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If $\alpha$, $\beta$ are the zeroes of polynomial $p(x) = 6x^2-5x - 3$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$.
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Solution: (a) $\alpha+\beta=\frac{5}{6}$, $\alpha\beta=-\frac{3}{6}$ (1 Mark)
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/6}{-3/6} = -\frac{5}{3}$ (1 Mark)
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{5/6}{-3/6} = -\frac{5}{3}$ (1 Mark)