Page 10 - Focus SPM KSSM Tg 4.5 - Add Maths

Additional Mathematics SPM Chapter 2 Quadratic Functions

The new quadratic equation is: POR = (α)(2α) = 25
x – (SOR)x + (POR) = 0 2
2
x – (4)x + (–17) = 0 2α = 25
2
2
2
Therefore, x – 4x – 17 = 0 25
2
α = 4
2
2
2
5
(b) The new roots are and . α = ± .............. b
α
b
2
2
2
New SOR = + Substitute b into :
α b
5
5
= 2b + 2α When α = , k = 6  
αb 2 2
= 2(b + α) 5 = 15 5 Form 4

αb When α = – , k = 6 – 2 
2
2
2   = –15
= 5 4 Therefore, the possible values of k are 15 and –15.
– 5
= –1
SPM Tips
2
2
New POR =    This example can also be solved by using α and α
1
b
α
= 4 as the roots of the equation. 2
αb
4
=
– 4 Try Questions 10 – 17 in ‘Try This! 2.1’
5
= –5
The new quadratic equation is: SPM Highlights
x – (SOR)x + (POR) = 0 The quadratic equation 4x + 5x – 6 = 0 has roots h and
2
2
x – (–1)x + (–5) = 0 h k
2
k. Form a quadratic equation with roots and .
2
2
Therefore, x + x – 5 = 0
2
Solution:
Try Question 9 in ‘Try This! 2.1’ 4 x + 5x – 6 = 0
2
3
5
x + x – = 0
2
6 4 5 2
One root of the quadratic equation 2x – kx + 25 = 0 SOR = h + k = – 4
2
is half of the other root of the equation. Find the POR = hk = – 3
possible values of k. 2 – 5
h k h + k 4 5
New SOR = + = = = –
Solution 2 2 2 2 8
2x – kx + 25 = 0 Divide each term of the – 3
2
equation by 2.
k
k
3
h
x – x + 25 = 0 New POR =    = hk = 4 2 = –
2
2
4
8
2
2 2
Assume one of the roots is α. The new equation is
Then, the other root is 2α. x – – 5  x + – 3  = 0


2
8

SOR = α + 2α = – – k  8 x + 5x – 3 = 0
8
2
2
3α = k
2
k = 6α .............
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