Question

For the equation 3x^2 + px + 3 = 0 , p>0, if one of the roots is square of the other, then p is equal to?

Answer 1

The value is p=3p=3.

Proof: Let a,a2a,a2 be the roots of the equation. Then you get

3x2+px+3=3(x−a)(x−a2)=3x2−3(a2+a)x+a3,3x2+px+3=3(x−a)(x−a2)=3x2−3(a2+a)x+a3,

So you first have to solve a3=1a3=1. This gives the three solutions (the three cube-roots of unity)

a1=1,a2=−12+i23–√,a3=−12−i23–√a1=1,a2=−12+i23,a3=−12−i23

which in turn give the possible values for pp

p1=−3(a21+a1)=−6p1=−3(a12+a1)=−6

p2=−3(a22+a2)=3p2=−3(a22+a2)=3

p3=−3(a23+a3)=3p3=−3(a32+a3)=3

p1p1 is your negative value, but p2=p3=3p2=p3=3 is positive and therefore p=3p=3 is the answer to the question.