11[a math cd]

1 © Penerbitan Pelangi Sdn. Bhd.

10. 4x2 − 1= (2x)2 − 12

= (2x − 1)(2x + 1)

11. x2 − 2x − 3= (x − 3)(x + 1)

x −3 −3x

x 1 x +x2 −3 −2x

12. m2 + 5m − 14= (m + 7)(m − 2)

m 7 7m

m –2 –2m +m2 –14 5m

13. p2 − 6p + 8= (p − 4)(p − 2)

p –4 –4p

p –2 –2p +p2 8 –6p

14. h2 + 8h + 15= (h + 5)(h + 3)

h 5 5h

h 3 3h +h2 15 8h

15. 2y2 + 5y − 3= (2y − 1)(y + 3)

2y –1 –y

y 3 6y +2y2 –3 5y

Paper 2

1. (x + 4)(x + 2)

= (x)(x) + (x)(2) + (4)(x) + (4)(2)= x2 + 2x + 4x + 8= x2 + 6x + 8

2. (x − 1)(x + 3)

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

3. (2x − 3)(x + 4)= (2x)(x) + (2x)(4) + (−3)(x) + (−3)(4)= 2x2 + 8x − 3x − 12= 2x2 + 5x − 12

4. (3x − 5)(x − 2)= (3x)(x) + (3x)(−2) + (−5)(x) + (−5)(−2)= 3x2 − 6x − 5x + 10= 3x2 − 11x + 10

5. (x + 5)(4x − 1)= (x)(4x) + (x)(−1) + (5)(4x) + (5)(−1)= 4x2 − x + 20x − 5= 4x2 + 19x − 5

6. 2x2 + 4 = 2(x2 + 2)

7. x2 − 3x = x(x − 3)

8. x2 − 1= x2 − 12

= (x − 1)(x + 1)

9. x2 − 9= x2 − 32

= (x − 3)(x + 3)

CHAPTER

11 Quadratic Expressions and Equations

CHAPTER

2

Mathematics SPM Chapter 11

© Penerbitan Pelangi Sdn. Bhd.

16. 2x2 + x − 15= (2x − 5)(x + 3)

2x –5 –5x

x 3 6x +2x2 –15 x

17. 3x2 − 5x − 12= (3x + 4)(x − 3)

3x 4 4x

x –3 –9x +3x2 –12 –5x

18. 3p2 − 11p + 6= (3p − 2)(p − 3)

3p –2 –2p

p –3 –9p +3p2 6 –11p

19. x2 = 4x − 3x2 − 4x + 3 = 0

20. 2h2 + 3h = 2h + 15 2h2 + 3h − 2h − 15 = 0 2h2 + h − 15 = 0

21. x(x + 1) = 12 x2 + x = 12 x2 + x − 12 = 0

22. p2 = 4(p − 2) = 4p − 8p2 − 4p + 8 = 0

23. 3y2 = 2(y − 1) + 7 3y2 = 2y − 2 + 73y2 − 2y − 5 = 0

24. 2m2 − 5–––––––3

= 3m

2m2 − 5 = 3 × 3m 2m2 − 5 = 9m 2m2 − 9m − 5 = 0

25. 2n2 + 5n––––––––1 + n = 2

2n2 + 5n = 2(1 + n) 2n2 + 5n = 2 + 2n 2n2 + 5n − 2 − 2n = 0 2n2 + 3n − 2 = 0

26. x2 − 4x = 0 x(x − 4) = 0x = 0 or x − 4 = 0 x = 4Hence, x = 0 or x = 4.

27. 2t2 = 3t 2t2 − 3t = 0 t(2t − 3) = 0t = 0 or 2t − 3 = 0 2t = 3

t = 3—2

Hence, t = 0 or t = 3—2.

28. (p − 2)(p + 3) = 0p − 2 = 0 or p + 3 = 0 p = 2 p = −3

Hence, p = 2 or p = −3.

29. y2 − 3y − 4 = 0 (y − 4)(y + 1) = 0

y – 4 –4y

y 1 y +y2 – 4 –3y

y − 4 = 0 or y + 1 = 0 y = 4 y = −1

Hence, y = 4 or y = −1.

30. 2k2 − 9k − 5 = 0(2k + 1)(k − 5) = 0

2k 1 k

k –5 –10k +2k2 –5 –9k

2k + 1 = 0 or k − 5 = 0 2k = −1 k = 5

k = − 1—2

Hence, k = − 1—2 or k = 5.

3



Paper 2

1. x(2x – 1) = 10 2x2 – x = 10 2x2 – x − 10 = 0 (2x − 5)(x + 2) = 0

2x –5 –5x

x 2 4x +2x2 –10 –x

2x − 5 = 0 or x + 2 = 0 2x = 5 x = −2

x = 5—2

Hence, x = 5—2 or x = −2.

2. 3x(3x + 2) – 1 = 6x 9x2 + 6x – 1 = 6x 9x2 + 6x – 6x = 1 9x2 = 1

x = 19

x = 13

or x = – 13

3. (x + 4)2 = 7x + 22 (x + 4)(x + 4) = 7x + 22 x(x + 4) + 4(x + 4) = 7x + 22 x2 + 4x + 4x + 16 = 7x + 22 x2 + 8x + 16 = 7x + 22 x2 + x – 6 = 0 (x + 3)(x – 2) = 0

x 3 3x

x –2 –2x +x2 –6 x

(x – 3) = 0 or (x – 2) = 0 x = –3 x = 2

Hence, x = –3 or x = 2.

4. 5x2 + 11x = 2(3 – x) 5x2 + 11x = 6 – 2x 5x2 + 11x + 2x – 6 = 0 5x2 + 13x − 6 = 0 (5x – 2)(x + 3) = 0

5x –2 –2x

x 3 15x +5x2 –6 13x

5x – 2 = 0 or x + 3 = 0

x = 25

x = –3

Hence, x = 25

or x = –3.

5. 2h2 – 8–––––––5 = 3h

2h2 − 8 = 5 × 3h 2h2 − 8 = 15h 2h2 − 15h − 8 = 0 (2h + 1)(h − 8) = 0

2h 1 h

h –8 –16h +2h2 –8 –15h

2h + 1 = 0 or h − 8 = 0 2h = −1 h = 8

h = − 1—2

Hence, h = − 1—2 or h = 8.

6. 5x + 152x

= x – 1

5x + 15 = 2x(x – 1) 5x + 15 = 2x2 – 2x 0 = 2x2 – 2x – 5x – 15 2x2 – 7x – 15 = 0 (2x + 3)(x – 5) = 0

2x 3 3x

x –5 –10x +2x2 –15 –7x

2x + 3 = 0 or x − 5 = 0 2x = −3 x = 5

x = − 3—2

Hence, x = − 3—2 or x = 5.

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7. 3x(x − 2)––––––––4

= 2 − x

3x(x − 2) = 4(2 − x) 3x2 − 6x = 8 − 4x 3x2 − 6x − 8 + 4x = 0 3x2 − 2x − 8 = 0 (3x + 4)(x − 2) = 0

3x 4 4x

x –2 –6x +3x2 –8 –2x

3x + 4 = 0 or x − 2 = 0 3x = −4 x = 2

x = − 4—3

Hence, x = − 4—3 or x = 2.

8. x2 – 2x – 11 = 3(x + 1) x2 – 2x – 11 = 3x + 3 x2 – 2x – 3x – 11 – 3 = 0 x2 – 5x – 14 = 0 (x – 7)(x + 2) = 0

x –7 –7x

x 2 2x +x2 –14 –5x

x – 7 = 0 or x + 2 = 0 x = 7 x = –2

Hence, x = 7 or x = –2.

Paper 2

1. p(p + 1) = 12 p2 + p = 12 p2 + p − 12 = 0(p + 4)(p − 3) = 0

p 4 4p

p –3 –3p +p2 –12 p

p + 4 = 0 or p − 3 = 0 p = −4 p = 3

Hence, p = −4 or p = 3.

2. n2 = 4(n + 3) n2 = 4n + 12 n2 − 4n − 12 = 0 (n − 6)(n + 2) = 0

n –6 –6n

n 2 2n +n2 –12 –4n

n − 6 = 0 or n + 2 = 0 n = 6 n = −2Hence, n = 6 or n = −2.

3. 2x2 = 7x − 3 2x2 − 7x + 3 = 0 (2x − 1)(x − 3) = 0

2x –1 –x

x –3 –6x +2x2 3 –7x

2x − 1 = 0 or x − 3 = 0 2x = 1 x = 3

x = 1—2

Hence, x = 1—2 or x = 3.

4. (y + 4)2 = 9 (y + 4)(y + 4) = 9 y2 + 4y + 4y + 16 = 9 y2 + 8y + 7 = 0 (y + 7)(y + 1) = 0

y 7 7y

y 1 y +y2 7 8y

y + 7 = 0 or y + 1 = 0 y = −7 y = −1Hence, y = −7 or y = −1.

5. 2y2 = y + 36 2y2 − y − 36 = 0(2y − 9)(y + 4) = 0

2y –9 –9y

y 4 8y +2y2 –36 –y

2y − 9 = 0 or y + 4 = 0 2y = 9 y = −4

y = 9—2Hence, y = 9—2

or y = −4.

5



6. (x + 3)2 = 2x + 21 x2 + 6x + 9 = 2x + 21 x2 + 6x + 9 – 2x – 21 = 0 x2 + 4x – 12 = 0 (x + 6)(x − 2) = 0

x 6 6x

x –2 –2x +x2 –12 4x

(x + 6) = 0 or (x − 2) = 0 x = −6 x = 2Hence, x = –6 or x = 2.

7. 3 – 2k2–––––––

5 = k

3 − 2k2 = 5k 0 = 2k2 + 5k − 3 2k2 + 5k − 3 = 0 (2k − 1)(k + 3) = 0

2k –1 –k

k 3 6k +2k2 –3 5k

2k − 1 = 0 or k + 3 = 0 2k = 1 k = −3

k = 1—2Hence, k = 1—2

or k = −3.

8. 5 − 23x = x + 5x2

0 = x + 5x2 − 5 + 23x 5x2 + 24x − 5 = 0(5x − 1)(x + 5) = 0

5x –1 –x

x 5 25x +5x2 –5 24x

5x − 1 = 0 or x + 5 = 0 5x = 1 x = −5

x = 1—5Hence, x = 1—5

or x = −5.

9. 4x(x – 2) = 21 4x2 – 8x = 21 4x2 – 8x – 21 = 0 (2x + 3)(2x − 7) = 0

2x 3 6x

2x –7 –14x +4x2 –21 –8x

(2x + 3) = 0 or (2x – 7) = 0 2x = −3 2x = 7 x = – 3

2 x = 7

2

Hence, x = − 32

or x = 7—2.

10. (2h + 1)(2h + 2) = 6 4h2 + 4h + 2h + 2 = 6 4h2 + 6h + 2 − 6 = 0 4h2 + 6h − 4 = 0 2h2 + 3h − 2 = 0 (2h − 1)(h + 2) = 0

2h –1 –h

h 2 4h +2h2 –2 3h

2h − 1 = 0 or h + 2 = 0 2h = 1 h = −2

h = 1—2

Hence, h = 1—2 or h = −2.

11. 2 + 9v − 2v2 = 4(1 + v) 2 + 9v − 2v2 = 4 + 4v 0 = 4 + 4v − 2 − 9v + 2v2

0 = 2v2 − 5v + 2 2v2 − 5v + 2 = 0 (2v − 1)(v − 2) = 0

2v –1 –v

v –2 –4v +2v2 +2 –5v

2v − 1 = 0 or v − 2 = 0 2v = 1 v = 2

v = 1—2

Hence, v = 1—2 or v = 2.

6



12. 5k + 2––––––3

= k2

5k + 2 = 3k2

0 = 3k2 − 5k − 2 3k2 − 5k − 2 = 0 (3k + 1)(k − 2) = 0

3k 1 k

k –2 –6k +3k2 –2 –5k

3k + 1 = 0 or k − 2 = 0 3k = −1 k = 2

k = − 1—3

Hence, k = − 1—3 or k = 2.

13. 4m2 – 2 –––––––7m

= 1

4m2 − 2 = 7m 4m2 − 7m − 2 = 0 (4m + 1)(m − 2) = 0

4m 1 m

m –2 –8m +4m2 –2 –7m

4m + 1 = 0 or m − 2 = 0 4m = −1 m = 2

m = − 1—4

Hence, m = – 1—4 or m = 2.

14. 2k = 2 – k–––––3k

2k × 3k = 2 − k 6k2 = 2 − k 6k2 + k − 2 = 0 (3k + 2)(2k − 1) = 0

3k 2 4k

2k –1 –3k +6k2 –2 k

3k + 2 = 0 or 2k − 1 = 0 3k = −2 2k = 1

k = − 2—3 k = 1—2

Hence, k = − 2—3 or k = 1—2

.

15. (a) Area of square ABCD = (m + 2)(m + 2)

(b) (m + 2)(m + 2) = 36 m2 + 2m + 2m + 4 = 36 m2 + 4m + 4 − 36 = 0 m2 + 4m − 32 = 0 (m − 4)(m + 8) = 0

m –4 –4m

m 8 8m +m2 –32 4m

m − 4 = 0 or m + 8 = 0 m = 4 m = −8 (Impossible)

Hence, m = 4.

16. (2t)2 + (2t − 1)2 = 52

4t2 + (2t − 1)(2t − 1) = 25 4t2 + (4t2 − 4t + 1) = 25 8t2 − 4t + 1 − 25 = 0 8t2 − 4t − 24 = 0 2t2 − t − 6 = 0 (2t + 3)(t − 2) = 0

2t 3 3t

t –2 –4t +2t2 –6 –t

2t + 3 = 0 or t − 2 = 0

t = − 3—2 (Impossible) t = 2

Hence, t = 2.

11[a math cd]

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