WAEC Further Mathematics 2018 Paper

Note: Given the sum of the roots and its product, we can get the equation using the formula:

$x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0$. This will be used later on in the course of our solution.

Given equation: $2x^{2} - 5x + 6 = 0; a = 2, b = -5, c = 6$.

$\alpha + \beta = \frac{-b}{a} = \frac{-(-5)}{2} = \frac{5}{2}$

$\alpha\beta = \frac{c}{a} = \frac{6}{2} = 3$

Given the roots of the new equation as $(\alpha + 1)$ and $(\beta + 1)$, their sum and product will be

$(\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = \frac{5}{2} + 2 = \frac{9}{2} = \frac{-b}{a}$

$(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 3 + \frac{5}{2} + 1 = \frac{13}{2} = \frac{c}{a}$

The new equation is given by: $x^{2} - \left(\frac{-b}{a}\right)x + \left(\frac{c}{a}\right) = 0$

= $x^{2} - \left(\frac{9}{2}\right)x + \frac{13}{2} = 2x^{2} - 9x + 13 = 0$