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If $p$ and $q$ are the zeroes of the quadratic polynomial $f(x) = 6x^2 + x - 2$, then find the value of $\frac{1}{p} + \frac{1}{q} - pq$.
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$p + q = -\frac{1}{6}$, $pq = -\frac{2}{6} = -\frac{1}{3}$
$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-\frac{1}{6}}{-\frac{1}{3}} = \frac{1}{2}$
$\frac{1}{p} + \frac{1}{q} - pq = \frac{1}{2} - (-\frac{1}{3}) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$
$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{-\frac{1}{6}}{-\frac{1}{3}} = \frac{1}{2}$
$\frac{1}{p} + \frac{1}{q} - pq = \frac{1}{2} - (-\frac{1}{3}) = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}$