WAEC 2011 · Paper 2 · Q2

(a)
If $\alpha$ and $\beta$ are the roots of the equation $2x^2 - 7x + 4 = 0$ , find the equation whose roots are $\dfrac\alpha\beta$ and $\dfrac\beta\alpha$ .

Worked solution (try it first)

For $2x^2 - 7x + 4 = 0$ : $\alpha + \beta = \frac72$ and $\alpha\beta = \frac42 = 2$ .
The sum of the new roots: $\dfrac\alpha\beta + \dfrac\beta\alpha = \dfrac{\alpha^2 + \beta^2}{\alpha\beta}$ .
The top: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$= \frac{49}{4} - 4$
$= \frac{33}{4}$ .
Divide by $\alpha\beta = 2$ : the new sum is $\frac{33}{8}$ .
The product of the new roots: $\dfrac\alpha\beta \times \dfrac\beta\alpha = 1$ .
The equation is $x^2 - (\text{sum})x + \text{product} = 0$ : $x^2 - \frac{33}{8}x + 1 = 0$ .
Multiply through by 8: $8x^2 - 33x + 8 = 0$ .