First slide
Theory of equations
Question

If the equation whose roots are the squares of the roots of the cubic x3ax2+bx1=0 is identical with the given cubic equation, then

Moderate
Solution

Given equation is x3ax2+bx1=0. If roots of the equation be α,β,γ, then

α2+β2+γ2=(α+β+γ)22(αβ+βγ+γα)=a22b 

α2β2+β2γ2+γ2α2=(αβ+βγ+γα)22αβγ(α+β+γ)=b22a α2β2γ2=1

So, the equation whose roots are α2,β2,γ2 is given by

x3a22bx2+b22ax1=0

It is identical to

x3ax2+bx1=0 a22b=a and b22a=b

Eliminating b, we get

a2a242a=a2a2

or   aa(a1)282(a1)=0or   aa32a2a6=0or   a(a3)a2+a+2=0  a=0 or a=3 or a2+a+2=0

which gives b = 0 or b = 3 or b2 + b + 2 = 0. So, a = b = 0 or a = b = 3 or a, b are roots of x2 + x + 2 = 0.

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