If $α, β$ are roots of $a x^{2} + b x + c = 0$ , then one root of the equation $a x^{2} - b x (x - 1) + c (x - 1)^{2} = 0$ is :

(\frac{α}{1 - α})

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(\frac{1 - β}{β})

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(\frac{α}{1 + α})

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(\frac{β}{1 + β})

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Solution

The correct options are
C $(\frac{α}{1 + α})$
D $(\frac{β}{1 + β})$
We have, $a x^{2} - b x^{2} + b x + c x^{2} - 2 c x + c = 0$
$(a - b + c) x^{2} + (b - 2 c) x + c = 0$
Sum of the roots (S)
$\frac{b - 2 c}{a - b + c} = \frac{(- \frac{b}{a} + \frac{2 c}{a})}{(1 - \frac{b}{a} + \frac{c}{a})}$
$S = \frac{α + β + 1 α β}{2 + α + β + α β} = \frac{α}{α + 1} + \frac{β}{β + 1}$
Product of the roots (P) $= \frac{c}{a - b + c}$
$\Rightarrow P = \frac{(\frac{c}{a})}{(1 - \frac{b}{c} + \frac{c}{a})}$
$= \frac{α β}{1 + α + β + α β} = \frac{α}{(α + 1)} \cdot \frac{β}{(β + 1)}$
Thus the roots are $\frac{α}{α + 1} a n d \frac{β}{β + 1}$

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