Let $α \neq β$ satisfy $α^{2} + 1 = 6 α, β^{2} + 1 = 6 β .$ Then, the quadratic equation whose roots are $\frac{α}{α + 1}, \frac{β}{β + 1}$ is

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$8 x^{2} - 8 x + 1 = 0$

$8 x^{2} + 8 x - 1 = 0$

$8 x^{2} + 8 x + 1 = 0$

$8 x^{2} - 8 x - 1 = 0$

answer is C.

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Detailed Solution

$\begin{array}{l} I f α^{2} + 1 = 6 α \Rightarrow α^{2} - 6 α + 1 = 0 \\ a n d β^{2} + 1 = 6 β \Rightarrow β^{2} - 6 β + 1 = 0 \\ ∴ α, β a r e r o o t s o f f (x) = x^{2} - 6 x + 1 = 0 \\ L e t \frac{α}{α + 1} = y \Rightarrow α = α y + y \\ \Rightarrow α (1 - y) = y \\ \Rightarrow α = \frac{y}{1 - y} \\ α i s r o o t s o f f (x) t h e n f (\frac{y}{1 - y}) = 0 \\ \Rightarrow {(\frac{y}{1 - y})}^{2} - 6 (\frac{y}{1 - y}) + 1 = 0 \\ \Rightarrow y^{2} - 6 y (1 - y) + {(1 - y)}^{2} = 0 \\ \Rightarrow y^{2} - 6 y + 6 y^{2} + 1 + y^{2} - 2 y = 0 \\ \Rightarrow 8 y^{2} - 8 y + 1 = 0 \\ ∴ Re q u i r e d E q u a t i o n i s 8 x^{2} - 8 x + 1 = 0 \end{array}$