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If $\alpha$, $\beta$ are zeroes of the polynomial $x^2 - 6x + 7$, then find the value of $4(\frac{1}{\alpha^2} + \frac{1}{\beta^2})$.
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(b) $\alpha + \beta = 6$ and $\alpha\beta = 7$ (1/2 Mark)
$4(\frac{1}{\alpha^2} + \frac{1}{\beta^2}) = 4(\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2})$
$= 4(\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha^2\beta^2})$
$= 4(\frac{36-14}{49})$ (1/2 Mark)
$= 4(\frac{22}{49})$
$= \frac{88}{49}$
(b) $\alpha + \beta = 6$ and $\alpha\beta = 7$ (1/2 Mark)
$4(\frac{1}{\alpha^2} + \frac{1}{\beta^2}) = 4(\frac{\alpha^2 + \beta^2}{\alpha^2\beta^2})$
$= 4(\frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha^2\beta^2})$
$= 4(\frac{36-14}{49})$ (1/2 Mark)
$= 4(\frac{22}{49})$
$= \frac{88}{49}$