Page 9 - Focus SPM KSSM Tg 4.5 - Add Maths

Additional Mathematics SPM Chapter 2 Quadratic Functions

2. One of the methods to form a quadratic equation Alternative Method
from the given roots of α and b is by expanding 2
the product of (x – α)(x – b) = 0. x = and x = 5
3
3. The second method is by finding the sum of 3x = 2 and x = 5
roots, (α + b) and the product of roots, (αb) 3x – 2 = 0 and x – 5 = 0
to be substituted into the quadratic equation
x – (SOR)x + (POR) = 0. (3x – 2)(x – 5) = 0
2
3x – 17x + 10 = 0
2
4. To determine the sum of roots and product of One root only means
roots of a quadratic equation (c) SOR = –4 + (–4) = –8 there are two equal
POR = –4 × (–4) = 16 roots.
Quadratic equation The quadratic equation is
ax + bx + c = 0 x – (SOR)x + (POR) = 0
2
2
where a ≠ 0 x – (–8)x + (16) = 0
2
Form 4
2
Divide each term of the x + 8x + 16 = 0
equation by a so that the Try Question 8 in ‘Try This! 2.1’
coefficient of x becomes 1.
2
c
b
x + x + = 0
2
a
a
5
Given α and b are the roots of a quadratic equation
5x – 2x – 4 = 0. Form a quadratic equation with
2
b roots
Sum of roots = – a (a) (5α + 1) and (5b + 1)
2
2
Product of roots = a c (b) α and b
Solution
Divide each term of the equation by 5, so that the
4 coefficient of x becomes 1.
2
2
4
Form a quadratic equation with the given roots. x – x – = 0
2
(a) 2 and –3 (b) 2 and 5 (c) –4 only 5 5
2
3 α + b = Given α and b are the
Solution 5 roots of the quadratic
equation.
(a) SOR = 2 + (–3) = –1 αb = – 4
POR = 2 × (–3) = –6 5
The quadratic equation is (a) The new roots are (5α + 1) and (5b + 1).
x – (SOR)x + (POR) = 0 New SOR = (5α + 1) + (5b + 1)
2
x – (–1)x + (–6) = 0 = 5(α + b) + 2
2
x + x – 6 = 0 2
2
= 5   + 2
5
2
(b) SOR = + 5 = 17 = 4
3 3
2
POR =   (5) = 10 New POR = (5α + 1)(5b + 1)
3
3
The quadratic equation is = 25αb + 5α + 5b + 1
= 25αb + 5(α + b) + 1
17
x –   x + 10 = 0 Multiply each term of = 25 – 4    + 1
2
2

the equation by 3.
+ 5
3
3
2
3x – 17x + 10 = 0 5 5
= –17
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