WAEC 2008 · Paper 2 · Q2

(a)
If $\alpha$ and $\beta$ are the roots of the equation $2x^2 + 5x - 6 = 0$ , find the equation whose roots are $(\alpha - 2)$ and $(\beta - 2)$ .

Worked solution (try it first)

From $2x^2 + 5x - 6 = 0$ : $\alpha + \beta = -\frac52$ and $\alpha\beta = -\frac62 = -3$ .
New sum: $(\alpha - 2) + (\beta - 2) = (\alpha + \beta) - 4$
$= -\frac52 - 4$
$= -\frac{13}{2}$ .
New product: $(\alpha - 2)(\beta - 2) = \alpha\beta - 2(\alpha + \beta) + 4$ .
Substitute: $-3 - 2\left(-\frac52\right) + 4$ is $-3 + 5 + 4$ , which is 6.
The equation is $x^2 - (\text{sum})x + \text{product} = 0$ : $x^2 + \frac{13}{2}x + 6 = 0$ .
Multiply by 2: $2x^2 + 13x + 12 = 0$ .