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Let α,β\alpha, \beta be the roots of x2−x+p=0x^2 - x + p = 0 and γ,δ\gamma, \delta be the roots of x2−4x+q=0x^2 - 4x + q = 0 where p and q are integers. If α,β,γ,δ\alpha, \beta, \gamma, \delta are in geometric progression then p+qp + q is

Solution

✅ Correct Option: 1

Solution figure for IPMAT Indore 2019 MCQ question 21 (Algebra):


Alternate solution

α,βα, β are roots of x2−x+p=0x² - x + p = 0, so α+β=1,αβ=pα + β = 1, αβ = p

γ,δγ, δ are roots of x2−4x+q=0,soγ+δ=4,γδ=qx² - 4x + q = 0, so γ + δ = 4, γδ = q

α,β,γ,δα, β, γ, δ are in GPGP

Let the GPGP be: a,ar,ar2,ar3a, ar, ar², ar³

Then: a+ar=1a + ar = 1 and ar2+ar3=4ar² + ar³ = 4

From these: a(1+r)=1a(1 + r) = 1 and ar2(1+r)=4ar²(1 + r) = 4

Dividing: r2=4r² = 4, so r=±2r = ±2

Since we need integer p,q,tryr=−2:p, q, try r = -2:

a(1−2)=1⟹a=−1a(1 - 2) = 1 ⟹ a = -1

GP:−1,2,−4,8GP: -1, 2, -4, 8

Therefore:

p=(−1)(2)=−2p = (-1)(2) = -2

q=(−4)(8)=−32q = (-4)(8) = -32

Answer: p+q=−34p + q = -34

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