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Question

x42x3+4x2+6x21=0 if two of its roots are equal in magnitude but opposite in sign.Then among the 4 roots the real roots be ±k.Then find k2 ?

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Solution

α+β=0butα+β+γ+δ=2
γ+δ=2.
Let αβ=pandγδ=q
Given equation is equivalent to
(x2+p)(x22x+q)=0.
Comparing the coefficients, we get
p+q=4,2p=6,pq=21
p=3,q=7
Thus equation reduced to,
(x23)(x22x+7)=0
Hence roots are ±3, and 1±i6.

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