Quadratic Equations Long Questions (Question 11 – 13)

Posted on May 25, 2020 by Myhometuition

Question 11:
Solve the following quadratic equation:

$\frac{5 x + 3 x^{2}}{1 - 2 x} = 6$

Solution:

$\begin{array}{l} \frac{5 x + 3 x^{2}}{1 - 2 x} = 6 \\ 5 x + 3 x^{2} = 6 - 12 x \\ 3 x^{2} + 5 x + 12 x - 6 = 0 \\ 3 x^{2} + 17 x - 6 = 0 \\ (3 x - 1) (x + 6) = 0 \\ 3 x - 1 = 0 or x + 6 = 0 \\ 3 x = 1 x = - 6 \\ x = \frac{1}{3} x = - 6 \end{array}$

Question 12:

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

Question 13 (4 marks):
Solve the following quadratic equation:

$- \frac{2}{3 x - 5} = \frac{x}{3 x - 1}$

Solution:

$\begin{array}{l} - \frac{2}{3 x - 5} = \frac{x}{3 x - 1} \\ - 2 (3 x - 1) = x (3 x - 5) \\ - 6 x + 2 = 3 x^{2} - 5 x \\ 3 x^{2} - 5 x + 6 x - 2 = 0 \\ 3 x^{2} + x - 2 = 0 \\ (3 x - 2) (x + 1) = 0 \\ 3 x - 2 = 0 or x + 1 = 0 \\ x = \frac{2}{3} or x = - 1 \end{array}$