If $α$ and $β$ are the roots of $2 X^{2} + X + 3 = 0,$ then the equation whose roots are $\frac{1 - α}{1 + α}$ and $\frac{1 - β}{1 + β}$ is

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$2 X^{2} + X + 3 = 0$

$2 X^{2} - X + 3 = 0$

$2 X^{2} + X - 3 = 0$

$2 X^{2} - X - 3 = 0$

answer is A.

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Detailed Solution

$\begin{array}{l} α, β a r e r o o t s o f 2 x^{2} + x + 3 = 0 t h e n \\ α + β = \frac{- 1}{2}, α β = \frac{3}{2} \\ s u m o f r o o t s = \frac{1 - α}{1 + α} + \frac{1 - β}{1 + β} \\ = \frac{(1 - α) (1 + β) + (1 - β) (1 + α)}{(1 + α) (1 + β)} \\ = \frac{1 - α + β - α β + 1 + α - β - α β}{1 + (α + β) + α β} \\ = \frac{2 - 2 α β}{1 + (α + β) + α β} \\ = \frac{2 - 3}{1 - \frac{1}{2} + \frac{3}{2}} \Rightarrow \frac{- 2}{4} \Rightarrow \frac{- 1}{2} \\ \Pr o d u c t o f r o o t s = (\frac{1 - α}{1 + α}) (\frac{1 - β}{1 + β}) \\ = \frac{1 - (α + β) + α β}{1 + (α + β) + α β} \\ = \frac{1 + \frac{1}{2} + \frac{3}{2}}{1 - \frac{1}{2} + \frac{3}{2}} \\ = \frac{2 + 1 + 3}{2 - 1 + 3} \Rightarrow \frac{6}{4} \Rightarrow \frac{3}{2} \\ Re q u i r e d E q u a t i o n x^{2} - (s u m o f r o o t s) x + \Pr o d u c t o f r o o t s = 0 \\ \Rightarrow x^{2} + \frac{1}{2} x + \frac{3}{2} = 0 \\ \Rightarrow 2 x^{2} + x + 3 = 0 \end{array}$