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Q.

If α, β, γ, δ are the roots of the equation x4Kx3+Kx2+Lx+M=0 where K, L, and M are real numbers, then the minimum value of α2+β2+γ2+δ2 is 

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a

2

b

1

c

-1

d

0

answer is B.

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Detailed Solution

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Roots of equation x4Kx3+Kx2+Lx+M=0 are α,β,γ,δ

Σα=K,Σαβ=KΣαβγ=L,αβγδ=Mα2+β2+γ2+δ2=(α+β+γ+δ)22Σαβ=K22K=(K1)21 α2+β2+γ2+δ2min=1

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