(a) Simplify \(\frac{x + 2}{x - 2} - \frac{x +... | WAEC SSCE - General Mathematics - 2009

Question 1 Report

(a) Simplify \(\frac{x + 2}{x - 2} - \frac{x + 3}{x - 1}\)

(b) The graph of the equation \(y = Ax^{2} + Bx + C\) passes through the point (0, 0), (1, 4) and (2, 10). Find the :

(i) value of C ; (ii) values of A and B ; (iii) co-ordinates of the other point where the graph cuts the x- axis.

Answer

(a)

Taking the LCM, \, \((x-2)(x-1)\),

\[\begin{aligned} \frac{x+2}{x-2}-\frac{x+3}{x-1} &=\frac{(x+2)(x-1)-(x+3)(x-2)}{(x-2)(x-1)}\\ &=\frac{(x^2+x-2)-(x^2+x-6)}{(x-2)(x-1)}\\ &=\frac{4}{(x-2)(x-1)}. \end{aligned}\]

Hence, \(\displaystyle \frac{x+2}{x-2}-\frac{x+3}{x-1}=\frac{4}{(x-2)(x-1)}\), where \(x\ne 1,2\).

(b)

Given \(y=Ax^2+Bx+C\) and that the graph passes through \((0,0)\):

\[0=A(0)^2+B(0)+C,\]

therefore,

\[\boxed{C=0}.\]

Using \((1,4)\) and \((2,10)\):

\[\begin{aligned} A+B&=4 \qquad \text{(1)}\\ 4A+2B&=10 \qquad \text{(2)} \end{aligned}\]

Twice (1) gives \(2A+2B=8\). Subtracting this from (2),

\[2A=2\quad\Rightarrow\quad A=1.\]

From \(A+B=4\),

\[1+B=4\quad\Rightarrow\quad B=3.\]

Thus the equation of the graph is

\[\boxed{y=x^2+3x}.\]

The plotted curve is shown below.

To find the other point where the graph cuts the \(x\)-axis, put \(y=0\):

\[\begin{aligned} x^2+3x&=0\\ x(x+3)&=0. \end{aligned}\]

Hence \(x=0\) or \(x=-3\). Since \((0,0)\) is already one intercept, the other point is

\[\boxed{(-3,0)}.\]

Read lesson note on Algebraic Expressions (WAEC)

Read lesson note on Quadratic Equations (WAEC)

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