02[anal add math cd]

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Additional Mathematics SPM Chapter 2

© Penerbitan Pelangi Sdn. Bhd.

1. (b), (c) and (d) are quadratic equations.

2. (a) 3x – 4 = x2

x2 – 3x + 4 = 0

(b) x(4 – x) = 5 4x – x2 = 5 x2 – 4x + 5 = 0

(c) (x – 1)(5 + x) = 2x 5x + x2 – 5 – x = 2x x2 + 4x – 5 – 2x = 0 x2 + 2x – 5 = 0

(d) x – 2 = 4xx + 1

(x – 2)(x + 1) = 4x x2 + x – 2x – 2 = 4x x2 – 5x – 2 = 0

(e) 5(x + 3)(2x – 1) = (x + 3)(4 – x) 5(2x2 – x + 6x – 3) = 4x – x2 + 12 – 3x 10x2 – 5x + 30x – 15 = 4x – x2 + 12 – 3x 10x2 + 25x – 15 = x – x2 + 12 10x2 + 25x – 15 – x + x2 – 12 = 0 11x2 + 24x – 27 = 0

3. (a) Substitute x = 1 into the expression, x2 – 2x + 1 = 12 – 2(1) + 1 = 0 Thus, x = 1 is a root.

(b) Substitute x = –2 into the expression, 5x2 – 3x = 5(–2)2 – 3(–2) = 20 + 6 =26(≠6) Thus, x = –2 is not a root.

(c) Substitute x = 2 into 3x2 and 4x + 4 respectively, 3x2 = 3(2)2

= 12

4x + 4 = 4(2) + 4 = 12 Since LHS = RHS, therefore x = 2 is a root.

4. (a) (x + 5) = 0 x = –5 Hence, x = –5 is a root.

(b) 2x – 1 = 0

x = 12

Hence, x = 12 is a root.

(c) When (1 – 3x) = 0

x = 13

When (x + 3) = 0 x = –3 Hence, x = 3 is not a root.

5. (a) x2 – 9 = 0 Try using the factors of 9, that is, 1, 9, –1, –9,

3, –3.

When x = 3 or x = –3, x2 – 9 = 0

Therefore, x = 3 and x = –3 are the roots.

Alternative Using improvement method,

x x2 – 9–1 –8–2 –5–3 01 –82 –53 0

Therefore, x = –3 and x = 3 are the roots.

CHAPTER

2 Quadratic Equations

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(b) x2 – 3x – 4 = 0 Try using the factor of 4, that is, 1, –1, 2, –2, 4, – 4.

When x = 1, x2 – 3x – 4 = 1 – 3 – 4 =–6≠0

When x = –1, x2 – 3x – 4 = 1 + 3 – 4 = 0

When x = 4, x2 – 3x – 4 = 42 – 3(4) – 4 = 0


(c) 3x2 – 3x – 6 = 0 x2 – x – 2 = 0 Try using the factors of 2, that is, 1, –1, 2, –2.

When x = 1, x2 – x – 2 = 1 – 1 – 2 =–2≠0

When x = –1, x2 – x – 2 = 1 + 1 – 2 = 0

When x = 2, x2 – x – 2 = 4 – 2 – 2 = 0


6. (a) 3x2 = x 3x2 – x = 0 x(3x – 1) = 0 x = 0 or 3x – 1 = 0

x = 1—3(b) x2 – 4 = 0 x2 = 4 x = ±AB4 = ±2

(c) x2 + 3x + 2 = 0 (x + 1)(x + 2) = 0 x + 1 = 0 or x + 2 = 0 x = –1 or x = –2

(d) 4x2 – 2x – 6 = 0 2x2 – x – 3 = 0 (2x – 3)(x + 1) = 0 2x – 3 = 0 or x + 1 = 0

x = 3—2 or x = –1

(e) 3x2 – 8 = 2x 3x2 – 2x – 8 = 0 (3x + 4)(x – 2) = 0 3x + 4 = 0 or x – 2 = 0 x = – 4—3 or x = 2

(f) (x – 1)(x + 2) = 2x x2 + 2x – x – 2 = 2x x2 – x – 2 = 0 (x – 2)(x + 1) = 0 x – 2 = 0 or x + 1 = 0 x = 2 or x = –1

(g) x + 3––––––2x – 1

= x + 3

x + 3 = (x + 3)(2x – 1) = 2x2 – x + 6x – 3 2x2 + 5x – 3 – x – 3 = 0 2x2 + 4x – 6 = 0 x2 + 2x – 3 = 0 (x + 3)(x – 1) = 0 x + 3 = 0 or x – 1 = 0 x = –3 or x = 1

7. (a) x2 + 4x = 1 x2 + 4x + 22 = 1 + 22

(x + 2)2 = 5 x + 2 = ±AB5 x = ±AB5 – 2 = AB5 – 2 or –AB5 – 2 = 0.2361 or – 4.236

(b) 2x2 + 4x – 3 = 0

x2 + 2x – 3—2 = 0

x2 + 2x = 3—2

x2 + 2x + 12 = 3—2 + 12

(x + 1)2 = 5—2

x + 1 = ±ABB5—2 x = ±ABB5—2 – 1

= ABB5—2 – 1 or –ABB5—2 – 1

= 0.5811 or –2.581

(c) (x – 1)(x – 2) = 1 x2 – 3x + 2 = 1 x2 – 3x = 1 – 2

x2 – 3x + 1 3—2 22 = –1 + 1 3—2 2

2

1x – 3—2 22 = 5—4

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x – 3—2 = ±ABB5—4 x = ±ABB5—4 + 3—2

= ABB5—4 + 3—2 or –ABB5—4 + 3—2 = 2.618 or 0.3820

(d) 2x – 1––––––––

1 + 11–––2 x

= –2––––––1 – 3x

(2x – 1)(1 – 3x) = –2 – 11x 2x – 6x2 – 1 + 3x = –2 – 11x 6x2 – 5x + 1 – 2 – 11x = 0 6x2 – 16x – 1 = 0 6x2 – 16x = 1

x2 – 16–––6 x = 1—6

x2 – 8—3 x = 1—6

x2 – 8—3 x + 1 4—3 22 = 1—6 + 1 4—3 2

2

1x – 4—3 22 = 1—6 + 16–––9

= 35–––18

x – 4—3 = ±ABBB35–––18

x = ±ABBB35–––18 + 4—3

= ABBB35–––18 + 4—3

or –ABBB35–––18 + 4—3 = 2.728 or –0.06110

8. (a) x2 + 4x = 1 x2 + 4x – 1 = 0 So, a = 1, b = 4 and c = –1

x = –b ± ABBBBBBb2 – 4ac–––––––––––––2a

= – 4 ± ABBBBBBBBB42 – 4(1)(–1)–––––––––––––––––2(1)

= – 4 ±ABB20––––––––2

= – 4 + ABB20–––––––––2

or – 4 – ABB20–––––––––2

= 0.236 or –4.236

(b) 2x2 + 4x – 3 = 0 So, a = 2, b = 4 and c = –3

x = –b ± ABBBBBBb2 – 4ac–––––––––––––2a

= – 4 ± ABBBBBBBBB42 – 4(2)(–3)––––––––––––––––––

2(2)

= – 4 ±ABB40––––––––4

= – 4 + ABB40–––––––––4

or – 4 – ABB40––––––––4

= 0.581 or –2.581

(c) (x – 1)(x – 2) = 1 x2 – 3x + 2 = 1 x2 – 3x + 1 = 0 So, a = 1, b = –3 and c = 1

x = –b ± ABBBBBBb2 – 4ac–––––––––––––2a

= –(–3) ± ABBBBBBBBBB(–3)2 – 4(1)(1)–––––––––––––––––––––2(1)

= 3 ± AB5–––––––2

= 3 + AB5–––––––2

or 3 – AB5––––––2

= 2.618 or 0.382

(d) 2x – 1–––––––––

1 + 11–––2 x

= –2––––––1 – 3x

(2x – 1)(1 – 3x) = –2 – 11x 2x – 6x2 – 1 + 3x = –2 – 11x 6x2 – 16x – 1 = 0 So, a = 6, b = –16 and c = –1

x = –b ± ABBBBBBb2 – 4ac–––––––––––––2a

= –(–16) ± ABBBBBBBBBBBB(–16)2 – 4(6)(–1)––––––––––––––––––––––––2(6)

= 16 ± ABBB280––––––––––12

= 16 + ABBB280––––––––––12

or 16 – ABBB280––––––––––12

= 2.728 or –0.061

9. (a) Sum of roots = 1 + 3 = 4

Product of roots = 1 × 3 = 3 Hence, the quadratic equation is x2 – 4x + 3 = 0.

(b) Sum of roots = –2 + 5 = 3

Product of roots = (–2)(5) = –10

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Hence, the quadratic equation is x2 – 3x + (–10) = 0 x2 – 3x – 10 = 0

(c) Sum of roots = (– 6) + (–1) = –7

Product of roots = (–6)(–1) = 6

Hence, the quadratic equation is x2 – (–7)x + 6 = 0 x2 + 7x + 6 = 0(d) Sum of roots = 1—2 + 7

= 15–––2

Product of roots = 1 1—2 2(7)

= 7—2 Hence, the quadratic equation is

x2 – 15–––2 x + 7—2 = 0

2x2 – 15x + 7 = 0

(e) Sum of roots = 4 + 4 = 8

Product of roots = 4 × 4 = 16

Hence, the quadratic equation is x2 – 8x + 16 = 0.

10. (a) x2 – 3x – 4 = 0

Therefore, sum of roots = 3 product of roots = – 4

(b) x2 + 8x + 1 = 0

Therefore, sum of roots = –8 product of roots = 1

(c) 2x2 – 6x – 7 = 0 x2 – 3x – 7—2 = 0

Therefore, sum of roots = 3 product of roots = – 7—2

(d) (x – 1)(x + 3) = 8 x2 + 2x – 3 – 8 = 0 x2 + 2x – 11 = 0

Therefore, sum of roots = –2 product of roots = –11

(e) x – 2––––––2x + 1 = x—5

5(x – 2) = x(2x + 1) 5x – 10 = 2x2 + x

2x2 – 4x + 10 = 0 x2 – 2x + 5 = 0

Therefore, sum of roots = 2 product of roots = 5

11. (a) 4x2 – 5x + 1 = 0 So, a = 4, b = –5 and c = 1

b2 – 4ac = (–5)2 – 4(4)(1) = 25 – 16 = 9 . 0 Hence, the two roots are distinct.(b) 3x2 + 2x + 6 = 0 So, a = 3, b = 2 and c = 6

b2 – 4ac = 22 – 4(3)(6) = 4 – 72 = – 68 , 0

Hence, there is no real roots.

(c) x2 + 4x + 4 = 0 So, a = 1, b = 4 and c = 4

b2 – 4ac = 42 – 4(1)(4) = 0 Hence, the two roots are equal.

(d) 5x – 8 = x2

x2 – 5x + 8 = 0 So, a = 1, b = –5 and c = 8

b2 – 4ac = (–5)2 – 4(1)(8) = 25 – 32 = –7 , 0

Hence, there is no real roots.

(e) (x – 3)(2x + 1) = 6x 2x2 – 5x – 3 – 6x = 0 2x2 – 11x – 3 = 0 So, a = 2, b = –11 and c = –3

b2 – 4ac = (–11)2 – 4(2)(–3) = 121 + 24 = 145 . 0

Hence, there are two different roots. (f) 2x – 1 = 4x––––––

3x + 5 (2x – 1)(3x + 5) = 4x 6x2 + 10x – 3x – 5 – 4x = 0 6x2 + 3x – 5 = 0 So, a = 6, b = 3 and c = –5

b2 – 4ac = 32 – 4(6)(–5) = 9 + 120 = 129 . 0

Hence, there are two different roots.

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12. 2x2 – kx + 2 = 0So, a = 2, b = –k and c = 2

Since the roots are equal, then b2 – 4ac = 0(–k)2 – 4(2)(2) = 0 k2 = 16 k = ±4

13. x2 – 3x – k = 0So, a = 1, b = –3 and c = –k

Since the roots are different,then b2 – 4ac . 0(–3)2 – 4(1)(–k) . 0 9 + 4k . 0 4k . –9 k . – 9—4

14. kx2 + 4x – 1 = 0So, a = k, b = 4 and c = –1

Since the roots are not real, then b2 – 4ac , 0

42 – 4k(–1) , 0 42 + 4k , 0 4k , –16 k , –4

15. kx2 + hx – 4 = 0So, a = k, b = h and c = –4

Since the roots are equal, then b2 – 4ac = 0 h2 – 4k(– 4) = 0 h2 + 16k = 0

16. 2x2 + px = k 2x2 + px – k = 0So, a = 2, b = p and c = –k

Since the roots are not real, then b2 – 4ac , 0 p2 – 4(2)(–k) , 0 p2 + 8k , 0

17. px2 – qx = 4 px2 – qx – 4 = 0So, a = p, b = –q and c = – 4

Since the roots are different, then b2 – 4ac . 0(–q)2 – 4(p)(– 4) . 0 q2 + 16p . 0

18. x2 – kx + 9 = 6x x2 – kx – 6x + 9 = 0x2 – (k + 6)x + 9 = 0So, a = 1, b = – (k + 6) and c = 9

Since the roots are equal, then b2 – 4ac = 0[–(k + 6)]2 – 4(1)(9) = 0 (k + 6)2 – 36 = 0 (k + 6)2 = 36 k + 6 = ±6 k = ±6 – 6 = 6 – 6 or –6 – 6 = 0 or –12

19. (x – 4)(2x + 3) = k2x2 + 3x – 8x – 12 – k = 0 2x2 – 5x – 12 – k = 0So, a = 2, b = –5 and c = –12 – k

Since the roots are real, then b2 – 4ac > 0 (–5)2 – 4(2)(–12 – k) > 0 25 + 96 + 8k > 0 121 + 8k > 0 8k > –121

k > – 121––––8

20. Given y = 4x – 1 ................................ 1and y = kx2 + 3x – 2 ....................... 2

Substitute 1 into 2, 4x – 1 = kx2 + 3x – 2kx2 + 3x – 4x – 2 + 1 = 0 kx2 – x – 1 = 0So, a = k, b = –1 and c = –1

Since the straight line intersects the curve at two different points, then b2 – 4ac . 0 (–1)2 – 4(k)(–1) . 0 1 + 4k . 0 4k . –1

k . – 1—4

21. Given y = hx – k ................................. 1and y = 4x2 – 5x + 6 ....................... 2

Substitute 1 into 2, hx – k = 4x2 – 5x + 6 4x2 – 5x – hx + 6 + k = 0 4x2 – (5 + h)x + 6 + k = 0So, a = 4, b = – (5 + h) and c = 6 + k

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Since the straight line does not intersect the curve,then b2 – 4ac , 0 [–(5 + h)]2 – 4(4)(6 + k) , 0 (5 + h)2 – 96 – 16k , 025 + 10h + h2 – 96 – 16k , 0 h2 + 10h – 16k , 96 – 25 h2 + 10h – 16k , 71

1. (2 – x)(x + 1) = 1—4 x(x – 5)

2x + 2 – x2 – x = 1—4 x2 – 5—4 x

x – x2 + 2 = 1—4 x2 – 5—4 x

1—4 x2 + x2 – 5—4 x – x – 2 = 0

5—4 x2 – 9—4 x – 2 = 0

Multiply both sides by 4,5x2 – 9x – 8 = 0So, a = 5, b = –9 and c = –8

x = –b ± ABBBBBBb2 – 4ac2a

= –(–9) ± ABBBBBBBBBBB(–9)2 – 4(5)(–8)2(5)

= 9 ± ABBB24110

= 9 + ABBB24110

or 9 – ABBB24110

= 2.452 or – 0.6524

2. 2x2 + ABpx = q – 12x2 + ABpx + 1 – q = 0So, a = 2, b = ABp and c = 1 – q

Since the equation has two equal roots,then b2 – 4ac = 0 (ABp)2 – 4(2)(1 – q) = 0 p – 8(1 – q) = 0 p – 8 + 8q = 0 8q = 8 – p

q = 8 – p

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3. Sum of roots = –5 + 23

= –15 + 23

= – 133

Product of roots = (–5)1 2—3 2 = – 10–––3Hence, the quadratic equation is

x2 – 1– 13–––3 2x + 1– 10–––3 2 = 0

x2 + 13–––3 x – 10–––3 = 0

Multiply both sides by 3,3x2 + 13x – 10 = 0

4. (a) (x – 1)(x + 2) = 3 x2 + 2x – x – 2 – 3 = 0 x2 + x – 5 = 0

(b) product of roots = –5

(c) a = 1, b = 1, c = –5 b2 – 4ac = 12 – 4(1)(–5) = 21 > 0

\ There are 2 different real roots.

5. 4nx2 + x + 4nx + n – 2 = 04nx2 + (1 + 4n)x + n – 2 = 0a = 4n, b = 1 + 4n, c = n – 2For two equal roots, b2 – 4ac = 0 (1 + 4n)2 – 4(4n)(n – 2) = 0 1 + 8n + 16n2 – 16n2 + 32n = 0 40n + 1 = 0 n = – 1

40

6. 3x2 – 4x + p – 1 = 0a = 3, b = –4, c = p – 1 b2 – 4ac , 0 (–4)2 – 4(3)(p – 1) , 0 16 – 12p + 12 , 0 28 – 12p , 0 28 , 12p

2812

, p

p . 73

7



1. Substitute x = 5 into 3x2 – px + 6 = 0,3(5)2 – p(5) + 6 = 0 75 – 5p + 6 = 0 5p = 81 p = 81

5

2. 2x2 + px + q = 0

x2 + p—2 x +

q—2 = 0

Sum of roots = – p—2

2 + (–3) = – p—2

–1 = – p—2

p = 2

Product of roots = q—2

2(–3) = q—2

q = –12

3. px2 + 2x = –px + q – 1 px2 + 2x + px + 1 – q = 0 px2 + (2 + p)x + 1 – q = 0

x2 + 1 2 + pp 2x + 1 1 – q

p 2 = 0

Sum of roots = – 1 2 + pp 2

1—2 + (– 4) = – 2—p – 1

2—p = 5—2

p = 4—5

Product of roots = 1 – q

p

1—2 (– 4) = 1 – q

45

–2 = (1 – q)1 54 2

= 54 – 5

4 q

54 q = 13

4

q = 135

4. (x – 1)(x + 2) = 3(x – 1) x2 + 2x – x – 2 = 3x – 3 x2 + x – 2 – 3x + 3 = 0 x2 – 2x + 1 = 0 (x – 1)2 = 0 x = 1

5. x – 4 = xx + 2

(x – 4)(x + 2) = xx2 + 2x – 4x – 8 = x x2 – 3x – 8 = 0So, a = 1, b = –3 and c = –8


= –(–3) ± ABBBBBBBBBBB(–3)2 – 4(1)(–8)2(1)

= 3 ± ABB412

= 3 + ABB412

or 3 – ABB412

= 4.702 or –1.702

6. 6—5 y = y2 – 1

Multiply both sides by 5, 6y = 5y2 – 55y2 – 6y – 5 = 0So, a = 5, b = –6 and c = –5


= –(– 6) ± ABBBBBBBBBBB(– 6)2 – 4(5)(–5)2(5)

= 6 ± ABBBBBBB36 + 10010

= 6 ± ABBB13610

= 6 + ABBB13610

or 6 – ABBB13610

= 1.766 or – 0.5662

7. x2 – 6x + 1 = (x2 – 6x + 32) – 32 + 1 Completing the square = (x – 3)2 – 8

Compare (x – 3)2 – 8 with (x + m)2 + n,therefore m = –3 and n = –8.

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8. x2 – 4x + 2 = 0 x2 – 4x + 22 – 22 + 2 = 0 (x – 2)2 – 2 = 0

Hence, a = 1, b = –2 and c = –2.

9. 3x2 – 6x – 1 = 0

x2 – 2x – 1—3 = 0

x2 – 2x + 12 – 12 – 1—3 = 0

(x – 1)2 – 1 – 1—3 = 0

(x – 1)2 – 4—3 = 0

Hence, a = 1, b = –1 and c = – 4—3 .

10. 2x2 + 4x + 1 = 0

x2 + 2x + 1—2 = 0

x2 + 2x + 12 – 12 + 1—2 = 0

(x + 1)2 – 1—2 = 0

2x2 + 4x + 1 = 8

(x + 1)2 – 1—2 = 8

(x + 1)2 = 8 + 1—2

= 172

(x + 1) = ±ABBB172

x = –1 + ABBB172

or –1 – ABBB172

= 1.915 or –3.915

11. Sum of roots = 1—3 + (–5)

= 1—3 – 5

= – 14–––3

Product of roots = 1 1—3 2(–5)

= – 5—3

Therefore, the quadratic equation is

x2 – 1– 143 2x + 1– 5—3 2 = 0

x2 + 14–––3 x – 5—3 = 0

3x2 + 14x – 5 = 0

12. 2x2 + 6x – 9 = 0 x2 + 3x – 9—2 = 0

(a) Sum of roots = –3(b) Product of roots = – 9—2

13. 2x2 – kx + h—2 = 0

x2 – k—2 x + h—4 = 0

Sum of roots = k—2

4 + (–5) = k—2

–1 = k—2 k = –2

Product of roots = h—4

4(–5) = h—4 h = –80

14. 2x2 + 4x – 7 = 0

x2 + 2x – 7—2 = 0

a + b = –2 and ab = – 7—2

Sum of the roots 2a and 2b = 2a + 2b = 2(a + b) = 2(–2) = – 4

Product of the roots 2a and 2b = (2a)(2b) = 4ab

= 41– 7—2 2 = –14

Hence, the quadratic equation is x2 – (– 4)x + (–14) = 0 x2 + 4x – 14 = 0

15. Let a and 3a are the roots of quadratic equation 2x2 – 2 = 8x – 4k2x2 – 8x + 4k – 2 = 0 x2 – 4x + 2k – 1 = 0

Sum of roots = 4 a + 3a = 4 4a = 4 a = 1

Product of roots = 2k – 1 a(3a) = 2k – 1 3a2 = 2k – 1 3(1)2 = 2k – 1 2k = 4 k = 2

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16. 3x2 – 5x – 2 = 0 (3x + 1)(x – 2) = 0 x = – 1—3 or 2

Since a . 0 and b , 0, then a = 2 and b = – 1—3

Sum of roots = (a – 1) + 1b + 3—4 2 = (2 – 1) + 1– 1—3 + 3—4 2 = 1 – 1—3 + 3—4

= 1712

Product of roots = (a – 1)1b + 3—4 2 = (2 – 1)1– 1—3 + 3—4 2 = (1)1 – 4 + 9

12 2 = 5

12

Hence, the quadratic equation is x2 – 1712 x + 5

12 = 0

12x2 – 17x + 5 = 0

17. x2 + (1 – p)x + 4 = 0So, a = 1, b = 1 – p and c = 4

Since the roots are equal,then b2 – 4ac = 0 (1 – p)2 – 4(1)(4) = 0 (1 – p)2 = 16 1 – p = ±4 –p = ±4 – 1 –p = 4 – 1 or – 4 – 1 p = –3 or 5

18. x2 – 2x = 9(2x – 5) – 5p = 18x – 45 – 5p

x2 – 2x – 18x + 45 + 5p = 0 x2 – 20x + 45 + 5p = 0So, a = 1, b = –20 and c = 45 + 5p

Since the roots are equal,then b2 – 4ac = 0(–20)2 – 4(1)(45 + 5p) = 0 400 – 180 – 20p = 0 220 – 20p = 0 –20p = –220

p = –220–20

= 11

19. 3px – 5 = (qx)2 – 1 3px – 5 = q2x2 – 1q2x2 – 3px – 1 + 5 = 0 q2x2 – 3px + 4 = 0So, a = q2, b = –3p and c = 4

Since the roots are equal,then b2 – 4ac = 0 (–3p)2 – 4q2(4) = 0 9p2 – 16q2 = 0 9p2 = 16q2

p2

q2 = 169

1 p—q 22 = 1 4—3 2

2

p—q = 4—3 p : q = 4 : 3

20. 4x2 – 5x + t + 2 = 0So, a = 4, b = –5 and c = t + 2

Since the roots are distinct,then b2 – 4ac . 0(–5)2 – 4(4)(t + 2) . 0 25 – 16t – 32 . 0 –16t . 7 t , – 7

16

21. (p – 1)x2 – 8x = 4(p – 1)x2 – 8x – 4 = 0So, a = p – 1, b = –8 and c = – 4

Since the roots are not real,then b2 – 4ac , 0 (–8)2 – 4(p – 1)(– 4) , 0 64 + 16p – 16 , 0 16p + 48 , 0 16p , – 48

p , – 48–––16 p , –3

22. Given y = 3x – k ................................ 1and y = 4 – x2 ................................. 2

Substitute 1 into 2, 3x – k = 4 – x2

x2 + 3x – k – 4 = 0So, a = 1, b = 3 and c = –k – 4

Since the straight line intersects the curve at two different points,then b2 – 4ac . 032 – 4(1)(–k – 4) . 0 9 + 4k + 16 . 0 4k + 25 . 0 4k . –25

k . – 254

10



23. Given y = 2x – 1 ................................ 1and y = x2 + p ................................. 2

Substitute 1 into 2, 2x – 1 = x2 + p x2 – 2x + 1 + p = 0So, a = 1, b = –2 and c = 1 + p

Since the straight line is a tangent to the curve,then b2 – 4ac = 0(–2)2 – 4(1)(1 + p) = 0 4 – 4 – 4p = 0 – 4p = 0 p = 0

24. x2 – px + q = 0So, a = 1, b = –p and c = q

Since the roots are equal,then b2 – 4ac = 0 (–p)2 – 4(1)(q) = 0 p2 – 4q = 0 .............................. 1Given q + p2 = 1 .............................. 2

2 – 1, 5q = 1 q = 1—5

Substitute q = 1—5 into 1,

p2 – 41 1—5 2 = 0

p2 – 4—5 = 0

p2 = 4—5

p = ±ABB4—5 = 0.8944 or –0.8944

25. (a) 4x – 6 + 3x2 = 0 3x2 + 4x – 6 = 0 So, a = 3, b = 4 and c = –6


= – 4 ± ABBBBBBBBB42 – 4(3)(– 6)2(3)

= – 4 ± ABB886

= – 4 + ABB886 or – 4 – ABB88

6 = 0.8968 or –2.230

(b) px2 + 2px + p = –3x px2 + 2px + 3x + p = 0 px2 + (2p + 3)x + p = 0 So, a = p, b = (2p + 3) and c = p

Since the roots are not real, then b2 – 4ac , 0 (2p + 3)2 – 4(p)(p) , 0 4p2 + 12p + 9 – 4p2 , 0 12p + 9 , 0 12p , –9 p , – 3—4

26. (a) x2 + px – 1—2 pq = qx

x2 + px – qx – 1—2 pq = 0

x2 + (p – q)x – 1—2 pq = 0

So, a = 1, b = p – q and c = – 1—2 pq

b2 – 4ac = (p – q)2 – 4(1)1– 1—2 pq2 = p2 – 2pq + q2 + 2pq = p2 + q2

Since p2 . 0 and q2 . 0 for all values of p and q, then p2 + q2 . 0 for all values of x. That is, b2 – 4ac . 0 for all values of x.

Hence, the quadratic equation has roots for all values of p and q.

(b) Given a and b are the roots of 3x2 – 8x + 2 = 0.

3x2 – 8x + 2 = 0

x2 – 8—3 x + 2—3 = 0

Sum of roots = a + b

= 8—3 Product of roots = ab

= 2—3

For the roots 2—a and 2—b ,

Sum of roots = 2—a + 2—b

= 2b + 2aab

= 2(b + a)ab

= 2 1 8—3 2

2—3 = 2 × 8—3 × 3—2 = 8

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Product of roots = 1 2—a 21 2—b 2 = 4

ab

= 42—3

= 4 × 3—2 = 6 Hence, the quadratic equation with roots 2—a and

2—b is x2 – 8x + 6 = 0.

27. (a) Given y + px – 1 = 0 y = 1 – px ................. 1 and x2 – 3x = y(y –3) x2 – 3x = y2 – 3y ................ 2

Substitute 1 into 2, x2 – 3x = (1 – px)2 – 3(1 – px) x2 – 3x = 1 – 2px + p2x2 – 3 + 3px p2x2 – x2 + 3x – 2px + 3px + 1 – 3 = 0 (p2 – 1)x2 + (3 + p)x – 2 = 0 So, a = p2 – 1, b = 3 + p and c = –2

Since the straight line touches the curve at only one point,

then b2 – 4ac = 0 (3 + p)2 – 4(p2 – 1)(–2) = 0 9 + p2 + 6p + 8p2 – 8 = 0 9p2 + 6p + 1 = 0 (3p + 1)2 = 0 3p + 1 = 0 p = – 1—3(b) 2x2 – 4x + 1 = 0

x2 – 2x + 1—2 = 0

Sum of roots = a + b = 2

Product of roots = ab

= 1—2 Sum of new roots = (a + 2) + (b + 2) = a + b + 4 = 2 + 4 = 6

Product of new roots = (a + 2)(b + 2) = ab + 2(a + b) + 4

= 1—2 + 2(2) + 4

= 8 1—2

= 172

172

q—2

q—2q—2q—2q—2

144

7—2

h—3

h3h2

92—9

a—2b—2

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(b) px2 + (p + 2)x = 4q + 10 px2 + (p + 2)x – 4q – 10 = 0

x2 + 1 p + 2 p 2x – 1 4q + 10

p 2 = 0

Sum of roots = – 1 p + 2 p 2

q + 1—p = – 1 p + 2 p 2

Multiply both sides by p, pq + 1 = –p – 2 pq + p = –3 .................................. 1

Product of roots = – 1 4q + 10 p 2

(q)1 1p 2 = – 1 4q + 10 p 2

q = – 4q – 10 5q = –10 q = –2

Substitute q = –2 into 1, p(–2) + p = –3 –p = –3 p = 3

31. (a) (h2 + 1)x2 + 2phx + p2 = 0 So, a = (h2 + 1), b = 2ph and c = p2

b2 – 4ac = (2ph)2 – 4(h2 + 1)(p2) = 4p2h2 – 4p2h2 – 4p2

= – 4p2

Since – 4p2 , 0 for all real non-zero p and p2 . 0, then b2 – 4ac , 0.

Therefore, the quadratic equation has no roots.

(b) x2 + (p + 1)2 = 3px – 2x x2 + 2x – 3px + (p + 1)2 = 0 x2 + (2 – 3p)x + (p + 1)2 = 0 So, a = 1, b = 2 – 3p and c = (p + 1)2

Since the equation has only one root, then b2 – 4ac = 0 (2 – 3p)2 – 4(1)(p + 1)2 = 0 4 – 12p + 9p2 – 4(p2 + 2p + 1) = 0 4 – 12p + 9p2 – 4p2 – 8p – 4 = 0 5p2 – 20p = 0 5p(p – 4) = 0 p = 0 or 4

x2 + (2 – 3p)x + (p + 1)2 = 0 When p = 4, x2 – 10x + 25 = 0 (x – 5)2 = 0 x = 5

a—2

b—2

a—2a—2a—6

b—2b—4

a—6b—4

a2

36b—4

112

116

13



32. (a) x2 + 2kx = k – 4 x2 + 2kx + 4 – k = 0 So, a = 1, b = 2k and c = 4 – k

Since x-axis is the tangent to the curve, then x has only one value.

Therefore, b2 – 4ac = 0 (2k)2 – 4(1)(4 – k) = 0 4k2 – 16 + 4k = 0 4k2 + 4k – 16 = 0 k2 + k – 4 = 0

k = –1 ± ABBBBBBBBBB(1)2 – 4(1)(– 4)

2(1)

= –1 ± ABBBBB1 + 162

= –1 ± ABB172

= –1 + ABB172

or –1 – ABB172

(b) 2x2 – 4x + 1 = 0

x2 – 2x + 1—2 = 0

Sum of roots = 2 a + b = 2

Product of roots = 1—2

ab = 1—2 Sum of the new roots = a2 + b2

= a2 + b2 + 2ab – 2ab = (a + b)2 – 2ab

= (2)2 – 21 1—2 2 = 4 – 1 = 3

Product of the new roots = a2b2

= (ab)2

= 1 1—2 22

= 1—4 Hence, the quadratic equation is

x2 – 3x + 1—4 = 0

4x2 – 12x + 1 = 0

1. 2x2 + 4x + 5 = 2(x2 + 2x) + 5 = 2(x2 + 2x + 12 – 12) + 5 = 2[(x + 1)2 – 1] + 5 = 2(x + 1)2 – 2 + 5 = 2(x + 1)2 + 3

2x2 + 4x + 5 = 21 2(x + 1)2 + 3 = 21 2(x + 1)2 = 18 (x + 1)2 = 9 x + 1 = ±3 x = ±3 – 1 = 3 – 1 or –3 – 1 = 2 or – 4

2. 7 – 6x – 3x2 = –3(x2 + 2x) + 7 = –3(x2 + 2x + 12 – 12) + 7 = –3[(x + 1)2 – 1] + 7 = –3(x + 1)2 + 3 + 7 = –3(x + 1)2 + 10

6 – 6x – 3x2 = 0 7 – 6x – 3x2 = 1 –3(x + 1)2 + 10 = 1 –3(x + 1)2 = –9 (x + 1)2 = 3 x + 1 = ±AB3 x = ±AB3 – 1 = AB3 – 1 or –AB3 – 1 = 0.7321 or –2.732

3. y = x2 + px – x – p

When the x-axis is the tangent to the curve, then b2 – 4ac = 0 for x2 + px – x – p = 0.That is, x2 + (p – 1)x – p = 0

b2 – 4ac = 0 (p – 1)2 – 4(1)(–p) = 0 p2 – 2p + 1 + 4p = 0 p2 + 2p + 1 = 0 (p + 1)2 = 0 p + 1 = 0 p = –1

4. x2 + ax + b = 0Sum of roots = – a q + 3q = – a 4q = – a q = – a—

4 ........................... 1

Product of roots = b q(3q) = b 3q2 = b ............................ 2

Substitute 1 into 2,

31– a—4 22 = b

3a2–––16

= b

3a2 = 16b

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5. x2 – ax = –2a x2 – ax + 2a = 0

Sum of roots = a p + q = a ................................. 1

Product of roots = 2a pq = 2a ......................... 2

Substitute 1 into 2,pq = 2(p + q)pq = 2p + 2q

6. 3x2 + p + 3x + px = 0 3x2 + (3 + p)x + p = 0

b2 – 4ac = (3 + p)2 – 4(3)(p) = (3 + p)2 – 12p = 9 + 6p + p2 – 12p = p2 – 6p + 9 = (p – 3)2

Since (p – 3)2 > 0 for all values of p,then b2 – 4ac > 0 for all values of p.

Therefore, equation 3x2 + p + 3x + px = 0 has roots for all values of p.

7. Substitute x = 0, y = 0 into y = ax2 + bx + c, \ c = 0y = ax2 + bxSubstitute x = 4, y = 8 into y = ax2 + bx, 8 = a(4)2 + b(4) 16a + 4b = 8 4a + b = 2 ....................................... 1

Given a + b + 4 = 0 a + b = – 4 ........................ 2

1 – 2, 3a = 6 a = 2

Substitute a = 2 into 2, 2 + b = –4 b = –6Therefore, a = 2, b = –6 and c = 0.

When y = 0, 2x2 – 6x = 0 2x(x – 3) = 0 x = 0 or 3

8. The quadratic equation is x2 – (–2 + p)x + (–2)(p) = 0 x2 – (p – 2)x – 2p = 0

Given product of roots = sum of roots –2p = –2 + p 3p = 2

p = 2—3

9. p2x2 + 2pqx + x2 + q2 = 0 (p2 + 1)x2 + 2pqx + q2 = 0

b2 – 4ac = (2pq)2 – 4(p2 + 1)(q2) = 4p2q2 – 4p2q2 – 4q2

= – 4q2

Since q is real non-zero number, then q2 . 0 for all values of q.Therefore, b2 – 4ac , 0 for all values of q.Hence, there is no real roots for all values of p and q.

10. (a) Sum of roots = p – 4 Product of roots = –4p

f (x) = x2 – (p – 4)x + (–4p) = x2 – (p – 4)x – 4p

(b) y = kf(x) = k[x2 – (p – 4)x – 4p]

Substitute x = 0 and y = 16 into the equation, 16 = k(– 4p) kp = – 4

When p = 2, k(2) = – 4 k = –2

11. y = x2 – 4x + cSince minimum point is above the x-axis,then b2 – 4ac , 0 (– 4)2 – 4(1)(c) , 0 16 – 4c , 0 – 4c , –16 c . 4

02[anal add math cd]

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