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$\alpha$ and $\beta$ are the zeroes of the polynomial $5x^2 – 16x - 10$. Find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
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$\alpha + \beta = \frac{16}{5}$, $\alpha \beta = -2$ (1 Mark)
$\therefore \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha+ \beta)^2 – 2\alpha\beta}{\alpha\beta}$ (1/2 Mark)
= $\frac{(\frac{16}{5})^2 + 4}{-2}$ (1/2 Mark)
= $\frac{\frac{256}{25} + 4}{-2} = \frac{356}{25 \times -2} = -\frac{356}{50}$ or $-\frac{178}{25}$ (1/2 Mark)
$\therefore \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{(\alpha+ \beta)^2 – 2\alpha\beta}{\alpha\beta}$ (1/2 Mark)
= $\frac{(\frac{16}{5})^2 + 4}{-2}$ (1/2 Mark)
= $\frac{\frac{256}{25} + 4}{-2} = \frac{356}{25 \times -2} = -\frac{356}{50}$ or $-\frac{178}{25}$ (1/2 Mark)