WAEC 2009 · Paper 2 · Q3

(a)

1. From $3x^2 + 4x - 5 = 0$: $\alpha + \beta = -\frac43$ and $\alpha\beta = -\frac53$.
2. Sum of the new roots: $\dfrac{1 + \alpha}{\beta} + \dfrac{1 + \beta}{\alpha} = \dfrac{\alpha + \alpha^2 + \beta + \beta^2}{\alpha\beta}$.
3. $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
   $= \frac{16}{9} + \frac{10}{3}$
   $= \frac{46}{9}$.
4. So the sum is $\dfrac{-\frac{12}{9} + \frac{46}{9}}{-\frac53} = \dfrac{34}{9} \times \left(-\dfrac35\right)$
   $= -\dfrac{34}{15}$.
5. Product of the new roots: $\dfrac{(1 + \alpha)(1 + \beta)}{\alpha\beta} = \dfrac{1 + (\alpha + \beta) + \alpha\beta}{\alpha\beta}$
   $= \dfrac{1 - \frac43 - \frac53}{-\frac53}$
   $= \dfrac{-2}{-\frac53}$
   $= \dfrac65$.
6. The equation is $x^2 - (\text{sum})x + \text{product} = 0$: $x^2 + \frac{34}{15}x + \frac65 = 0$.
7. Multiply by 15: $15x^2 + 34x + 18 = 0$.