WAEC 2011 · Paper 2 · Q7

(a)
Divide $\dfrac{x^2 - 4}{x^2 + x}$ by $\dfrac{x^2 - 4x + 4}{x + 1}$ .
(b)
The diagram shows the graphs of $y = ax^2 + bx + c$ and $y = mx + k$ where $a$ , $b$ , $c$ , $m$ and $k$ are constants. Use the graph(s) to: (i) find the roots of the equation $ax^2 + bx + c = mx + k$ ; (ii) determine the values of $a$ , $b$ and $c$ using the coordinates of points $L(-1, 0)$ , $M(0, 2)$ and $N(2, 0)$ , and hence write down the equation of the curve; (iii) determine the line of symmetry of the curve $y = ax^2 + bx + c$ .
Separate values with commas, e.g. 3, −2

The curve and the line; read off where they cross.

Worked solution (try it first)

To divide by a fraction, multiply by it turned upside down: $\frac{x^2 - 4}{x^2 + x} \times \frac{x + 1}{x^2 - 4x + 4}$ .
Factorise each part: $x^2 - 4 = (x + 2)(x - 2)$ , $x^2 + x = x(x + 1)$ and $x^2 - 4x + 4 = (x - 2)^2$ .
So the product is $\frac{(x + 2)(x - 2)}{x(x + 1)} \times \frac{x + 1}{(x - 2)^2}$ .
Cancel $(x + 1)$ and one $(x - 2)$ to get $\frac{x + 2}{x(x - 2)}$ .

The roots of $ax^2 + bx + c = mx + k$ are the $x$ -values where the line meets the curve.
From the graph, these are $x = -1.5$ and $x = 1.5$ .

The line of symmetry is halfway between the points where the curve crosses the $x$ -axis, $x = -1$ and $x = 2$ : $x = \frac{-1 + 2}{2} = 0.5$ .