If $α, β$ are the roots of ${ax}^{2} + 2 bx + c = 0$ and $α + δ, β + δ$ are the roots of ${Ax}^{2} + 2 Bx + C = 0$ then $\frac{b^{2} - ac}{B^{2} - AC} =$

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$a / A$

$A / a$

${(a / A)}^{2}$

${(A / a)}^{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} I f α, β a r e r o o t s o f a x^{2} + 2 b x + c = 0 \\ α + β = \frac{- 2 b}{a}, α β = \frac{c}{a} \\ {(α - β)}^{2} = {(α + β)}^{2} - 4 α β \\ = \frac{4 b^{2}}{a^{2}} - 4 (\frac{c}{a}) \\ = \frac{4}{a^{2}} (b^{2} - a c) \\ α + δ, α + β a r e r o o t s o f A x^{2} + 2 B x + C = 0 \\ (α + δ) + (α + β) = \frac{- 2 B}{A}, (α + δ) (α + β) = \frac{C}{A} \\ ∴ {[(α + δ) - (β - δ)]}^{2} = {[(α + δ) + (δ + β)]}^{2} - 4 (α + δ) (β + δ) \\ = \frac{4 B^{2}}{A^{2}} - 4 \frac{C}{A} \\ {(α - β)}^{2} = \frac{4}{A^{2}} (B^{2} - 4 A C) \\ ∴ \frac{\frac{4}{a^{2}} (b^{2} - a c)}{\frac{4}{A^{2}} (B^{2} - A C)} = 1 \\ \Rightarrow \frac{b^{2} - a c}{B^{2} - A C} = {(\frac{a}{A})}^{2} \end{array}$