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

m = 5 or m = 4

Question 6:

Solve the equation:

\frac{- 6 y - 2}{y} = \frac{7}{1 - y}

Solution:

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

Question 7:

Solve the equation:

\frac{4 m}{7} = m (8 m - 9)

Solution:

\begin{array}{l} \frac{4 m}{7} = m (8 m - 9) \\ 4 m = 7 m (8 m - 9) \\ 4 m = 56 m^{2} - 63 m \\ 56 m^{2} - 63 m - 4 m = 0 \\ 56 m^{2} - 67 m = 0 \\ m (56 m - 67) = 0 \\ m = 0 or 56 m - 67 = 0 \\ m = \frac{67}{56} \end{array}

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

n = 5 or n = – 12 (not accepted)

When n = 5,

Length of AB = 5 + 7 = 12 cm