An algebra problem by sumantra chatterjee
Algebra
Level
pending
If the roots of the equation ax^4+ax^3+bx^2+cx+d=0 are in GP then
c^2=(a^2)×(b^2)
a^2=b
b^2=ac
c^2=(a^2)×d
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1 solution
Let p, q, r, s be the roots of the equation. Then pqrs=d and since p, q, r, s are in GP hence d=(ps)^2 as ps=qr. Now [(p+q+r+s)/ps]=(-a/ps) or (1/p+1/q+1/r+1/s)=(-a/ps). Again 1/p,1/q,1/r,1/s are the roots of equation dx^4+cx^3+bx^2+ax+1=0. Therefore 1/p+1/q+1/r+1/s=(-c/d). So (-a/ps)=(-c/d). So [a^2/(ps)^2]=c^2/d^2. So a^2/d=c^2/d^2. So finally a^2×d=c^2.