# 1.6.2 Quadratic Equations, SPM Paper 2 (Long Questions)

Question 5:
Solve the equation:
(m + 2)(m – 4) = 7(m – 4).

Solution:
(m + 2)(m – 4) = 7(m – 4)
m2– 4m + 2m – 8 = 7m – 28
m2– 9m + 20 = 0
(m – 5)(m – 4) = 0
= 5 or m = 4

Question 6:
Solve the equation:
$\frac{-6y-2}{y}=\frac{7}{1-y}$

Solution:

$\begin{array}{l}\frac{-6y-2}{y}=\frac{7}{1-y}\\ \left(-6y-2\right)\left(1-y\right)=7y\\ -6y+6{y}^{2}-2+2y-7y=0\\ 6{y}^{2}-11y-2=0\\ \left(6y+1\right)\left(y-2\right)=0\\ 6y+1=0\text{or}y=2\\ y=-\frac{1}{6}\end{array}$

Question 7:
Solve the equation:
$\frac{4m}{7}=m\left(8m-9\right)$

Solution:

Question 8:
Diagram above shows a rectangle ABCD.
(a) Express the area of ABCD in terms of n.
(b) Given the area of ABCD is 60 cm2, find the length of AB.

Solution:
(a)
Area of ABCD
= (n + 7) × n
= (n2+ 7n) cm2

(b)
Given the area of ABCD = 60
n2+ 7n = 60
n2+ 7n – 60 = 0
(n – 5) (n + 12) = 0
= 5 or    n = – 12 (not accepted)

When n = 5,
Length of AB = 5 + 7 = 12 cm