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If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = 6x^2 + 11x - 10$, find the value of $\frac{\alpha}{\beta} + \frac{\beta}{\alpha}$.
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Here $\alpha+\beta = -\frac{11}{6}$ and $$\begin{aligned}& \alpha\beta = -\frac{10}{6} \\ & \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} \\ & = \frac{(-\frac{11}{6})^2-2\times(-\frac{10}{6})}{-\frac{10}{6}} \\ & = \frac{\frac{121}{36}+\frac{20}{6}}{-\frac{10}{6}} = \frac{\frac{121+120}{36}}{-\frac{10}{6}} = \frac{241}{36} \times -\frac{6}{10} = -\frac{241}{60}\end{aligned}$$