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If $α, β, γ, δ \in R$ satisfy $\frac{{(α + 1)}^{2} + {(β + 1)}^{2} + {(γ + 1)}^{2} + {(δ + 1)}^{2}}{α + β + γ + δ} = 4$
If biquadratic equation $a_{0} x^{4} + a_{1} x^{3} + a_{2} x^{2} + a_{3} x + a_{4} = 0$ has the roots $(α + \frac{1}{β} - 1), (β + \frac{1}{γ} - 1), (γ + \frac{1}{δ} - 1), (δ + \frac{1}{α} - 1) .$ Then the value of $\frac{a_{2}}{a_{0}}$ is:

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$6$

none of these

$4$

$- 4$

answer is C.

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Detailed Solution

$α = 1, β = 1, γ = 1, δ = 1 (a s) (α - 1)^{2} + (β - 1)^{2} + (γ - 1)^{2} + (δ - 1)^{2} = 0$

$∴$ The roots of given equation is equal to 1.

$S_{2} = \frac{a_{2}}{a_{0}} = 6$

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