Page 15 - Focus SPM KSSM Tg 4.5 - Add Maths

Additional Mathematics SPM Chapter 2 Quadratic Functions

Solution The condition for no real roots is
2
3x + 4 = 8x + 2px 2 b – 4ac  0
2
2
3x – 2px – 8x + 4 = 0 4 – 4(2 – m)(–3)  0
2
2
(3 – 2p)x – 8x + 4 = 0 16 + 24 – 12m  0
2
12m . 40
a = 3 – 2p, b = –8, c = 4 10
m .
For two different real roots, 3
b – 4ac . 0 Try Question 4 in ‘Try This! 2.2’
2
(–8) – 4(3 – 2p)(4) . 0
2
64 – 48 + 32p . 0 13
16 + 32p . 0
2
32p . –16 Given the quadratic equation (n – 2m)x – 4mx + m = 0,
16 where m ≠ 0, has two equal roots. Find the relation
p . –
32 between m and n.
Form 4
1
p . – Solution
2
(n – 2m)x – 4mx + m = 0
2
Try Question 2 in ‘Try This! 2.2’ a = n – 2m, b = –4m, c = m
For two equal real roots,
b – 4ac = 0
2
2
11 (–4m) – 4(n – 2m)(m) = 0
16m – 4mn + 8m = 0
2
2
2
2
The quadratic equation x – 4kx + (k + 3) = 0, where k 24m – 4mn = 0
2
is a constant and k . 0, has two equal roots. Find the 4m(6m – n) = 0
value of k. m = 0 or 6m – n = 0
Solution Since m ≠ 0, therefore 6m = n
x – 4kx + (k + 3) = 0 m = n
2
2
a = 1, b = –4k, c = (k + 3) 2 6
The condition for two equal roots is Try Questions 5 – 8 in ‘Try This! 2.2’
b – 4ac = 0
2
(–4k) – 4(1)(k + 3) = 0 SPM Highlights
2
2
16k – 4(k + 6k + 9) = 0
2
2
2
16k – 4k – 24k – 36 = 0 The quadratic equation (1 + 2q)x – 4qx + 2q – 4 = 0 has
2
2
12k – 24k – 36 = 0 two equal roots. Find the value of q.
2
k – 2k – 3 = 0 Solution:
2
(k + 1)(k – 3) = 0 (1 + 2q)x – 4qx + 2q – 4 = 0
2
k + 1 = 0 or k – 3 = 0 a = 1 + 2q, b = –4q, c = 2q – 4
k = –1 k = 3 For two equal real roots,
Since k . 0, therefore k = 3. b – 4ac = 0
2
2
(–4q) – 4(1 + 2q)(2q – 4) = 0
16q – 8q + 16 – 16q + 32q = 0
2
2
Try Question 3 in ‘Try This! 2.2’ 24q = –16
2
q = –
3
12
The quadratic equation (2 – m)x = 3 – 4x has no real Try This! 2.2
2
roots. Find the range of values of m.
1. Find the discriminant for each of the following
Solution quadratic equations. Hence, determine the type of
(2 – m)x = 3 – 4x roots for each of the quadratic equation.
2
2
2
(2 – m)x + 4x – 3 = 0 (a) 9x – 12x + 4 = 0 (b) 5x – 3x + 6 = 0
2
2
2
a = 2 – m, b = 4, c = –3 (c) 2x + x – 3 = 0 (d) x = 4x + 7
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