If $α, β$ are the zeros of the polynomial $f (x) = x^{2} - 3 x + 6,$ then find the value of $\frac{1}{α} + \frac{1}{β} + α^{2} + β^{2} - 2 α β$

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Solution

$α + β = - (- 3) = 3$
$α β = 6$
$(α + β)^{2} - 4 α β = (α - β)^{2} = 9 - 24 = - 15$
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{3}{6} = \frac{1}{2}$
$\frac{1}{α} + \frac{1}{β} + (α - β)^{2}$
$\frac{1}{2} - 15$
$⟹ - \frac{29}{2}$

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