$\begin{array}{l} L e t α, β b e t h e r o o t s o f x^{2} - c + p = 0 a n d γ, δ b e t h e r o o t s o f x^{2} - 4 x + q = 0. I f α, β, γ, δ a r e i n G . P, t h e n t h e I n t e g r a l v a l u e s o f p a n d q, r e s p e c t i v e l y a r e \end{array}$

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$- 2, - 32$

$- 2, 3$

$- 6, 3$

$- 6, - 32$

answer is A.

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Detailed Solution

$\begin{array}{l} α a n d β a r e r o o t s o f x^{2} - x + p a n d γ, δ a r e r o o t s o f x^{2} - 4 x + q \\ L e t α = a, β = a r, γ = a r^{2}, δ = a r^{3} \\ ∴ α + β = 1 \Rightarrow a (1 + r) = 1 - - (1) \\ a n d γ + δ = 4 \Rightarrow a r^{2} (1 + r) = 4 - - (2) \\ \frac{(2)}{(1)} \Rightarrow \frac{a r^{2} (1 + r)}{a (1 + r)} = 4 \\ \Rightarrow r^{2} = 4 \Rightarrow r \pm 2 \\ f r o m (1) a (1 \pm 2) = 1 \\ \Rightarrow a = - 1, \frac{1}{3} \\ N o w α β = p \Rightarrow a^{2} r = p \\ \Rightarrow p = - 2, \frac{2}{9} \\ a n d γ δ = q \Rightarrow a^{2} r^{5} = q \\ \Rightarrow q = - 32, \frac{32}{9} \\ ∴ p, q a r e - 2, - 32 \end{array}$