If $α, β$ are the roots of ${ax}^{2} + bx + c = 0$ then ${(aα + b)}^{- 2} + {(aβ + b)}^{- 2} =$

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$\frac{b^{2} - 2 ac}{a^{2} c^{2}}$

$\frac{b^{2} - 2 ac}{c^{4}}$

$\frac{b^{2} - 2 ac}{a^{4}}$

$\frac{b^{2} - 2 ac}{a^{3} c}$

answer is C.

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Detailed Solution

$\begin{array}{l} I f α, β a r e r o o t s o f a x^{2} + b x + c = 0 t h e n \\ α + β = \frac{- b}{a}, α β = \frac{c}{a} \\ a n d \\ a α^{2} + b α + c = 0, a β^{2} + b β + c = 0 \\ \Rightarrow a α + b = \frac{- c}{α}, \Rightarrow a β + b = \frac{- c}{a} \\ ∴ {(a α + b)}^{- 2} + {(a β + b)}^{- 2} = {(\frac{- c}{α})}^{- 2} + {(\frac{- c}{β})}^{- 2} \\ = \frac{α^{2}}{c^{2}} + \frac{β^{2}}{c^{2}} \\ = \frac{{(α + β)}^{2} - 2 α β}{c^{2}} \\ = \frac{\frac{b^{2}}{a^{2}} - \frac{2 c}{a}}{c^{2}} = \frac{b^{2} - 2 a c}{a^{2} c^{2}} \end{array}$