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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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Sol. $f(x) = 6x^2 + 11x - 10$
$\alpha + \beta = -\frac{11}{6}$ and $\alpha\beta = -\frac{10}{6}$
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$
$= \frac{(-\frac{11}{6})^2 - 2(-\frac{10}{6})}{-\frac{10}{6}}$
$= -\frac{241}{60}$
$\alpha + \beta = -\frac{11}{6}$ and $\alpha\beta = -\frac{10}{6}$
$\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$
$= \frac{(-\frac{11}{6})^2 - 2(-\frac{10}{6})}{-\frac{10}{6}}$
$= -\frac{241}{60}$