# 2.10.3 Quadratic Equations, SPM Practice (Paper 2)

2.10.3 Quadratic Equations, SPM Practice (Paper 2)

Question 5:
Given 3t and (t – 7) are the roots of the quadratic equation 4x2 – 4x + m = 0 where m is a constant.
(a)  Find the values of t and m.
(b)  Hence, form the quadratic equation with roots 4t and 2t + 6.

Solutions:
(a)
Given 3t and (t – 7) are the roots of the quadratic equation 4x2 – 4x + m = 0
a = 4, b = – 4, c = m
Sum of roots = $-\frac{b}{a}$
3t + (t– 7) = $-\frac{-4}{4}$
3t + t– 7 = 1
4t = 8
t = 2

Product of roots = $\frac{c}{a}$
3t (t– 7) = $\frac{m}{4}$
4 [3(2) (2 – 7)] = m ← (substitute t = 2)
4 [3(2) (2 – 7)] = m
4 (–30) = m
m = –120

(b)
t = 2
4t = 4(2) = 8
2t + 6 = 2(2) + 6 = 10

Sum of roots = 8 + 10 = 18
Product of roots = 8(10) = 80

Using the formula, x2– (sum of roots)x + product of roots = 0
Thus, the quadratic equation is,
x2 – 18x + 80 = 0