Chapter 17 Solutions y3f6t

Chapter 17 Exercise 17.1 Q. 1. x2 + 10x + 21 = 0

Checking solutions x2 − 3x + 2 = 0

(x + 3)(x + 7) = 0 x+3=0

OR

(1)2 − 3(1) + 2 = 0

x+7=0

x = −3 OR

3−3=0

x = −7

0=0

Checking solutions x2

+ 10x + 21 = 0

True

∴ x = 1 is a solution.

(−3)2 + 10(−3) + 21 = 0

x2 − 3x + 2 = 0

9 − 30 + 21 = 0

(2)2 − 3(2) + 2 = 0

30 − 30 = 0

4−6+2=0

0=0

6−6=0

True

0=0

∴ x = −3 is a solution.

True

∴ x = 2 is a solution.

x2 + 10x + 21 = 0 (−7)2 + 10(−7) + 21 = 0

Q. 3. x2 − 5x = 0

49 − 70 + 21 = 0

x(x − 5) = 0

70 − 70 = 0 0=0

x=0

OR

x−5=0 x=5

True Checking solutions

∴ 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

x2 − 5x = 0 x2 − 5x = 0 (5)2 − 5(5) = 0 (0)2 − 5(0) = 0 0−0=0 25 − 25 = 0 True 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 (−4)2 − (−4) − 20 = 0 16 + 4 − 20 = 0 20 − 20 = 0 0=0 True ∴ x = −4 is a solution

x2 − x − 20 = 0 (5)2 − (5) − 20 = 0 25 − 5 − 20 = 0 25 − 25 = 0 0=0 True ∴ x = 5 is a solution

Active Maths 2 (Strands 1–5): Ch 17 Solutions

1

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 5 x = −__ 3

x = −1

Checking solutions 3x2 + 8x + 5 = 0 52 5 3 −__ + 8 −__ + 5 = 0 3 3 25 40 3 × ___ – ___ + 5 = 0 93 3

( )

3x2 + 8x + 5 = 0

( )

3(−1)2 + 8(−1) + 5 = 0 3−8+5=0 8−8=0

0=0 40 15 − ___ + ___ = 0 3 3 True 40 40 ___ ___ − =0 ∴ x = –1 is a solution 3 3 0=0 True 5 __ ∴ x = − is a solution 3 (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.) 25 ___ 3

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

7q = − 1 OR 1 q = −__ 7

2

q+1=0 q = −1


Checking solutions 7q2 + 8q + 1 = 0

7q2 + 8q + 1 = 0

12 1 7 −__ + 8 −__ + 1 = 0 7 7 8 1 ___ __ − +1=0 7 7 49 8 __ 1 __ 7 __ − + =0 7 7 7 8 __ 8 __ − =0 7 7 0=0

7(−1)2 + 8(−1) + 1 = 0

( )

( ) ( )

7−8+1=0 8−8=0 0=0 True ∴ q = –1 is a solution

True

1 __ ∴ q = − is a solution 7

Q. 10. 4x2 − 16x = 0

Q. 15. 81 − 4x2 = 0

(÷ 4)

x2 − 4x = 0

(9)2 − (2x)2 = 0

x(x − 4) = 0

(9 + 2x)(9 − 2x) = 0

x = 0 OR

x−4=0

9 + 2x = 0

x=4 Q. 11.

3y2

2x = −9 OR

− 2y − 1 = 0

3y = −1 1 y = −__ 3

OR

y−1=0

OR

y=1

−2x = −9 9 x = __ 2

Q. 16. 5a2 = 2a 5a2 − 2a = 0 a(5a − 2) = 0

Q. 12. 3x2 + 20x − 7 = 0

a = 0 OR

(3x − 1)(x + 7) = 0 3x − 1 = 0

OR

3x = 1 1 x = __ 3

OR

5a − 2 = 0 5a = 2 2 a = __ 5

x+7=0 x = −7 Q. 17. 100x2 − 25 = 0 4x2 = 1

(3x + 1)(x + 6) = 0 3x + 1 = 0

OR

3x = −1 OR 1 x = −__ 3 Q. 14. −2x2 + 11x = 0

(÷ 25)

4x2 − 1 = 0

Q. 13. 3x2 + 19x + 6 = 0 x+6=0 x = −6

(× −1)

2x2 − 11x = 0

1 x2 = __ 4 __ 1 x = ± __ 4

√

1 x = ±__ 2

Q. 18. 2x2 + 2x − 12 = 0

x (2x − 11) = 0 x = 0 OR

9 − 2x = 0

9 x = −__ OR 2 9 x = ±__ 2

(3y + 1)(y − 1) = 0 3y + 1 = 0

OR

2x − 11 = 0 2x = 11 11 x = ___ 2

(÷ 2)

x2 + x − 6 = 0 (x − 2)(x + 3) = 0 x − 2 = 0 OR x = 2 OR

x+3=0 x = −3


3

Q. 19. 25x2 − 25 = 0 x2

Q. 25. 3x2 − 24x + 36 = 0

(÷25)

−1=0

x2 − 8x + 12 = 0

x2 = 1

(x − 2)(x − 6) = 0

__

x = ±√ 1 x = ±1 Q. 20. 2b2 − b = 10

x − 2 = 0 OR

x−6=0

x = 2 OR

x=6

Q. 26. 5y2 − 125 = 0

2b2 − b − 10 = 0

5y2 = 125

(2b − 5)(b + 2) = 0 2b − 5 = 0 OR

y2 = 25

b+2=0

2b = 5 OR 5 b = __ 2

b = −2

y = ±5 Q. 27. (2x + 1)(x − 3) − 4 = 0 2x2 − 6x + x − 3 − 4 = 0

Q. 21. 2x2 − 3x = 0

2x2 − 5x − 7 = 0

x(2x − 3) = 0 x = 0 OR

(2x − 7)(x + 1) = 0

2x − 3 = 0

2x − 7 = 0

2x = 3 3 x = __ 2

2x = 7

3x2 − 9x + 4x − 12 − 10 = 0

(2x + 1)(2x + 3) = 0 2x + 7 = 0

(3x − 11)(x + 2) = 0 OR

x = −2

2x + 1 = 0

OR 25x − 14 = 0 25x = 14 14 x = ___ 25

OR

2x = −1 OR 1 x = −__ OR 2

(25x + 14)(25x − 14) = 0

3x + 2 = 0 3x = −2 2 x = −__ 3

Q. 30. 9x2 − 7x − 2 = 0 (9x + 2)(x − 1) = 0 9x + 2 = 0

Q. 24. 5y2 = 7y

OR

x−1=0

9x = −2 2 x = −__ 9

− 7y = 0

y(5y − 7) = 0 y = 0 OR 5y − 7 = 0

Q. 31. −6x2 + 5x + 1 = 0

5y = 7 7 y = __ 5

6x2


(× −1)

− 5x − 1 = 0

(6x + 1)(x − 1) = 0 6x + 1 = 0

OR

6x = −1 OR 1 x = −__ 6

4

2x = −3 3 x = −__ 2

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

(25x)2 − (14)2 = 0

25x = −14 OR 14 x = −___ OR 25

2x + 3 = 0

Q. 29. 6x2 + 7x + 2 = 0

Q. 23. 625x2 − 196 = 0

25x + 14 = 0

OR

2x = −1 OR 1 x = −__ OR 2

x+2=0

3x = 11 OR 11 x = ___ 3

x+1=0

Q. 28. 4x2 + 8x + 3 = 0

3x2 − 5x − 22 = 0 3x − 11 = 0

OR

7 x = __ 2

Q. 22. (3x + 4)(x − 3) = 10

5y2

(÷ 3)

x−1=0 x=1

Q. 32. 4x2 − 13x + 3 = 0

Checking solutions (non calculator working shown)

(4x − 1)(x − 3) = 0 4x − 1 = 0 OR

x−3=0

4x = 1 OR 1 x = __ 4

x=3

10m2 + 29m − 21 = 0

()

Q. 33. 10x2 − 37x + 7 = 0

18 ___

(5x − 1)(2x − 7) = 0 5x − 1 = 0 OR

2x − 7 = 0

5x = 1 OR 1 x = __ OR 5

2x = 7 7 x = __ 2

5

∴m=

(÷5) Q. 34. 80a2 − 245 = 0 2 16a − 49 = 0 16a2 = 49 49 a2 = ___ 16 ___ 49 a = ± ___ 16 7 __ a=± 4 Q. 35.

is a solution

( )

( )

245 ____ 2

203 42 − ____ − ___ = 0 2 2 245 ____ 2

+ 16b + 15 = 0

2b + 3 = 0

2b + 5 = 0

OR

2b = −3 OR 3 b = −__ OR 2 Q. 36.

5

2b = −5 5 b = −__ 2

− 6 + 13c = 0

True

∴ m = − is a solution 2 1 Q. 38. __x2 + 3x + 5 = 0 4 x2 + 12x + 20 = 0

(× 4)

x = −10

(× 10) Q. 39. x2 − 2.4x − 6.4 = 0 2 10x − 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 8 x = −__ 5

c+2=0

8c = 3 OR 3 c = __ 8

c = −2

Q. 37. 21 = 10m2 + 29m 10m2 + 29m − 21 = 0 (5m − 3)(2m + 7) = 0 5m = 3 OR 3 m = __ OR 5

0=0

7 __

x = −2 OR

(8c − 3)(c + 2) = 0

5m − 3 = 0 OR

245 − ____ = 0 2

(x + 10)(x + 2) = 0

8c2 + 13c − 6 = 0 8c − 3 = 0 OR

True

72 7 10 −__ + 29 −__ − 21 = 0 2 2 49 203 105 × ___ − ____ − 21 = 0 42 2

(2b + 3)(2b + 5) = 0

8c2

3 __

87 105 + ___ − ____ = 0 5 5 105 ____ 105 ____ − =0 5 5 0=0

10m2 + 29m − 21 = 0

√

4b2

()

32 3 10 __ + 29 __ − 21 = 0 5 5 2 87 9 10 × ____ + ___ − 21 = 0 5 25 5

2m + 7 = 0 2m = –7 7 m = −__ 2

5 x2 x Q. 40. __ + __ − __ = 0 3 2 6

(× 6)

2x2 + 3x − 5 = 0 (2x + 5)(x − 1) = 0 2x + 5 = 0 OR x − 1 = 0 2x = −5 OR x=1 5 x = −__ 2


5

3x2 ___

Q. 41.

5

6x2

11x − ____ − 1 = 0 10

(× 10)

Q. 3.

− 11x − 10 = 0

Q. 42.

2x − 5 = 0

3x = −2 OR

2x = 5

2 x = −__ OR 3

5 x = __ 2

1 2 ____ 17x __ 1 __ x − + =0 8

2

= 25

___

x=

(× 8)

4x − 1 = 0 OR

x−4=0

4x = 1 OR

x=4

1 __

Q. 4.

4

Exercise 17.2 x2 + 8x + 15________ =0

−b ± √b2 − 4ac x = _______________ 2a

a = 1, b = 8, c = 15 b2 − 4ac = (8)2 − 4(1)(15) = 64 − 60 = 4 __

x=

Q. 2.

−8 ± √ 4 _________ 2×1

−8 + 2 x = _______ OR 2

−8 − 2 x = _______ 2

−6 x = ___ 2

OR

−10 x = ____ 2

∴ x = −3

OR

x = −5

x2 + 7x + 10________ =0

−b ± √b2 − 4ac x = _______________ 2a a = 1, b = 7, c = 10 b2 − 4ac = (7)2 − 4(1)(10) = 9 __

x=

6

−7 ± √9 _________

2×1 −7 + 3 x = _______ OR 2

−7 − 3 x = _______ 2

∴ x = −2

x = −5

OR

2a

= 1 + 24

(4x − 1)(x − 4) = 0

Q. 1.

2

b2 − 4ac = (1)2 − 4(1)(−6)

4x2 − 17x + 4 = 0

x=

−b ± √ b − 4ac _______________

a = 1, b = 1, c = −6

OR

2

________

x=

(3x + 2)(2x − 5) = 0 3x + 2 = 0

x2 + x − 6 = 0


−1 ± √ 25 __________

2×1 −1 + 5 x = _______ OR 2

−1 − 5 x = _______ 2

4 x = __ 2

OR

−6 x = ___ 2

∴x=2

OR

x = −3

2x2 − 13x +________ 20 = 0

−b ± √ b2 − 4ac x = _______________ 2a a = 2, b = −13, c = 20 b2 − 4ac = (−13)2 − 4(2)(20) = 169 − 160 =9 __

x=

−(−13) ± √9 _____________ 2×2

13 − 3 x = _______ 4 10 ___ OR x = 4 5 __ x=4 OR x = 2 Checking solutions

13 + 3 x = _______ 4 16 ___ x= 4

OR

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 52 5 2 __ − 13 __ + 20 = 0 2 2

()

()

12_12 − 32_12 + 20 = 0 32_12 − 32_12 = 0

∴x=

5 __ 2

0 = 0 True is a solution.

Q. 5.

2x2 − 7x + 5________ =0

−b ± √ b2 − 4ac x = _______________ 2a a = 2, b = −7, c = 5 b2 − 4ac = (−7)2 − 4(2)(5) = 9 __

–(–7) ± √ 9 __________

7±3 ______

= 4 2(2) 7−3 7+3 x = ______ OR x = ______ 4 4 5 __ x= OR x = 1 2

x=

Q. 6.

4x2 − 17x +________ 13 = 0

−b ± √ b2 − 4ac x = _______________ 2a a = 4, b = −17, c = 13 b2 − 4ac = (−17)2 − 4(4)(13) = 289 − 208 = 81 x=

x=

2×7

−25 − 17 −25 + 17 x = _________ OR x = _________ 14 14 −8 −42 x = ___ OR x = ____ 14 14 4 x = −__ OR 7

3x2 − 4x − 7 = 0

________

x=

−b ± √ b − 4ac _______________ 2

2a

a = 3, b = −4, c = −7 = 16 + 84

17 − 9 x = _______ 8

26 x = ___ 8

OR

8 x = __ 8

OR

x=1

13 ___ 4

5x2 − 18x + 9 = 0

________

a = 5, b = −18, c = 9 b2 − 4ac = (−18)2 − 4(5)(9) = 324 − 180 = 144

x = −3

Q. 9. 3x2 = 4x + 7

2×4

−b ± √ b2 − 4ac x = _______________ 2a

____

−(−18) ± √ 144 x = _______________ 2×5 18 − 12 18 + 12 x = ________ OR x = ________ 10 10 30 6 OR x = ___ x = ___ 10 10 3 x=3 OR x = __ 5 Q. 8.

____

−25 ± √ 289 ____________

b2 − 4ac = (−4)2 − 4(3)(−7)

−(−17) ± √ 81 ______________

17 + 9 x = _______ OR 8

x= Q. 7.

___

a = 7, b = 25, c = 12 b2 − 4ac = (25)2 − 4(7)(12) = 625 − 336 = 289

7x2 + 25x + 12 = 0

________

−b ± √ b2 − 4ac x = _______________ 2a

= 100

____

x=

−(−4) ± √ 100 ______________ 2×3

4 + 10 x = _______ OR 6

4 − 10 x = _______ 6

14 x = ___ 6

OR

−6 x = ___ 6

7 x = __ 3

OR

x = −1

Q. 10. 4x2 = 4x + 8 ∴ 4x2 − 4x − 8 = 0

________

−b ± √ b2 − 4ac x = _______________ 2a a = 4, b = −4, c = −8 b2 − 4ac = (−4)2 − 4(4)(−8) = 16 + 128 = 144

____

x=

−(−4) ± √ 144 ______________ 2×4

4 + 12 x = _______ OR 8

4 − 12 x = _______ 8

16 x = ___ 8

OR

−8 x = ___ 8

x=2

OR

x = −1


7

___

Q. 11. 23x + 20 = −6x2

x=

6x2 + 23x + 20 = 0

2

2

___

2a

__

__

∴x=

−6 + 4√ 2 __________

___

2×6

Q. 2.

+ 7x = 4 + 7x − 4 = 0

x=

−b ± √ b − 4ac _______________

x2 + 8x + 5 = 0

= 64 − 20 = 44 ∴x=

___

−8 ± √44 __________

2 ×___1 −8 + √ 44 x = __________ OR 2

(–4)

___

x=

−8 −√ 44 _________ 2

Give answer to 3 decimal places

2

x = −0.683|3... OR x = −7.316|6...

2a

∴ x = −0.683

a = 2, b = 7, c = −4 Q. 3.

b2 − 4ac = (7)2 − 4(2)(−4)

x = −7.317

OR

y2 − 2y − 29 = 0

= 49 + 32

a = 1, b = −2, c = −29

= 81

b2 − 4ac = (−2)2 − 4(1)(−29)

___

−7 ± √81 x = __________ 2×2 −7 + 9 _______ 4

2 __ 4

1 __ 2

= 4 + 116

OR

x=

= 120

−7 − 9 _______

∴y=

4

OR

x=

OR

x = −4

Q. 1. x2 + 6x + 1 = 0

4

− 4ac =

____

____

=

2 – √ 120 _________ 2

Give answer to 2 d.p.

Q. 4.

a = 1, b = 6, c = 1 (6)2

____

−(−2) ± √ 120 ______________

2×1 2 ± √ 120 y = _________ OR 2

−16 ____

Exercise 17.3

− 4(1)(1)

y = 6.47|7... OR

y = −4.47|7

y = 6.48

y = −4.48

a = 1, b = −5, c = −28 b2 − 4ac = (−5)2 − 4(1)(−28) = 25 + 112 = 137

___

−6 ± √ 32 __________ 2×1


OR

x2 − 5x − 28 = 0

= 36 − 4 = 32

8

__

b2 − 4ac = (8)2 − 4(1)(5)

________

∴x=

2

a = 1, b = 8, c = 5

(×2)

2x2

b2

x=

x = −3 + 2√2 OR x = −3 − 2√ 2

2x2

x=

OR

2

__

−23 ± √49 ___________

−23 − 7 −23 + 7 x = ________ OR x = ________ 12 12 −30 −16 OR x = ____ x = ____ 12 12 −5 −4 OR x = ___ x = ___ 3 2

x=

__

−6 − 4√ 2 __________

Dividing by 2 gives

= 49

x=

__

= 4√ 2

= 529 − 480

Q. 12. x2 + x = 2 2

2

= √ 16 × √ 2

b2 − 4ac = (23)2 − 4(6)(20)

7 __

x=

OR

Note √ 32 = √ 16 × 2

−b ± √ b − 4ac _______________

a = 6, b = 23, c = 20

x=

−6 − √ 32 __________

_______

___

________

x=

___

−6 + √ 32 __________

∴x=

____

−(−5) ± √ 137 ______________ 2×1

____

x=

Q. 5.

____

____

5 + √ 137 _________

x=

OR

2

5 − √137 _________

x=

2

4


x = 8.3|5... OR

x = −3.3|5...

x = 4.723|1... OR x = −2.223|1...

x = 8.4

x = −3.4

x = 4.723

OR

2x2 − 11x − 9 = 0

Q. 8.

x = −2.223

OR

4q2 − q − 13 = 0

a = 2, b = −11, c = −9

a = 4, b = −1, c = −13

b2

b2 − 4ac = (−1)2 − 4(4)(−13)

− 4ac =

(−11)2

− 4(2)(−9)

= 121 + 72

= 1 + 208

= 193

= 209

____

−(−11) ± √ 193 _______________

2×2 11 + √ 193 x = __________ OR 4 ____

∴q=

____

11 − √ 193 x = __________ 4


x = 6.223|1... OR

x = −0.723|1...

x = 6.223

x = −0.723

OR

____

−(−1) ± √209 ______________

2×4 ____ 1 − √ 209 1 + √209 _________ _________ OR q = q= 8 8

____


∴ q = 1.9|3... OR q = 1.9

3x2 − 10x + 4 = 0

Q. 9.

q = −1.6|8... q = −1.7

OR

5x2 + 4x − 5 = 0

a = 3, b = −10, c = 4

a = 5, b = 4, c = −5

b2 − 4ac = (−10)2 − 4(3)(4)

b2 − 4ac = (4)2 − 4(5)(−5)

∴x=

= 100 − 48

= 16 + 100

= 52

= 116

___

____

−(−10) ± √ 52 ______________

2×3 ___ 10 − √ 52 10 + √ 52 _________ _________ OR x = x= 6 6 Give answer in surd form ___

___

_______

√52 = √4 × 13 ___ __ = √4 × √ 13 ___

= 2√13

___

___

10 + 2√ 13 ∴ x = __________ OR 6

10 − 2√ 13 x = __________ 6

Dividing by 2 gives ___

x= Q. 7.

x=

OR

4

5 − √193 _________


∴x=

Q. 6.

____

5 + √ 193 _________

5 + √ 13 ________ 3

∴x=

−4 ± √ 116 ___________

2×5 ____ −4 − √ 116 −4 + √ 116 ___________ ___________ OR x = x= 10 10 Give answer to 2 d.p. x = 0.67|7... OR

x = −1.47|7...

∴ x = 0.68

x = −1.48

x=

a = 8, b = −5, c = −11 b2 − 4ac = (−5)2 − 4(8)(−11) = 25 + 352 = 377

5 − √13 ________ 3

OR

Q. 10. 8x2 − 5x − 11 = 0

___

OR

____

____

∴x=

−(−5) ± √377 ______________

a = 2, b = −5, c = −21

2×8 ____ 5 − √377 5 + √ 377 _________ _________ OR x = x= 16 16

b2 − 4ac = (−5)2 − 4(2)(−21)


2x2

− 5x − 21 = 0

= 25 + 168= 193 ____

−(−5) ± √193 ∴ x = ______________ 2×2

____

x = 1.526|0... OR x = −0.901|0... x = 1.526

OR x = −0.901


9

__

Q. 11.

9a2

− 8a − 24 = 0

∴x=

2 ×__2 __ −4 + √ 8 −4 − √ 8 _________ _________ x= OR x = 4 4

a = 9, b = −8, c = −24 b2 − 4ac = (−8)2 − 4(9)(−24)

∴a=

−4 ± √8 _________

= 64 + 864

Give answer in surd form

= 928

Note: √ 8 = √ 4 × 2

__

____

__

−(−8) ± √ 928 ______________

2×9 8 + √ 928 a = _________ OR 18 ____

__

= 2√ 2

____

__

8 − √928 a = _________ 18

a = 2.13|6... OR

a = −1.24|7...

a = 2.14

a = −1.25

OR

∴x=

−4 + 2√ 2 __________ 4

x= 4x2

−2 + √ 2 _________ 2

∴x=

−(−9) ± √21 ∴ x = _____________ 2×3 ___ ___ 9 − √ 21 9 + √21 ________ ________ OR x = x= 6 6

2×4 ___ −10 − √20 10 + √20 _________ ___________ OR x = x= 8 8 Give answer in surd form ___

__

__

= 2√ 5 ∴x=

−10 + 2√5 ___________

____

−5 + √ 5 _________

b = −0.673

Q. 14. 2x2 + 4x + 1 = 0 a = 2, b = 4, c = 1 − 4(2)(1)

= 16 − 8 = 8

10

4

__

OR

x=

−5 − √ 5 _________ 4

a = 1, b = −12, c = 14 b2 − 4ac = (−12)2 − 4(1)(14) = 144 − 56 = 88

b = 0.945|6... OR b = −0.672|9...

− 4ac =

8

Q. 16. x2 − 12x + 14 = 0


(4)2

OR x =

__

x=

−(−3) ± √ 317 ∴ b = ______________ 2 × 11 ____ ____ 3 − √317 3 + √317 _________ _________ OR b = b= 22 22

8

__

−10 −2√5 __________

Dividing by 2 gives

= 9 + 308

OR

__

= √4 × √5

b2 − 4ac = (−3)2 − 4(11)(−7)

b = 0.946

______

Note: √ 20 = √ 4 × 5

__

a = 11, b = −3, c = −7

= 317

___

−10 ± √20 ___________ ___


Q. 13. 11b2 − 3b − 7 = 0

2

= 100 − 80 = 20

x = 0.7

OR x =

b2 − 4ac = (10)2 − 4(4)(5)

___

OR

__

−2 − √ 2 _________

+ 10x + 5 = 0

= 81 − 60

x = 2.3

4

a = 4, b = 10, c = 5

b2 − 4ac = (−9)2 − 4(3)(5)

x = 0.7|3...

OR x =

__

a = 3, b = −9, c = 5

x = 2.2|6... OR

__

−4 − 2√2 __________

Dividing by 2 gives

Q. 15.

Q. 12. 3x2 − 9x + 5 = 0

= 21

__

= √4 × √2


b2

______


∴x=

___

−(−12) ± √88 ______________

2___ ×1 12 + √88 x = _________ OR 2

___

x=


12 − √ 88 _________ 2

___

_______

Note: √88 = √ 4 × 22


___

__

___

___

___

x=

2

12 − 2√ 22 __________ 2

Dividing by 2 gives ___

∴ x = 6 + √ 22

___

x = 6 − √22

OR

Q. 17. 5x2 − 16x + 9 = 0 − 4ac =

(−16)2

− 4(5)(9)

= 256 − 180 = 76

__

∴x=

Give answer to 1 d.p. x = 2.5

OR

x = 0.7|2... 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

Dividing by 2 gives __

x=

Give answer to 2 d.p. OR

__

x=

OR

4

9x2

−1 − √5 _________ 4

=0

9x2 − 10x + 2 = 0 a = 9, b = −10, c = 2 = 100 − 72 = 28

___

−(−10) ± √ 28 x = ______________ 2× 9 ___ ___ 10 − √28 10 + √ 28 _________ _________ OR x = x= 18 18 Give answer in surd form ___

______

___

__

Note: √ 28 = √4 × 7

__

√28 = √4 × √7 __ = 2√7 __ __ 10 + 2√ 7 10 − 2√ 7 _________ _________ ∴x= OR x = 18

x = 5 + √7 ___________ 9

18

OR

__

x=

5 − √7 _______ 9

Q. 21. 3x = 20 − 8x2 8x2 + 3x − 20 = 0 a = 8, b = 3, c = −20

x = 0.04|9...

b2 − 4ac = (3)2 − 4(8)(−20)

x = 0.05

= 9 + 640 = 649

Q. 19. 4x2 = 1 − 2x

____

+ 2x − 1 = 0

a = 4, b = 2, c = −1 b2 − 4ac = (2)2 − 4(4)(−1) = 4 + 16 = 20

−1 + √ 5 _________

__

−(−21) ± √ 381 ∴ x = _______________ 2 × 15 ____ ____ 21 − √ 381 21 + √ 381 __________ __________ OR x = x= 30 30

∴ x = 1.35

8

Dividing by 2 gives

____

x = 1.35|0... OR

OR x =

8

___

∴ x = 2.4|7... OR

__

−2 − 2√ 5 __________

b2 − 4ac = (−10)2 − 4(9)(2)

−(−16) ± √ 76 ∴ x = ______________ 2×5 ___ ___ 16 − √ 76 16 + √ 76 _________ _________ OR x = x= 10 10

4x2

−2 + 2√5 __________

Q. 20. 2 − 10x +

a = 5, b = −16, c = 9 b2

= 2√5

___

OR x =

__

__

= 2√ 22 12 + 2√ 22 __________

__

Note √ 20 = √ 4 × √ 5

= √ 4 × √ 22

___

x=

−3 ± √ 649 ___________

2 ____ ×8 ____ −3 − √649 −3 + √ 649 ___________ ___________ OR x = x= 16 16 Give answer to 3 d.p. ∴ x = 1.404|7... OR x = −1.779|7...

−2 ± √ 20 ∴ x = 1.405 OR x = −1.780 ∴ x = __________ 2× 4 ___ ___ −2 − √ 20 −2 + √20 __________ __________ OR x = x= 8 8 Active Maths 2 (Strands 1–5): Ch 17 Solutions

11

Q. 22. 15x2 − 14x = 9

b2 − 4ac = (−2)2 − 4(5)(−11)

15x2 − 14x − 9 = 0

= 4 + 220

a = 15, b = −14, c = −9

= 224

b2

− 4ac =

(−14)2

− 4(15)(−9)

= 196 + 540 = 736

____

−(−14) ± √736 ∴ x = _______________ 2 × 15 ____ ____ 14 − √ 736 14 + √ 736 __________ __________ OR x = x= 30 30 Give answer in surd form

Note: √ 736 = √16 × 46 ___

___

= √ 16 × √46

x = −1.29|6...

∴ x = 1.70

x = −1.30

OR

a = 16, b = 40, c = 3

___

= 4√46

___

b2 − 4ac = (40)2 − 4(16)(3)

___

14 + 4√46 14 − 4√46 ∴ x = __________ OR x = __________ 30 ___ 30 ___ 7 − 2√ 46 7 + 2√46 x = _________ OR x = _________ 15 15

= 1,600 − 192 = 1,408 ______ –40 ± √1,408 _____________ x= 2 × 16 ______ ______ −40 − √1,408 −40 + √ 1,408 ______________ ______________ OR x = x= 32 32

Q. 23. 8x2 = 3 − 8x 8x2 + 8x − 3 = 0 a = 8, b = 8, c = −3


b2 − 4ac = (8)2 − 4(8)(−3)

x = −0.07|7... OR x = −2.42|2...

= 64 + 96

∴ x = −0.08

= 160

____

x=

−8 ± √160 ___________

2____ ×8 ____ −8 − √160 −8 + √ 160 ___________ ___________ OR x = x= 16 16 ____

b2 − 4ac = (6)2 − 4(5)(−60) = 36 + 1,200

___

= √ 16 × √10 ___

= 4√ 10

___

∴x=

−8 + 4√ 10 ___________

OR x = 16 Dividing by 4 gives

x=

4

___

−8 −4√ 10 __________

___

−2 + √ 10 __________

OR x =

(× 15)

a = 5, b = 6, c = −60

________

Note: √ 160 = √ 16 × 10 ___

x = −2.42

OR

x2 2x Q. 26. __ + ___ − 4 = 0 5 3 5x2 + 6x − 60 = 0

Give answer as a surd

16

= 1,236 ______ −6 ± √1,236 _____________ ∴x= ______ 2 × 5 ______ −6 − √ 1,236 −6 + √ 1,236 _____________ _____________ OR x = x= 10 10

___

−2 − √10 __________ 4

Q. 24. 5x2 = 2x + 11 5x2 − 2x − 11 = 0 a = 5, b = −2, c = −11

12

x = 1.69|6... OR

3 Q. 25. 2x2 + 5x + __ = 0 (× 8) 8 16x2 + 40x + 3 = 0

________

____

____

−(−2) ± √224 ∴ x = ______________ 2×5 ____ ____ 2 − √ 224 2 + √224 _________ _________ OR x = x= 10 10 Give answer to 2 d.p.


Give answer to 3 d.p. x = 2.915|6... OR x = −4.115|6... x = 2.916

OR

x = −4.116

3 4 1 (× 90) Q. 27. __x2 = __ x + __ 5 2 9 40x2 = 135x + 18

Exercise 17.4 Q. 1.

40x2 − 135x − 18 = 0

Solve x2 + 9x + 20 = 0

a = 40, b = −135, c = −18

(x + 4)(x + 5) = 0

b2

x+4=0

− 4ac =

(−135)2

− 4(40)(−18)

OR

2)2

3)2

Q. 28. (x +

− 2(2x −

x = −0.13

(y + 1)2 + 9(y + 1) + 20 = 0

Q. 2.

+ 4x + 4 −

2[4x2

− 12x + 9] = 10

+ 28x − 24 = 0

Q. 3.

b2 − 4ac = (−28)2 − 4(7)(24)

−(–28) ± √ 112 ∴ x = ______________ 2×7 ____ ____ 28 − √ 112 28 + √112 __________ __________ OR x = x= 14 14

__

Q. 4.

__

= 4√ 7 __

__

28 + 4√ 7 28 − 4√ 7 ∴ x = _________ OR x = _________ 14 14 Dividing by 2 gives __

7

OR

x = −4 OR

x=8

∴ 4t = −4 OR

4t = 8

t = −1 OR

t=2

Solve x2 − 2x − 3 = 0

__

x=

14 − 2√7 _________ 7

OR

x−3=0

x = −1 OR

x=3

(t + 1)2 − 2(t + 1) − 3 = 0

_______

= √ 16 × √ 7

x=

x−8=0

Use this to solve

Give answer as surd.

14 + 2√ 7 _________

OR

x+1=0

____

___

Solve x2 − 4x − 32 = 0

(x + 1)(x − 3) = 0

= 784 − 672

Note: √112 = √ 16 × 7

y = −6

(4t)2 − 4(4t) − 32 = 0

(× −1)

a = 7, b = −28, c = 24

____

y = −5 OR

Use this to solve

7x2 − 28x + 24 = 0

= 112

y + 1 = −5

x+4=0

= 10

x2 + 4x + 4 − 8x2 + 24x − 18 − 10 = 0 −7x2

∴ y + 1 = −4 OR

(x + 4)(x − 8) = 0

(x + 2)(x + 2) − 2(2x − 3)(2x − 3) = 10 x2

x = −5

Use this to solve

Give answer to 2 d.p. x = 3.50

x+5=0

x = −4 OR

= 21,105_______ −(−135) ± √ 21,105 x = ___________________ _______2 × 40 _______ 21,105 135 − 135 + √ √21,105 x = ______________ OR x = ______________ 80 80 ∴ x = 3.50|3... OR x = −0.12|8...

OR

∴ t + 1 = −1 OR

t+1=3

t = −2 OR

t=2

(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

2 a = −__ 3

OR

a=2


13

Q. 5.

Solve 2x2 + 7x − 4 = 0

Solve x2 + 2x − 24 = 0

Q. 6.

(2x − 1)(x + 4) = 0

(x − 4)(x + 6) = 0

2x − 1 = 0 OR

x − 4 = 0 OR

x+4=0

2x = 1 OR 1 x = __ 2

x = −4

∴x=4

(

)

24 − 2(2y + 2) − (2y + 2)2 = 0

)

12 1 2 2p − __ + 7 2p − __ − 4 = 0 2 2 1 __ 1 1 __ ∴ 2p − = OR 2p − __ = −4 2 2 2 2p = 1 OR p= Q. 7.

1 __ 2

i.e. × −1 and rearrange order (2y + 2)2 + 2(2y + 2) − 24 = 0

2p = −3

2

7 __

p=−

OR

∴ 2y + 2 = 4 OR

1 __

4

OR

2y = −8

y=1

OR

y = −4

(i) Solve 2x2 − 9x − 18 = 0 2x + 3 = 0

OR

x−6=0

2x = −3 OR

x=6

3 x = −__ 2 (ii) Hence, solve 2(2a2 − 2)2 − 9(2a2 − 2) − 18 = 0 −3 ∴ 2a2 – 2 = ___ 2 1 __ 2 2a = 2 1 2 a = __ 4 __ 1 ∴ a = ± __ 4 1 __ a=± 2 1 __ 1 __ ∴a= ,− ,2 2 2

√

OR 2a2 − 2 = 6 OR

2a2 = 8

OR

a2 = 4 __

OR

a = ±√4

OR

a = ±2

OR

−2

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

14

OR

3q2 − 5q = 8

3q2 − 5q + 2 = 0

OR

3q2 − 5q − 8 = 0

(3q − 2)(q − 1) = 0

OR

(3q − 8)(q + 1) = 0


2y + 2 = −6

2y = 2

(2x + 3)(x − 6) = 0

Q. 8.

x = −6

OR

Hence, solve

Hence, solve

(

x+6=0

3q – 2 = 0

OR

q − 1 = 0 OR

3q − 8 = 0 OR

q+1=0

3q = 2 OR 2 __ q= 3 8 2 __ ∴ q = , 1, __, −1 3 3

q = 1 OR

3q = 8 OR 8 q = __ 3

q = −1

OR

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: ________

−b ± √b2 − 4ac x = _______________ 2a ___ 32 −(−2) ± √ gives x = _____________ 2×1 ___ ___ 2 − √ 32 2 + √ 32 ________ ________ OR x = ∴x= 2 2 x = 3.82|8...

OR

x = −1.82|8...

x = 3.83

OR

x = −1.83

b2 − 4ac = (−2)2 − 4(5)(−15) = 4 + 300

____

2 − √ 304 x = _________ 10

x = 1.9|4... OR

x = −1.5|4...

x = 1.9

x = −1.5

OR

(ii) Hence, solve

(

)

OR

2p = 3.17

p = 4.415 OR

p = 1.585

−3 × − 8 = 249 −3 −8 = −119

3x(4x − 1) − 2(4x − 1) = 0 (3x − 2)(4x − 1) = 0 3x − 2 = 0 OR

4x − 1 = 0

3x = 2 OR 2 x = __ OR 3

4x = 1 1 x = __ 4

Hence, solve

(

)

2 a a 5 __ + 1 − 2 __ + 1 − 15 = 0 2 2 a a ∴ __ + 1 = 1.9 OR __ + 1 = −1.5 2 2 a a __ = 0.9 OR __ = −2.5 2 2

a = 1.8

2p = 8.83

2p − 5 = −1.83

12x2 − 3x − 8x + 2 = 0

____

2×5

2 + √304 x = _________ OR 10

OR

∴ 12x2 − 11x + 2 = 0

−(−2) ± √ 304 ______________ ____

∴ 2p − 5 = 3.83

Factors of 24 that add to give –11:

a = 5, b = −2, c = −15

∴x=

(2p − 5)2 − 2(2p − 5) − 7 = 0

Q. 11. Solve 12x2 − 11x + 2 = 0

Q. 10. (i) Solve to 1 decimal place: 5x2 − 2x − 15 = 0

= 304

Hence, solve

OR

a = −5

12(3y2 + y)2 − 11(3y2 + y) + 2 = 0 2 1 ∴ 3y2 + y = __ OR 3y2 + y = __ 4 3 2 __ 2 Solving 3y + y = (× 3) 3 9y2 + 3y = 2 9y2 + 3y − 2 = 0


15

Q. 6. Pair of roots 8, 0

(3y − 1)(3y + 2) = 0 3y − 1 = 0 OR 3y = 1 OR 1 y = __ OR 3 Solving 3y2 + y = 12y2 + 4y = 1

1 __ 4

3y = −2 2 y = −__ 3

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

(6y − 1)(2y + 1) = 0 6y − 1 = 0 OR

∴ x2 − 8x = 0

(× 4)

12y2 + 4y − 1 = 0

x2 − (0 − 6)x + (0 × −6) = 0

2y + 1 = 0

6y = 1 OR 1 y = __ OR 6 1 __ 2 1 1 __ ∴ y = , − , __, −__ 3 3 6 2

2y = −1 1 y = −__ 2

Exercise 17.5 Q. 1.

x2 − (8 + 0)x + (8 × 0) = 0

3y + 2 = 0

x2 + 6x = 0 Q. 9. Pair of roots ±5 x = ±5 x2 = 25 x2 − 25 = 0

1 2 Q. 10. Pair of roots −__, __ 3 7 1 2 x = −__ OR x = __ 7 3

Pair of roots 1, 2 x2 −

3x = −1

of Product x+( =0 ( Sum roots ) of roots )

3x + 1 = 0 OR

21x2 − 6x + 7x − 2 = 0 21x2 + x − 2 = 0

Pair of roots 5, 6


∴ x2 − (5 + 6)x + (5 × 6) = 0

Q. 11. Roots of x2 + px + q = 0 are 3 and −2 x2 − (3 − 2)x + (3 × − 2) = 0

x2 − 11x + 30 = 0 Q. 3.

x2 − x − 6 = 0

Pair of roots 4, −2 x2 −


∴ x2 − (4 − 2)x + (4 × −2) = 0

∴ p = −1 and q = −6 Q. 12. Roots of x2 + bx + c = 0 are 0 and −4

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

16

7x − 2 = 0

3x(7x − 2) + 1(7x − 2) = 0

x2 − 3x + 2 = 0

x2 −

7x = 2

∴ (3x + 1)(7x − 2) = 0

∴ x2 − (1 + 2)x + (1 × 2) = 0

Q. 2.

OR


x = 0 and x = −4 x = 0 and x + 4 = 0 ∴ x(x + 4) = 0 x2 + 4x = 0 ∴ b = 4 and c = 0

Q. 13. Given roots the same:

Q. 4.

x2 − (a + a)x + (a × a) = 0 i.e.

x2

a2

− 2ax +

Area of rectangle = length × width ∴ x(x + 2) + (2x + 3)(x + 1) + x(x) = 189 x2 + 2x + 2x2 + 2x + 3x + 3 + x2 = 189

=0

but x2 − 8x + c = 0

4x2 + 7x − 186 = 0

∴ 2a = 8 hence a = 4

4x2 − 24x + 31x − 186 = 0

∴c=4×4

4x(x − 6) + 31(x − 6) = 0 (4x + 31) (x − 6) = 0

c = 16

4x + 31 = 0

Exercise 17.6 x2 + 3x = 88 Q. 5.

(x + 11)(x − 8) = 0 OR

x−8=0

x = −11 OR

x=8

x=6

Perimeter of rectangle = 160 m (i) P = 2 × length + 2 × width If x = length then 160 = 2x + 2 × width

Since x is a positive number, the number is 8.

160 − 2x = 2 × width

160 − 2x _________

x = 1st number

= width 2 ∴ width = 80 − x

x + 2 = 2nd number x(x + 2) = 399

(ii) Area of rectangle = 1536 m2

x2 + 2x − 399 = 0

Area of rectangle = l × w

(x − 19)(x + 21) = 0

∴ x(80 − x) = 1,536

x − 19 = 0

80x − x2 = 1,536

OR

x + 21 = 0

x = 19 OR

x = −21

If x = 19 then x + 2 = 21

(× −1)

x2 − 80x = −1,536 i.e. x2 − 80x + 1,536 = 0

19 × 21 = 399 as required

(x − 32)(x − 48) = 0

x = −21 rejected as −21 ∉ N

∴ x − 32 = 0

∴ The numbers are 19 and 21 Q. 3.

4x = −31 OR

∴x=6

x2 + 3x – 88 = 0

Q. 2.

x−6=0

31 x = −___ 4 31 ___ x = − rejected as length is positive 4

Q. 1. x = positive number

∴ x + 11 = 0

OR

OR

x − 48 = 0

x = 32 OR

x = 48

If x = 32 then length = 32

Let x = the number

and width = 80 − 32

x2 + 15 = 8x x2 − 8x + 15 = 0

= 48

(x − 3)(x − 5) = 0

If x = 48 then length = 48

x − 3 = 0 OR

x−5=0

x = 3 OR

x=5

The number could be 3 OR

and width = 80 − 48 = 32 5

∴ The dimensions of the rectangle are 32 m and 48 m.


17

Q. 6.

(i) Area of rectangle = l × w

Area = (x + 5)(16 − 2x)

∴ x(x + 5) = 14

= 16x − 2x2 + 80 − 10x

x2 + 5x − 14 = 0

= −2x2 + 6x + 80

(x − 2)(x + 7) = 0 x − 2 = 0 OR

(ii)

x+7=0

x = 2 OR

x + 5 + 2x x

x = −7

x

16 – 2x + 2x

x = −7 rejected as length is positive

x

∴ x = 2 cm (ii) Area of parallelogram = b × h

Garden + Path (total area covered)

∴ (2x − 5)(x + 3) = 21

Area = (3x + 5)(16)

2x2 + 6x − 5x − 15 − 21 = 0

= 48x + 80

2x2 + x − 36 = 0

Area of path only

(2x + 9)(x − 4) = 0 2x + 9 = 0 x=−

9 __ 2

OR

x−4=0

OR

x=4

= 48x + 80 − (−2x2 + 6x + 80) = 48x + 80 + 2x2 − 6x − 80 = 2x2 + 42x

9 x = −__ rejected as length is positive 2

(iii) Given area of path = 67.5 m 2x2 + 42x = 67.5

∴ x = 4 cm

2x2 + 42x − 67.5 = 0

1 (iii) Area of triangle = __ bh 2 1 ∴ __(4x + 3)(2x − 7) = 34.5 (×2) 2 (4x + 3)(2x − 7) = 69

a = 2, b = 42, c = −67.5 b2 − 4ac = (42)2 − 4(2)(−67.5) = 2,304 ______

8x2 − 28x + 6x − 21 − 69 = 0 8x2 − 22x − 90 = 0 4x2

x=

(÷2)

4x (x − 5) + 9 (x − 5)=0 OR

x–5=0

Q. 8.

Garden (i)

−42 ± 48 = _________ 4

x = 1.5

∴ x = 1.5 m

9 x = −__ OR x=5 4 9 x = −__ rejected as length is positive 4 ∴ x = 5 cm

Q. 7.

4

x = −22.5 rejected, as length is positive

4x2 − 20x + 9x − 45 = 0 (4x + 9)(x − 5) = 0

−42 ± √2,304 ______________

x = −22.5 OR

− 11x − 45 = 0

4x + 9 = 0

x

(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

x+5

g = −2(0)2 + 20(0) + 7 = 7 1991 t = 1 16 – 2x

18


g = −2(1)2 + 20(1) + 7 = 25

1992 t = 2

0 = (n + 1)(n − 4)

g = −2(2)2 + 20(2) + 7 = 39

n+1=0

1993 t = 3

OR

n−4=0

n = −1 OR

n=4

g = −2(3)2 + 20(3) + 7 = 49

Since n ≥ 0, n = −1 rejected.

1994 t = 4

∴n=4

g = −2(4)2 + 20(4) + 7 = 55

Q. 10.

1995 t = 5 −2(5)2

g=

(i) Perimeter = 46 m Length = x

+ 20(5) + 7 = 57

Perimeter = 2l + 2w

1996 t = 6 g=

−2(6)2

∴ 2x + 2w = 46

+ 20(6) + 7 = 55

2w = 46 − 2x

1997 g = 49

w = 23 − x

1998 g = 39

width = 23 − x

1999 g = 25 1990 to 1994 = 175,000 games

(ii)

x–8 23 – x

1995 to 1999 = 225,000 games ∴ 1995 to 1999 more games were sold. (iii) 25,000 games sold

x–8

∴ −2t2 + 20t + 7 = 25 −2t2 + 20t − 18 = 0 2t2 t2

− 20t + 18 = 0

(÷ 2)

= 3x − 16 width = 23 − x + x − 8 + x − 8

∴ 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 + 20(11) + 7

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=

=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

g = −15 i.e. −15,000 games and thus the formula is no longer valid

n2

x–8

length = x + x − 8 + x − 8

(t − 1)(t − 9) = 0

∴g=

x–8

Total area of path and pool:

− 10t + 9 = 0

–2(11)2

x

2x2 − 9x − 126 = 0 (2x − 21)(x + 6) = 0 2x − 21 = 0 OR 2x = 21 OR

x+6=0

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

− 3n − 4 Active Maths 2 (Strands 1–5): Ch 17 Solutions

19

Q. 11. C(x) = 9 + 16x − 0.5x2

Since t ≥ 0 ∴ t = 4 seconds.

C = cost in euros

It will take 4 seconds to hit the ground.

x = no. of chairs produced (i) If x = 10

Q. 13. N(t) = 16t2 − 200t + 625

C = 9 + 16(10) − 0.5(10)2

(i) When t = 0

C = 119

N = 16(0)2 − 200(0) + 625

costs €119

= 625 625 spores present

(ii) 9 + 16x −0.5x2 = 135 −0.5x2 + 16x −126 = 0 x2

(ii) At t = 3

− 32x + 252 = 0

N = 16(3)2 − 200(3) + 625

(x − 14)(x − 18) x = 14 OR

= 169

x = 18

169 spores after 3 minutes

Answer: 14 chairs

(iii) 625 spores present initially

(iii) If x = 15

84% of 625 = 525 spores killed

C = 9 + 16(15) −

0.5(15)2

∴ 100 spores left

C = €136.50

∴ 16t2 − 200t + 625 = 100

Chairs sold at €30 each

16t2 − 200t + 525 = 0

∴ Profit = €30 × 15 − €136.50

a = 16, b = −200, c = 525

= €313.50

b2 − 4ac = (−200)2 − 4(16)(525) = 6,400

Q. 12. h(t) = −5t2 + 18t + 8

Using the quadratic formula: ______ −(−200) ± √6,400 t = _________________ 2 × 16 200 − 80 200 + 80 t = _________ OR t = _________ 32 32

(i) When t = 0 h = −5(0)2 + 18(0) + 8 h=8 Object fired from height of 8 m

t = 8_34

(ii) Average speed in 1st second h = 21

(iv) The laboratory technician may have recommended not releasing the fungicide for commercial use as 3_43 minutes may be too long to wait to kill 84% of the spores initially present.

Distance 21 − 8 Average Speed: ________ = _______ 1 time ∴ Average speed = 13 m/s (iii) When h = 0 −5t2 + 18t + 8 = 0 Q. 14.

(i) Area of field

5t2 − 20t + 2t − 8 = 0

= (8x + 6)(13x + 11)

5t(t − 4) + 2(t − 4) = 0

= 104x2 + 88x + 78x + 66

(5t + 2)(t − 4) = 0

= 104x2 + 166x + 66

5t + 2 = 0 t=−

20

t = 3_34

∴ The shortest time is 3_34 minutes

when t = 1, h = −5(1)2 + 18(1) + 8

5t2 − 18t − 8 = 0

OR

2 __ 5

OR

t−4=0

OR

t=4


b2 − 4ac = (−9)2 − 4(2)(−1)

(ii) Area covered by barn = x(x + 5) =

x2

= 81 + 8 = 89

+ 5x

Using the quadratic formula: ___

(iii) Area unbuilt = 1,476

m2

x=

∴ 104x2 + 166x + 66 − (x2 + 5x) = 1,476 103x2 + 161x − 1,410 = 0

−(−9) ± √89 _____________

2___ ×2 9 + √ 89 x = ________ OR 4

___

x=

9 − √ 89 ________ 4

a = 103, b = 161, c = −1,410

x = 4.608... OR x = −0.108...

b2 − 4ac = (161)2 − 4(103)(−1,410)

Since x ≥ 0 reject x = −0.108 x = 4.608...m

= 606,841 Using the quadratic formula: ________ −161 ± √ 606,841 x = _________________ 2 × 103 −161 − 779 −161 + 779 x = ____________ OR x = ____________ 206 206 x = 3 OR

x = −4.56...

as x > 0, x = −4.56... rejected ∴x=3 Dimensions of the field for x = 3 are:

∴ 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

8x + 6 = 8(3) + 6 = 30 m

45x2 − 42x + 49

and 13x + 11 = 13(3) + 11

= 64x2 − 336x + 441

= 50 m Dimensions of field: 50 m by 30 m

b2 − 4ac = (−294)2 − 4(19)(392)

y = height above ground (m)

= 56,644

x = horizontal distance (m)

Using the quadratic formula: _______ −(−294) ± √ 56,644 x = ___________________ 2 _______ × 19 56,644 294 + √ x = ______________ 38

(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 a = 2, b = −9, c = −1

0 = 19x2 − 294x + 392 a = 19, b = −294, c = 392

Q. 15. y = −2x2 + 9x + 1

⇒ −2x2 + 9x + 1 = 0

(× 100)

OR _______ 294 − √56,644 x = ______________ 38 28 x = 14 OR x = ___ 19 28 If x = ___ then 19 28 11 3x − 7 = 3 × ___ − 7 = −2__ 19 19

(

)

which is invalid, as length cannot 28 be negative ∴ x = __ is rejected 19 ∴ x = 14 Active Maths 2 (Strands 1–5): Ch 17 Solutions

21

(ii) 2x2 − 5x + 9 = 0

(ii) Two shorter sides (7x − 2)2 + (2x − 8)2

a = 2, b = −5, c = 9

= 49x2 − 28x + 4 + 4x2 − 32x + 64

b2 − 4ac = (−5)2 − 4(2)(9)

= 53x2 − 60x + 68

= 25 − 72

Longest side (hypotenuse)

= −47

(6x + 2)2 = 36x2 + 24x + 4

Since the discriminant b2 − 4ac is negative, the equation has no real solutions, i.e. we cannot evaluate _____ √ −47 .

If it is a right-angled triangle 53x2 − 60x + 68 = 36x2 + 24x + 4 17x2 − 84x + 64 = 0

(iii) A possible correction to the equation is to change +9 to −9,

a = 17, b = −84, c = 64 b2 − 4ac = (−84)2 − 4(17)(64)

so that b2 − 4ac = 25 + 72

= 2,704 Using the quadratic formula ______ −(−84) ± √ 2,704 x = ________________ ______ 2 × 17 ______ 2,704 84 − √ 2,704 84 + √ ____________ ____________ OR x = x= 34 34 16 ___ x = 4 OR x = 17

= 97 which is positive. Q. 18.

27x

( )

(27x)2 + (36x)2 = 602 729x2 + 1,296x2 = 3,600 2,025x2 = 3,600 3,600 x2 = ______ 2,025 ______ 3,600 x = ______ 2,025 60 x = ___ 45

∴ Abdul has indeed made a mistake.

√

___

4 ± √ 76 ________

6 ________ −b ± √ b2 − 4ac x = _______________ 2a ∴ 2a = 6 ⇒ a = 3

x = 1_13 hours x = 1 hour 20 mins

−b = 4 ⇒ b = −4

Started at 08:50 then 60 km apart after 1 hour 20 mins

b2 − 4ac = 76 ∴

(−4)2

− 4(3)(c) = 76 16 − 12c = 76 −12c = 60 c = −5

∴ time = 10:10 9x2 5x Q. 19. y = −___ + ___ + 11 20 2 y = height above sea level (m)

The quadratic equation of the form ax2

+ bx + c = 0

where a = 3, b = −4, c = 5 is 3x2 − 4x − 5 = 0

22

E

Where x = time in hours

2x − 8 = 2(4) − 8 = 0 16 16 2 If x = ___ then 2 ___ − 8 = −6__ 17 17 17 We cannot have a negative length side or a side of length zero.

(i) x =

60 km

36x

If x = 4, then one of the sides

Q. 17.

N


x = horizontal distance from cliff face (m) (i) When x = 0 5 −9 y = ___(0)2 + __(0) + 11 20 2 Pebble is 11 m high

(ii) When x = 3 5 −9 y = ___(3)2 + __(3) + 11 20 2 9 y = 14__ 20

9 Height of pebble is 14__ 20 m (or 14.45 m)

(iii) When y = 0 5 −9 2 __ ___ x + x + 11 = 0 20

(× −20)

2

9x2 − 50x − 220 = 0 a = 9, b = −50, c = −220 − 4(9)(−220) = 10,420 b2 − 4ac = (−50)2 _______ −(−50) ± √10,420 ∴ x = _________________ 2×9 _______ _______ 10,420 50 − √ 10,420 50 + √ _____________ _____________ OR x = x= 18 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). Active Maths 2 (Strands 1–5): Ch 17 Solutions

23

Q. 21.

(i)

(ii) If R = €3,000 1 (× −2) 80x − __x2 = 3,000 2 2 −160x + x = −6,000

y x

x

x

x2 − 160x + 6,000 = 0

y

(x − 60)(x − 100) = 0

y + y + x + x + x = 200

∴ x − 60 = 0 OR

2y + 3x = 200

x = 60 OR

2y = 200 − 3x

(

(iii) c(x) = 10x + 50 where c = cost

)

If x = 25 then c = 10 × 25 + 50 c = 300

200x − 3x2 = __________ 4

Cost is €300 to produce 25 units

(iii) Area of field = 833_13 m2 200x − 3x2 ∴ __________ = 833_13 4 10,000 200x − 3x2 = _______ 3 600x − 9x2

9x2

(iv) If c = €750 (× 4)

then 10x + 50 = 750 10x = 700

(× 3)

x = 70

= 10,000

∴ 70 units manufactured

− 600x + 10,000 = 0

(v) Profit = sales revenue − cost

9x2 − 300x − 300x + 10,000 = 0

If profit is €1,950 then 1 80x − __x2 − (10x + 50) = 1,950 2 1 2 __ − x + 70x − 50 − 1,950 = 0 2 1 2 __ − x + 70x − 2,000 = 0 (×−2) 2

3x(3x − 100) − 100(3x − 100) = 0 (3x − 100)(3x − 100) = 0 ∴ 3x − 100 = 0 3x = 100 x = 33_13 _ 1

If x = 333 and y = then y =

∴ y = 50

2

x2 − 140x + 4,000 = 0

200 − 3x _________

( )

200 − 3 33_13 ____________

(x − 40)(x − 100) = 0

2

1 The value of x is 33__ m and the 3 value of y is 55 m

∴ x = 40 OR

R = 80(50) −

Q. 23. h = 22 + 60t − 10t2

1 __

(50)2

2 R = 4,000 − 1,250 R = 2,750 Revenue of €2,750

24

x = 100

40 units must be made

1 Q. 22. R(x) = 80x − __ x2 2 x = no. of furniture units (i) When x = 50

x = 100

60 units must be sold

200 − 3x y = _________ 2 (ii) Area of one of the fields y 200 − 3x 1 x × __ = x × __ × _________ 2 2 2

x − 100 = 0


t = time in seconds h = height in metres (i) When h = 0 22 + 60t − 10t2 = 0

(×−1)

10t2

(÷ 2)

− 60t − 22 = 0

5t2 − 30t − 11 = 0 a = 5, b = −30, c = −11

b2 − 4ac = (−30)2 − 4(5)(−11) = 900 + 220

i.e. h = 22 + 60(3) − 10(3)2

= 1,120 Using the quadratic formula ______ −(−30) ± √ 1,120 t = ________________ 2×5 ______ ______ 1,120 1,120 30 − 30 + √ √ t = ____________ OR t = ____________ 10 10 t = 6.346... OR

The maximum height occurs at t = 3 seconds

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)

= 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

height

3 time

–0.346

6.346

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 (iii)

x2

x = −10

− 25 = 0

x2 = 25

___

x = ±√ 25 x = ±5

(iv) x2 − 7x = 0 x(x − 7) = 0 ∴ x = 0 OR

x=7 Active Maths 2 (Strands 1–5): Ch 17 Solutions

25

(b) Using quadratic formula ________ −b ± √ b2 − 4ac x = _______________ 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 −5 + √ 37 x = __________ OR 2

x = 0.54|1...

___

x=

−5 − √ 37 __________ 2

x = −5.54|1...

OR

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 7 + √ 53 x = ________ OR 2 ___

___

x=

x = 7.14|0... OR

7 − √ 53 ________ 2

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 − 4(1)(−2) = 24 b2 − 4ac = (−4)2 ___ −(−4) ± √ 24 ∴ x = _____________ 2 ___ ×1 ___ 4 − √ 24 4 + √ 24 ________ ________ OR x = x= 2 2 x = 4.44|9... OR x = 4.45 OR

x = −0.44|9...

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

26


t=3

Q. 2. (a)

(i) Solve 2x2 − 3x = 0 x(2x − 3) = 0 x = 0 OR

2x = 3 3 x = __ 2

(ii) 10x2 + x − 2 = 0 (2x + 1)(5x − 2) = 0 2x + 1 = 0 OR 5x − 2 = 0 2 −1 ___ OR x = __ x= 5 2 (iii) 9x2 − 4 = 0 4 x2 = __ 9 __

√

4 x = ± __ 9 2 x = ±__ 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 ___________

OR

4 ___ −4 + √22 x = __________ 2

OR

x=

___

−8 − 2√22 ___________

4 ___ −4 − √22 x = __________ 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 11 + √101 x = __________ OR 10

____

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 1 + √73 x = ________ OR 12

___

x=

1 − √ 73 ________ 12


27

Q. 3. (a)

(i) Given roots 2 and 3 Product Sum of x2 − x + ________ = 0 roots of roots

(

(

)

)

∴ 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 5 + √ 53 x = ________ OR 2

___

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 ___ 10 + √ 92 x = _________ OR 4

x = 4.897|9... OR ∴ x = 4.898

28

OR

___

x=

10 − √ 92 _________ 4

x = 0.102|0... x = 0.102 to 3 d.p.


(iii) 3 − x − 3x2 = 0 3x2

(×−1)

+x−3=0

a = 3, b = 1, c = −3 b2 − 4ac = (1)2 − 4(3)(−3) = 37 ___

∴x=

−1 ± √ 37 __________ 2×3

___

x=

(c)

(i)

___

−1 + √37 __________

OR

x=

−1 − √ 37 __________

6 x = 0.847|1...

OR

6 x = –1.180|4...

x = 0.847

OR

x = −1.180 to 3 d.p.

21x2

+ 2x − 3 = 0

(7x + 3)(3x − 1) = 0 ∴ 7x + 3 = 0 OR

3x − 1 = 0

3 x = −__ OR 7

1 x = __ 3

(ii) 5x2 + 6 = 13x 5x2 − 13x + 6 = 0 (5x − 3)(x − 2) = 0 5x − 3 = 0 OR 3 OR ∴ x = __ 5

x−2=0 x=2

(iii) 5(x2 − 4) = 2(x − 10) 5x2 − 20 = 2x − 20 5x2 − 2x = 0 x(5x − 2) = 0 ∴ x = 0 OR Q. 4. (a)

2 x = __ 5

(i) x2 – ax – b = 0 with roots 7 and −2 Sum of Product x2 − x+ =0 roots of roots

(

) (

)

∴ x2 − (7 − 2)x + (7 × −2) = 0 x2 − 5x − 14 = 0

∴

∴x+1=0

OR

x = −1 OR

x − 11 = 0 x = 11

Hence solve 3t − 1 = −1 OR

x2

+ ax + b = 0 with roots −8 and −2

x2

(x + 1)(x − 11) = 0

(3t − 1)2 − 10(3t − 1) − 11 = 0

∴ a = 5 and b = 14 (ii)

(b) Solve x2 − 10x − 11 = 0

3t − 1 = 11

3t = 0

OR

3t = 12

t=0

OR

t=4

− (−8 − 2)x + (−8 × −2) = 0

x2

− (−10x) + 16 = 0

x2 + 10x + 16 = 0 ∴ a = 10 and b = 16


29

(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___ ___ −2 − √ 44 −2 + √ 44 __________ __________ OR x = x= 2 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 Q. 5.

t = −5.3 to 1 d.p.

N(t) = −2t2 + 11t + 13

(iv) The formula only gives valid values between t = 0 and t = 6_21. Outside these times negative answers are given, e.g. if t = 7

(i) When t = 0 N = −2(0)2 + 11(0) + 13 N = 13

N = −2(7)2 + 11(7) + 13

∴ 13,000 bacteria present initially

∴ N = −8 rejected, as we cannot have −8,000 bacteria present

(ii) When N = 25 Q. 6.

−2t2 + 11t + 13 = 25 −2t2

+ 11t − 12 = 0

(× −1)

∴ 1 − x = side length of large square

2t2 − 11t + 12 = 0

(1 − x)2 = 2(x)2

2t2 − 8t − 3t + 12 = 0

(1 − x)(1 − x) = 2x2

2t(t − 4) − 3(t − 4) = 0

1 − 2x + x2 − 2x2 = 0

(2t − 3)(t − 4) = 0

−x2 − 2x + 1 = 0

∴ 2t − 3 = 0 OR t − 4 = 0 3 t=4 t = __ OR 2 As t is time in hours, there will be 25,000 bacteria present at 1_12 hours and 4 hours.

x2 + 2x − 1 = 0

(iii) When N = 0 −2t2 + 11t + 13 = 0

(×−1)

2t2 − 11t − 13 = 0 ∴ 2t − 13 = 0

OR

(× −1)

a = 1, b = 2, c = −1 b2 − 4ac = (2)2 − 4(1)(−1) = 8 __

∴x=

−2 ± √ 8 _________

2 ×__1 −2 + √ 8 x = _________ OR 2 x = 0.414|2... OR

__

x=

−2 − √ 8 _________ 2

x = −2.414...

Since x > 0 reject x = −2.414...

(2t − 13)(t + 1) = 0

∴ x = 0.414 to 3 d.p. t+1=0

13 t = −1 t = ___ OR 2 ∴ No bacteria at 6_12 hours. (t = −1 rejected, as time cannot be negative)

30

Let x = side length of small square


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.

Q. 7.

2x(x − 3) + 9(x − 3) = 0

x = width of rectangle

(2x + 9)(x − 3) = 0

(i) length = 2x + 3

∴ 2x + 9 = 0

(ii) Area of rectangle = length × width ∴ Area = (2x + 3)x = 2x2 + 3x

∴

x−3=0

−9 x = ___ OR 2

x=3

since x > 0 then x = 3

(iii) Area = 27 m2 2x2

OR

width of rectangle = 3 m

+ 3x = 27

length of rectangle = 2 × 3 + 3

2x2 + 3x − 27 = 0 2x2 Q. 8.

=9m

− 6x + 9x − 27 = 0

(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 ______

−(−8) ± √ 14.64 x = _______________ 2______ ×2 8 + √ 14.64 x = ___________ OR 4

______

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 Active Maths 2 (Strands 1–5): Ch 17 Solutions

31

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

Q. 10. Area of garden = 6x2 + 11x − 10

(iii) 10 + 36t −5t2 = 52.75 =0 5t2 − 36t + 42.75 ___________________ +36 ± √(−36)2 − 4(5)(42.75) t = ___________________________ 2(5) ____ 36 ± √ 441 36 ± 21 = __________ = ________ 10 10 t = 1.5 months or 5.7 months

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 = 6x 6x2 + 11x − 20 6x

(iv) When t = 4

2x = 8

v = 10 + 36(4) − 5(4)2 v = 74 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. t2 Q. 11. d = 3t + __ 4 d = distance in m t = number of seconds (i) When t = 1 12 d = 3(1) + __ 4 d = 3_14 3.25 m travelled in 1 second

32

∴x=4

i.e. €74,000


If x = 10 3x − 2 = 3(10) − 2

If x = 4 3x − 2 = 3(4) − 2

= 28 2x + 5 = 2(10) + 5

= 10 2x + 5 = 2(4) + 5

= 25

= 13

Original dimensions of the garden are 25 m by 28 m OR 10 m by 13 m

(ii) When d = 14 t2 3t + __ = 14 4 12t + t2 = 56

(× 4)

t2 + 12t − 56 = 0 a = 1, b = 12, c = −56 b2 − 4ac = (12)2 − 4(1)(−56) = 368 ____

t=

−12 ± √ 368 ____________

2 × 1____ −12 + √ 368 t = ____________ 2

t = 3.5|91...

____

−12 − √ 368 ____________

OR

t=

OR

t = −15.591...

2

Since t ≥ 0, t = −15.591... rejected ∴ t = 3.6 seconds to 1 d.p. distance travelled (iii) Average speed = _______________ time taken when d = 32 t2 (× 4) ⇒ 3t + __ = 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____ ____ −12 − √ 656 −12 + √ 656 ____________ ____________ OR t = t= 2 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 x2

(÷ 2)

− 263x + 15,960 = 0 Active Maths 2 (Strands 1–5): Ch 17 Solutions

33

a = 1, b = −263, c = 15,960 b2 − 4ac = (−263)2 − 4(1)(15,960) = 5,329 ______

−(−263) ± √5,329 x = _________________ 2 ______ ×1 263 + 5,329 √ x = _____________ OR 2 x = 168

______

x=

263 − √5,329 _____________ 2

x = 95

OR

The lengths of the sides of the triangle are 193 cm, 168 cm and 95 cm. Q. 13. h(t) = 54t − 5t2

Q. 14. x = width of garden

h = height in m

House

t = time in seconds x

(i) At t = 3

x

h = 54(3) − 5(3)2

y

h = 117

80 metres of fencing

Height will be 117 m

(i) y + x + x = 80

(ii) When launched t = 0 ∴ h = 54(0) − 5(0)2

y = 80 − 2x (ii) Area = l × w

h=0

= (80 − 2x)x

Height is zero metres

= 80x − 2x2

(iii) When h = 0 ⇒ 54t − 5t2 = 0

(×−1)

5t2 − 54t = 0 t(5t − 54) = 0 ∴ t = 0 OR

54 t = ___ 5

t = 10_45 It will take 10.8 seconds for the rocket to return to ground.

(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 ______

80 ± √2,500 _____________

x=

2(2)

x = 32.5 OR

80 ± 50 = ________ 4

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

34


Q. 15.

x y

x y

x y

x

x

y x

Length of fencing = 6x + 4y Length of fencing = 800 m ∴ 6x + 4y = 800 4y = 800 − 6x 3 y = 200 − __x 2

(

)

3 Area = 200 − __x (3x) = 10,200 2 9x2 600x − ___ – 10,200 = 0 (× −2) 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 ________

−(−1,200) ± √ 705,600 x = _____________________ 2×9 1,200 − 840 1,200 + 840 x = ____________ OR x = ____________ 18 18 x = 113_13 OR

x = 20

3 If x = 20 then y = 200 − __(20) 2 y = 170 3 _ 1 If x = 1133 then y = 200 − __ 113_13 2 y = 30

(

)

∴ Possible sets of lengths and widths of each plot are 20 m and 170 m or 30 m and 113_13 m. Q. 16.

(a)

1

8

7

x

2

3

8m

x

4

10 m

6

5

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 Active Maths 2 (Strands 1–5): Ch 17 Solutions

35

(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)

2x + 8 x

x

x

8m

2x + 10

10 m

(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 ______ −36 ± √2,304 ______________ ∴x= 2×4 −36 + 48 ∴ x = _________ OR 8 x = 1_12

OR

−36 − 48 x = _________ 8 x = −10_12

Since x > 0, x = –10_12 rejected

∴ x = 1_12

The width of the path is 1.5 m.

36


(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: Sum of lengths: Shorter length Larger length

p

p+q

_____

_________

q

p

p (b) x = __ q 1 (c) 1 + __ x

1 (j) ______ ≈ 0.618 1.618 1 So, 1 + ______ ≈ 1.618 1.618 So, the golden ratio minus its reciprocal equals 1.

p+q p ______ = (d) __ q p q p __ __ = 1 + q p 1 __ x=1+ x x≠0 x2 = x + 1

OR 1 x = 1 + __ x (from part (d))

x2 − x − 1 = 0 __

(e) x =

So, the golden ratio minus its reciprocal equals 1.

1 ± √5 _______ 2

x ≈ −0.618 OR

(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. 1 (i) −__ x1 = 1.618 = x2 = golden ratio (to three decimal places)

x ≈ 1.618

(k) 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

(f) x1 = −0.618 x2 = 1.618 (g) x2


37

(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 in the Fibonacci sequence are approximately in golden ratio to each other the further we move into the sequence.

38


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