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Question: Let α, β be the roots of \(x^{2} - x + p = 0\) and γ, δ be root of \(x^{2} - 4x + q = 0\). If α, β, ...

Let α, β be the roots of x2−x+p=0x^{2} - x + p = 0 and γ, δ be root of x2−4x+q=0x^{2} - 4x + q = 0. If α, β, γ, δ are in G.P., then the integral value of p and q respectively are

A

– 2, – 32

B

– 2, 3

C

– 6, 3

D

– 6, – 32

Answer

– 2, – 32

Explanation

Solution

α+β=1\alpha + \beta = 1, αβ=p\alpha\beta = p, γ+δ=4\gamma + \delta = 4, γδ=q\gamma\delta = q

Since α, β, γ, δ are in G.P.

r=β/α=γ/β=δ/γr = \beta/\alpha = \gamma/\beta = \delta/\gamma

α+αr=1\alpha + \alpha r = 1 ⇒ α(1+r)=1\alpha(1 + r) = 1, α(r2+r3)=4\alpha(r^{2} + r^{3}) = 4 ⇒ α.r2(1+r)=4\alpha.r^{2}(1 + r) = 4

So r2=4r^{2} = 4 ⇒ r=±2r = \pm 2

If r=2r = 2, α+2α=1\alpha + 2\alpha = 1 ⇒ α=1/3\alpha = 1/3 and r=−2r = - 2, α−2α=1\alpha - 2\alpha = 1

⇒ α=−1\alpha = - 1

But p=αβ∈Ip = \alpha\beta \in I ∴ r=−2,α=−1r = - 2,\alpha = - 1

∴p=−2p = - 2, q=α2r5=1(−2)5=−32q = \alpha^{2}r^{5} = 1( - 2)^{5} = - 32