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If $\alpha$, $\beta$ are the zeroes of polynomial $p(x) = -9x^2 - 6x + 1$. Find the value of $\alpha^2 + \beta^2$.
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Solution: (a) $\alpha + \beta = -\frac{-6}{9}$, $\alpha\beta = -\frac{1}{9}$ (1)
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$= \frac{2}{3}$ (1)
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$
$= \frac{2}{3}$ (1)