ∴ 2t = 6
t = 1 OR t = 3
27Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 2. (a) (i) Solve
2x2 − 3x = 0
x(2x − 3) = 0
x = 0 OR 2x = 3
x = 3 __ 2
(ii) 10x2 + x − 2 = 0
(2x + 1)(5x − 2) = 0
2x + 1 = 0 OR 5x − 2 = 0
x = −1 ___ 2 OR x = 2 __ 5
(iii) 9x2 − 4 = 0
x2 = 4 __ 9
x = ± √__
4 __ 9
x = ± 2 __ 3
(b) (i) 2x2 + 8x − 3 = 0
a = 2, b = 8, c = −3
b2 − 4ac = (8)2 − 4(2)(−3) = 64 + 24 = 88
∴ x = −8 ± √___
88 __________ 2 × 2
Note: √___
88 = √_______
4 × 22 = 2 √___
22
∴ x = −8 + 2 √___
22 ___________ 4 OR x = −8 − 2 √___
22 ___________ 4
x = −4 + √___
22 __________ 2 OR x = −4 − √___
22 __________ 2
(ii) 5x2 − 11x + 1 = 0
a = 5, b = −11, c = 1
b2 − 4ac = (−11)2 − 4(5)(1) = 101
∴ x = −(−11) ± √
____ 101 _______________ 2 × 5
x = 11 + √____
101 __________ 10 OR x = 11 − √____
101 __________ 10
(iii) 6x2 − x − 3 = 0
a = 6, b = −1, c = −3
b2 − 4ac = (−1)2 − 4(6)(−3) = 73
∴ x = −(−1) ± √
___ 73 _____________ 2 × 6
x = 1 + √___
73 ________ 12 OR x = 1 − √___
73 ________ 12
28 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 3. (a) (i) Given roots 2 and 3
x2 − ( Sum of roots ) x + ( Product ________
of roots ) = 0
∴ x2 – (2 + 3)x + (2 × 3) = 0
x2 − 5x + 6 = 0
(ii) Given roots −4 and 2
x2 − (−4 + 2)x + (−4 × 2) = 0
x2 − (−2x) − 8 = 0
x2 + 2x − 8 = 0
(iii) Given roots 3 and −5
x2 − (3 − 5)x + (3 × −5) = 0
x2 − (−2x) − 15 = 0
x2 + 2x − 15 = 0
(iv) Given roots 2 and 0
x2 − (2 + 0)x + (2 × 0) = 0
x2 − 2x = 0
(b) (i) x2 = 5x + 7
x2 − 5x − 7 = 0
a = 1, b = −5, c = −7
b2 − 4ac = (−5)2 − 4(1)(−7) = 53
∴ x = −(−5) ± √
___ 53 _____________ 2 × 1
x = 5 + √___
53 ________ 2 OR x = 5 − √___
53 ________ 2
x = 6.140|0... OR x = −1.140|0... x = 6.140 OR x = −1.140 to 3 d.p.
(ii) 2x2 + 1 = 10x
2x2 − 10x + 1 = 0
a = 2, b = −10, c = 1
b2 − 4ac = (−10)2 − 4(2)(1) = 92
∴ x = −(−10) ± √
___ 92 ______________ 2 × 2
x = 10 + √___
92 _________ 4 OR x = 10 − √___
92 _________ 4
x = 4.897|9... OR x = 0.102|0...
∴ x = 4.898 OR x = 0.102 to 3 d.p.
29Active Maths 2 (Strands 1–5): Ch 17 Solutions
(iii) 3 − x − 3x2 = 0 (×−1) 3x2 + x − 3 = 0 a = 3, b = 1, c = −3 b2 − 4ac = (1)2 − 4(3)(−3) = 37
∴ x = −1 ± √___
37 __________ 2 × 3
x = −1 + √___
37 __________ 6 OR x = −1 − √___
37 __________ 6
x = 0.847|1... OR x = –1.180|4... x = 0.847 OR x = −1.180 to 3 d.p.
(c) (i) 21x2 + 2x − 3 = 0
(7x + 3)(3x − 1) = 0
∴ 7x + 3 = 0 OR 3x − 1 = 0
x = − 3 __ 7 OR x = 1 __ 3
(ii) 5x2 + 6 = 13x
5x2 − 13x + 6 = 0
(5x − 3)(x − 2) = 0
5x − 3 = 0 OR x − 2 = 0
∴ x = 3 __ 5 OR x = 2
(iii) 5(x2 − 4) = 2(x − 10)
5x2 − 20 = 2x − 20
5x2 − 2x = 0
x(5x − 2) = 0
∴ x = 0 OR x = 2 __ 5
Q. 4. (a) (i) x2 – ax – b = 0 with roots 7 and −2
x2 − ( Sum of roots ) x + ( Product of roots
) = 0
∴ x2 − (7 − 2)x + (7 × −2) = 0
x2 − 5x − 14 = 0
∴ a = 5 and b = 14
(ii) x2 + ax + b = 0 with roots −8 and −2
∴ x2 − (−8 − 2)x + (−8 × −2) = 0 x2 − (−10x) + 16 = 0 x2 + 10x + 16 = 0 ∴ a = 10 and b = 16
(b) Solve x2 − 10x − 11 = 0 (x + 1)(x − 11) = 0 ∴ x + 1 = 0 OR x − 11 = 0 x = −1 OR x = 11 Hence solve (3t − 1)2 − 10(3t − 1) − 11 = 0 3t − 1 = −1 OR 3t − 1 = 11 3t = 0 OR 3t = 12
t = 0 OR t = 4
30 Active Maths 2 (Strands 1–5): Ch 17 Solutions
(c) (i) Solve to 1 d.p. x2 + 2x − 10 = 0 a = 1, b = 2, c = −10 b2 − 4ac = (2)2 − 4(1)(−10) = 44
∴ x = −2 ± √___
44 __________ 2 × 1
x = −2 + √___
44 __________ 2 OR x = −2 − √___
44 __________ 2
x = 2.3|1... OR x = −4.3|1... x = 2.3 OR x = −4.3 to 1 d.p. (ii) Hence solve
(t + 1)2 + 2(t + 1) − 10 = 0
t + 1 = 2.3 OR t + 1 = −4.3
∴ t = 1.3 OR t = −5.3 to 1 d.p.
Q. 5. N(t) = −2t2 + 11t + 13
(i) When t = 0 N = −2(0)2 + 11(0) + 13 N = 13 ∴ 13,000 bacteria present initially
(ii) When N = 25
−2t2 + 11t + 13 = 25 −2t2 + 11t − 12 = 0 (× −1) 2t2 − 11t + 12 = 0 2t2 − 8t − 3t + 12 = 0 2t(t − 4) − 3(t − 4) = 0 (2t − 3)(t − 4) = 0 ∴ 2t − 3 = 0 OR t − 4 = 0
t = 3 __ 2 OR t = 4
As t is time in hours, there will be 25,000 bacteria present at 1 1 _ 2 hours and 4 hours.
(iii) When N = 0
−2t2 + 11t + 13 = 0 (×−1)
2t2 − 11t − 13 = 0
(2t − 13)(t + 1) = 0
∴ 2t − 13 = 0 OR t + 1 = 0
t = 13 ___ 2 OR t = −1
∴ No bacteria at 6 1 _ 2 hours. (t = −1 rejected, as time cannot be negative)
(iv) The formula only gives valid values between t = 0 and t = 6 1 _ 2 .Outside these times negative answers are given, e.g. if t = 7
N = −2(7)2 + 11(7) + 13
∴ N = −8 rejected, as we cannot have −8,000 bacteria present
Q. 6. Let x = side length of small square
∴ 1 − x = side length of large square (1 − x)2 = 2(x)2
(1 − x)(1 − x) = 2x2
1 − 2x + x2 − 2x2 = 0 −x2 − 2x + 1 = 0 (× −1) x2 + 2x − 1 = 0 a = 1, b = 2, c = −1 b2 − 4ac = (2)2 − 4(1)(−1) = 8
∴ x = −2 ± √__
8 _________ 2 × 1
x = −2 + √__
8 _________ 2 OR x = −2 − √__
8 _________ 2
x = 0.414|2... OR x = −2.414...
Since x > 0 reject x = −2.414...
∴ x = 0.414 to 3 d.p.
Side length small square = 0.414 m to 3 d.p.
Side length large square
= 1 − 0.414
= 0.586 m to 3 d.p.
31Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 7. x = width of rectangle
(i) length = 2x + 3
(ii) Area of rectangle = length × width
∴ Area = (2x + 3)x
= 2x2 + 3x
(iii) Area = 27 m2
∴ 2x2 + 3x = 27
2x2 + 3x − 27 = 0
2x2 − 6x + 9x − 27 = 0
2x(x − 3) + 9(x − 3) = 0
(2x + 9)(x − 3) = 0
∴ 2x + 9 = 0 OR x − 3 = 0
x = −9 ___ 2 OR x = 3
since x > 0 then x = 3
width of rectangle = 3 m
length of rectangle = 2 × 3 + 3
= 9 m
Q. 8. (i) 20 m length of wire
Perimeter of one square
= 4(x + 0.5)
= 4x + 2
∴ Perimeter of other square
= 20 − (4x + 2)
= 18 − 4x
∴ Side length of other square
= 18 − 4x _______ 4
= 4.5 − x
(ii) (x + 0.5)2 + (4.5 − x)2 = 14.33
x2 + x + 0.25 + 20.25 − 9x + x2 − 14.33 = 0
∴ 2x2 − 8x + 6.17 = 0
a = 2, b = −8, c = 6.17
b2 − 4ac = (−8)2 − 4(2)(6.17) = 14.64
x = −(−8) ± √
______ 14.64 _______________ 2 × 2
x = 8 + √______
14.64 ___________ 4 OR x = 8 − √______
14.64 ___________ 4
x = 2.95|6... OR x = 1.04|3
x = 2.96 OR x = 1.04 to 2 d.p.
We can use either value of x to find the side length of each square.
If x = 2.96 then side lengths are
x + 0.5 = 2.96 + 0.5 = 3.46
similarly
4.5 − x = 4.5 − 2.96
= 1.54
∴ Side lengths are 1.54 m and 3.46 m
32 Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 9. v(t) = 10 + 36t − 5t2
v = value of investment (‘000s) t = time in months (i) When t = 0 v = 10 + 36(0) − 5(0)2
v = 10 Original value of investment
€10,000 (ii) When t = 3 v = 10 + 36(3) − 5(3)2
v = 73 Increase: €73,000 − €10,000 = €63,000 Investment increased by €63,000
(iii) 10 + 36t −5t2 = 52.75
5t2 − 36t + 42.75 = 0
t = +36 ± √
___________________ (−36)2 − 4(5)(42.75) ___________________________ 2(5)
= 36 ± √____
441 __________ 10 = 36 ± 21 ________ 10
t = 1.5 months or 5.7 months
(iv) When t = 4
v = 10 + 36(4) − 5(4)2
v = 74 i.e. €74,000
When t = 5
v = 10 + 36(5) − 5(5)2
v = 65 i.e. €65,000
After 4 months the value of the investment starts to decline, dropping from €74,000 at 4 months to €65,000 at 5 months. This is why the investor chooses to sell after 4 months.
Q. 10. Area of garden = 6x2 + 11x − 10
6x2 + 11x − 10 = 6x2 + 15x − 4x − 10
= 3x(2x + 5) − 2(2x + 5)
= (3x − 2)(2x + 5)
Therefore, the original dimensions of the garden are 3x – 2 and 2x + 5
Area = l × w
Comparing the new area and the original area:
(3x − 2 + 2)(2x + 5 − 2)
= [6x2 + 11x − 10] − 10
3x(2x + 3) = 6x2 + 11x − 20
6x2 + 9x = 6x2 + 11x − 20
−2x = −20
∴ x = 10
OR
(3x − 2 − 2)(2x + 5 + 2)
= [6x2 + 11x − 10] − 10
(3x − 4)(2x + 7) = 6x2 + 11x − 20
6x2 + 21x − 8x − 28 = 6x2 + 11x − 20
2x = 8
∴ x = 4
If x = 10 If x = 43x − 2 = 3(10) − 2
= 28
3x − 2 = 3(4) − 2
= 102x + 5 = 2(10) + 5
= 25
2x + 5 = 2(4) + 5
= 13
Original dimensions of the garden are 25 m by 28 m OR 10 m by 13 m
6666xxxx666 2222 6x6 2 6666xxxx666 2222 6x6 2
Q. 11. d = 3t + t2 __ 4
d = distance in m
t = number of seconds
(i) When t = 1
d = 3(1) + 12 __ 4
d = 3 1 _ 4
3.25 m travelled in 1 second
33Active Maths 2 (Strands 1–5): Ch 17 Solutions
(ii) When d = 14
3t + t2 __ 4 = 14 (× 4)
12t + t2 = 56
t2 + 12t − 56 = 0
a = 1, b = 12, c = −56
b2 − 4ac = (12)2 − 4(1)(−56) = 368
t = −12 ± √____
368 ____________ 2 × 1
t = −12 + √____
368 ____________ 2 OR t = −12 − √____
368 ____________ 2
t = 3.5|91... OR t = −15.591...
Since t ≥ 0, t = −15.591... rejected
∴ t = 3.6 seconds to 1 d.p.
(iii) Average speed = distance travelled _______________ time taken
when d = 32
⇒ 3t + t2 __ 4 = 32 (× 4)
12t + t2 = 128
t2 + 12t − 128 = 0
a = 1, b = 12, c = −128
b2 − 4ac = (12)2 − 4(1)(−128) = 656
∴ t = −12 ± √____
656 ____________ 2 × 1
t = −12 + √____
656 ____________ 2 OR t = −12 − √____
656 ____________ 2
t = 6.8|0... OR t = −18.80...
Since t ≥ 0, t = −18.80... rejected
∴ t = 6.8 seconds to 1 d.p.
∴ Average speed = 32 m/6.8
= 4.7 m/s to 1 d.p.
Q. 12.
193 x
263 – x
Using Pythagoras’ theorem
x2 + (263 − x)2 = 1932
x2 + 69,169 − 526x + x2 − 37,249 = 0
2x2 − 526x + 31,920 = 0 (÷ 2)
x2 − 263x + 15,960 = 0
34 Active Maths 2 (Strands 1–5): Ch 17 Solutions
a = 1, b = −263, c = 15,960
b2 − 4ac = (−263)2 − 4(1)(15,960) = 5,329
x = −(−263) ± √
______ 5,329 _________________ 2 × 1
x = 263 + √
______ 5,329 _____________ 2 OR x =
263 − √______
5,329 _____________ 2
x = 168 OR x = 95
The lengths of the sides of the triangle are 193 cm, 168 cm and 95 cm.
Q. 13. h(t) = 54t − 5t2
h = height in m
t = time in seconds
(i) At t = 3
h = 54(3) − 5(3)2
h = 117
Height will be 117 m
(ii) When launched t = 0
∴ h = 54(0) − 5(0)2
h = 0
Height is zero metres
(iii) When h = 0
⇒ 54t − 5t2 = 0 (×−1)
5t2 − 54t = 0
t(5t − 54) = 0
∴ t = 0 OR t = 54 ___ 5
t = 10 4 _ 5
It will take 10.8 seconds for the rocket to return to ground.
Q. 14. x = width of garden
House
xx
y
80 metres of fencing
(i) y + x + x = 80
y = 80 − 2x
(ii) Area = l × w
= (80 − 2x)x
= 80x − 2x2
(iii) 80x − 2x2 = 487.5
2x2 − 80x + 487.5 = 0
a = 2, b = −80, c = 487.5
b2 − 4ac = (−80)2 −4(2)(487.5)
= 2,500
x = 80 ± √
______ 2,500 _____________ 2(2) = 80 ± 50 ________ 4
x = 32.5 OR x = 7.5
Answer 1: width = 32.5 m length = 80 − 2(32.5)
= 15 m
Answer 2: width = 7.5 m length = 80 − 2(7.5)
= 65 m
35Active Maths 2 (Strands 1–5): Ch 17 Solutions
Q. 15. x x x
x x x
yyyy
Length of fencing = 6x + 4y
Length of fencing = 800 m
∴ 6x + 4y = 800
4y = 800 − 6x
y = 200 − 3 __ 2 x
Area = ( 200 − 3 __ 2 x ) (3x) = 10,200
600x − 9x2 ___ 2 – 10,200 = 0 (× −2)
9x2 − 1,200x + 20,400 = 0
a = 9, b = −1,200, c = 20,400
b2 − 4ac = (−1,200)2 − 4(9)(20,400) = 705,600
x = −(−1,200) ± √
________ 705,600 _____________________ 2 × 9
x = 1,200 + 840
____________ 18 OR x = 1,200 − 840
____________ 18
x = 113 1 _ 3 OR x = 20
If x = 20 then y = 200 − 3 __ 2 (20)
y = 170
If x = 113 1 _ 3 then y = 200 − 3 __ 2 ( 113 1 _ 3 ) y = 30
∴ Possible sets of lengths and widths of each plot are 20 m and 170 m or 30 m and 113 1 _ 3 m.
Q. 16. (a) x1
7
2
6
3
4810 m
8 m
5
x
Eight sections numbered 1 to 8
Areas of 1, 3, 5 and 7 are the same at x × x
Areas of 2 and 6 are the same at 8 × x
36 Active Maths 2 (Strands 1–5): Ch 17 Solutions
(b) Areas of 4 and 8 are the same at 10 × x
∴ Overall area of path
= 4(x2) + 2(8x) + 2(10x)
= 4x2 + 16x + 20x
= 4x2 + 36x
Area of plot = 143 m2
Area of garden = 80 m2
∴ Area of path = 63 m2
Kevin’s equation is
Area of path + area of garden = total area
(4x2 + 36x) + 80 = 143
Simplified:
4x2 + 36x − 63 = 0
(c)
xx
10 m
8 m
2x + 8
2x + 10
x
(d) Area of plot = 143 m2
From Elaine’s diagram total area = (2x + 8)(2x + 10)
∴ (2x + 8)(2x + 10) = 143
4x2 + 20x + 16x + 80 − 143 = 0
4x2 + 36x − 63 = 0
(e) Solving 4x2 + 36x − 63 = 0
a = 4, b = 36, c = −63
b2 − 4ac = (36)2 − 4(4)(−63)
= 2,304
∴ x = −36 ± √
______ 2,304 ______________ 2 × 4
∴ x = −36 + 48 _________ 8 OR x = −36 − 48 _________ 8
x = 1 1 _ 2 OR x = −10 1 _ 2
Since x > 0, x = –10 1 _ 2 rejected
∴ x = 1 1 _ 2
The width of the path is 1.5 m.
37Active Maths 2 (Strands 1–5): Ch 17 Solutions
(f) Tony’s method of guess and check.
If x = 1 then the overall dimensions of the plot would have been 8 + 1 + 1 = 10
and 10 + 1 + 1 = 12
Area: 10 × 12 = 120 m2 which is smaller than 143 m2 so the path must be greater than 1 m
If x = 2 then the dimensions
are 8 + 2 + 2 = 12
and 10 + 2 + 2 = 14
giving an area of 12 × 14 = 168 m2
which is too big
If x = 1.5 then the dimensions
are 8 + 1.5 + 1.5 = 11
and 10 + 1.5 + 1.5 = 13
giving 11 × 13 = 143 m2 is the correct area
(g) For simple whole numbers like a path of 1 or 2 m, Tony’s method is quick and easy.
However, for large numbers or decimal lengths, Elaine’s method is the most direct.
Q. 17. (a) Larger length: Shorter length
Sum of lengths: Larger length
p _____
q
p + q _________
p
(b) x = p
__ q
(c) 1 + 1 __ x
(d) p
__ q = p + q
______ p
p
__ q = 1 + q __ p
x = 1 + 1 __ x x ≠ 0
x2 = x + 1
x2 − x − 1 = 0
(e) x = 1 ± √__
5 _______ 2
x ≈ −0.618 OR x ≈ 1.618
(f) x1 = −0.618
x2 = 1.618
(g) x2
(h) Correct to the nearest thousandth, the golden ratio is equal to 1.618. So, if one quantity is 61.8% larger than another quantity, then the two quantities are in golden ratio to each other.
(i) − 1 __ x1 = 1.618 = x2 = golden ratio
(to three decimal places)
(j) 1 ______ 1.618 ≈ 0.618
So, 1 + 1 ______ 1.618 ≈ 1.618
So, the golden ratio minus its reciprocal equals 1.
OR
x = 1 + 1 __ x (from part (d))
So, the golden ratio minus its reciprocal equals 1.
(k) 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
38 Active Maths 2 (Strands 1–5): Ch 17 Solutions
(l) (Term 2) ÷ (Term 1) 1.000(Term 3) ÷ (Term 2) 2.000(Term 4) ÷ (Term 3) 1.500(Term 5) ÷ (Term 4) 1.667(Term 6) ÷ (Term 5) 1.600(Term 7) ÷ (Term 6) 1.625(Term 8) ÷ (Term 7) 1.615(Term 9) ÷ (Term 8) 1.619
(Term 10) ÷ (Term 9) 1.618
(m) Consecutive terms in the Fibonacci sequence are approximately in golden ratio to each other the further we move into the sequence.
