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