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$\alpha, \beta$ are zeroes of the polynomial $p(x) = 3x^2 - 6x - 5$. Find the value of $\frac{1}{\alpha^2} + \frac{1}{\beta^2}$.
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Sol. $\alpha + \beta = 2, \alpha \beta = -\frac{5}{3}$ (I) (1 Mark)
$\therefore \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2} = \frac{4 + \frac{10}{3}}{\frac{25}{9}}$ (II) ($\frac{1}{2}$ Mark)
$= \frac{66}{25}$ (III) ($\frac{1}{2}$ Mark)
$\therefore \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2\beta^2} = \frac{(\alpha+\beta)^2-2\alpha\beta}{(\alpha\beta)^2} = \frac{4 + \frac{10}{3}}{\frac{25}{9}}$ (II) ($\frac{1}{2}$ Mark)
$= \frac{66}{25}$ (III) ($\frac{1}{2}$ Mark)