02[anal add math cd]

1
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 =
4x
x + 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 =
1
2
Hence, x =
1
2
is a root.
(c) When (1 – 3x) = 0
x =
1
3
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 0
1 –8
2 –5
3 0
Therefore, x = –3 and x = 3 are the roots.
CHAPTER
2 Quadratic Equations

2
(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 2
2
= –1 + 1 3
—
2 2
2
1x –
3
—
2 2
2
=
5
—
4

3
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 2
2
=
1
—
6
+ 1 4
—
3 2
2
1x –
4
—
3 2
2
=
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(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

4
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
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
x2
–
15
–––
2
x +
7
—
2
= 0
2x2
– 15x + 7 = 0
(e) Sum of roots = 4 + 4
= 8
Product of roots = 4 × 4
= 16
– 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
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
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.

5
12. 2x2
– kx + 2 = 0
So, 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 = 0
So, 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 = 0
So, 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 = 0
So, a = k, b = h and c = –4
then b2
– 4ac = 0
h2
– 4k(– 4) = 0
h2
+ 16k = 0
16. 2x2
+ px = k
2x2
+ px – k = 0
So, a = 2, b = p and c = –k
then b2
– 4ac , 0
p2
– 4(2)(–k) , 0
p2
+ 8k , 0
17. px2
– qx = 4
px2
– qx – 4 = 0
So, 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 = 0
x2
– (k + 6)x + 9 = 0
So, a = 1, b = – (k + 6) and c = 9
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) = k
2x2
+ 3x – 8x – 12 – k = 0
2x2
– 5x – 12 – k = 0
So, 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.................................1
and y = kx2
+ 3x – 2........................2
Substitute 1 into 2,
4x – 1 = kx2
+ 3x – 2
kx2
+ 3x – 4x – 2 + 1 = 0
kx2
– x – 1 = 0
So, 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..................................1
and y = 4x2
– 5x + 6........................2
hx – k = 4x2
– 5x + 6
4x2
– 5x – hx + 6 + k = 0
4x2
– (5 + h)x + 6 + k = 0
So, a = 4, b = – (5 + h) and c = 6 + k

6
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 , 0
25 + 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 = 0
So, a = 5, b = –9 and c = –8
x =
–b ± ABBBBBBb2
– 4ac
2a
=
–(–9) ± ABBBBBBBBBBB(–9)2
– 4(5)(–8)
2(5)
=
9 ± ABBB241
10
=
9 + ABBB241
10
or
9 – ABBB241
10
= 2.452 or – 0.6524
2. 2x2
+ ABpx = q – 1
2x2
+ ABpx + 1 – q = 0
So, 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
8
3. Sum of roots = –5 +
2
3
=
–15 + 2
3
= –
13
3
Product of roots = (–5)1 2
—
3 2
= –
10
–––
3
x2
– 1–
13
–––
3 2x + 1–
10
–––
3 2 = 0
x2
+
13
–––
3
x –
10
–––
3
= 0
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 = 0
4nx2
+ (1 + 4n)x + n – 2 = 0
a = 4n, b = 1 + 4n, c = n – 2
For 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 = 0
a = 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
28
12
, p
p .
7
3

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 + p
p 2x + 1
1 – q
p 2 = 0
Sum of roots = – 1
2 + p
p 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
4
5
–2 = (1 – q)1 5
4 2
=
5
4
–
5
4
q

5
4
q =
13
4
q =
13
5
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 =
x
x + 2
(x – 4)(x + 2) = x
x2
+ 2x – 4x – 8 = x
x2
– 3x – 8 = 0
So, a = 1, b = –3 and c = –8
x =
–b ± ABBBBBBb2
– 4ac
2a
=
–(–3) ± ABBBBBBBBBBB(–3)2
– 4(1)(–8)
2(1)
=
3 ± ABB41
2
=
3 + ABB41
2
or
3 – ABB41
2
= 4.702 or –1.702
6.
6
—
5
y = y2
– 1
6y = 5y2
– 5
5y2
– 6y – 5 = 0
So, a = 5, b = –6 and c = –5
x =
–b ± ABBBBBBb2
– 4ac
2a
=
–(– 6) ± ABBBBBBBBBBB(– 6)2
– 4(5)(–5)
2(5)
=
6 ± ABBBBBBB36 + 100
10
=
6 ± ABBB136
10
=
6 + ABBB136
10
or
6 – ABBB136
10
= 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.

8
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
=
17
2
(x + 1) = ±ABBB17
2
x = –1 + ABBB17
2
or –1 – ABBB17
2
= 1.915 or –3.915
11. Sum of roots =
1
—
3
+ (–5)
=
1
—
3
– 5
= –
14
–––
3
—
3 2(–5)
= –
5
—
3
Therefore, the quadratic equation is
x2
– 1–
14
3 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
x2
– (– 4)x + (–14) = 0
x2
+ 4x – 14 = 0
15. Let a and 3a are the roots of quadratic equation
2x2
– 2 = 8x – 4k
2x2
– 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

9
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
=
17
12
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
–
17
12
x +
5
12
= 0
12x2
– 17x + 5 = 0
17. x2
+ (1 – p)x + 4 = 0
So, a = 1, b = 1 – p and c = 4
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 = 0
So, a = 1, b = –20 and c = 45 + 5p
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 = q2
x2
– 1
q2
x2
– 3px – 1 + 5 = 0
q2
x2
– 3px + 4 = 0
So, a = q2
, b = –3p and c = 4
then b2
– 4ac = 0
(–3p)2
– 4q2
(4) = 0
9p2
– 16q2
= 0
9p2
= 16q2

p2
q2 =
16
9
1
p
—
q 2
2
= 1 4
—
3 2
2

p
—
q
=
4
—
3
p : q = 4 : 3
20. 4x2
– 5x + t + 2 = 0
So, 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 = 0
So, a = p – 1, b = –8 and c = – 4
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.................................1
and y = 4 – x2
..................................2
3x – k = 4 – x2
x2
+ 3x – k – 4 = 0
So, a = 1, b = 3 and c = –k – 4
Since the straight line intersects the curve at two
different points,
then b2
– 4ac . 0
32
– 4(1)(–k – 4) . 0
9 + 4k + 16 . 0
4k + 25 . 0
4k . –25
k . –
25
4

10
23. Given y = 2x – 1.................................1
and y = x2
+ p..................................2
2x – 1 = x2
+ p
x2
– 2x + 1 + p = 0
So, 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 = 0
So, a = 1, b = –p and c = q
then b2
– 4ac = 0
(–p)2
– 4(1)(q) = 0
p2
– 4q = 0...............................1
Given 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
x =
–b ± ABBBBBBb2
– 4ac
2a
=
– 4(3)(– 6)
2(3)
=
– 4 ± ABB88
6
=
– 4 + ABB88
6
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
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 + 2a
ab
=
2(b + a)
ab
=
2 1 8
—
3 2
2
—
3
= 2 ×
8
—
3
×
3
—
2
= 8

11
—
a 21 2
—
b 2
=
4
ab
=
4
2
—
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
x2
– 3x = (1 – px)2
– 3(1 – px)
x2
– 3x = 1 – 2px + p2
x2
– 3 + 3px
p2
x2
– 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
=
17
2
17
2
q
—
2
q
—
2
q
—
2
q
—
2
q
—
2
14
4
7
—
2
h
—
3
h
3
h2
9
2
—
9
a
—
2
b
—
2

12
(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 1
p 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
)
= 4p2
h2
– 4p2
h2
– 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
—
2
a
—
2
a
—
6
b
—
2
b
—
4
a
—
6
b
—
4
a2
36
b
—
4
1
12
1
16

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 + 16
2
=
–1 ± ABB17
2
=
–1 + ABB17
2
or
–1 – ABB17
2
(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 = a2
b2
= (ab)2
= 1 1
—
2 2
2
=
1
—
4
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 = 0
Sum of roots = – a
q + 3q = – a
4q = – a
q = –
a
—
4
............................1
Product of roots = b
q(3q) = b
3q2
= b.............................2
31–
a
—
4
2
2
= b

3a2
–––
16
= b
3a2
= 16b

14
5. x2
– ax = –2a
x2
– ax + 2a = 0
Sum of roots = a
p + q = a..................................1
Product of roots = 2a
pq = 2a..........................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 = 0
y = ax2
+ bx
Substitute 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 = –6
Therefore, 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. p2
x2
+ 2pqx + x2
+ q2
= 0
(p2
+ 1)x2
+ 2pqx + q2
= 0
b2
– 4ac = (2pq)2
– 4(p2
+ 1)(q2
)
= 4p2
q2
– 4p2
q2
– 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 + c
Since minimum point is above the x-axis,
then b2
– 4ac , 0
(– 4)2
– 4(1)(c) , 0
16 – 4c , 0
– 4c , –16
c . 4

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