The cubic equation x³ + ax+ a = 0 has three roots x₁, x₂ and x₃ with x₁ ≤ x₂ ≤ x₃, where a is real and non zero, such that:

x₁²/ x₂ + x₂²/ x₃ + x₃²/ x₁ = 8

Determine x₁, x₂ and x₃

(In reply to agree, but . . . by xdog)

x^3+ax+a=0; assume that a=-8; then x^3-8x-8=0 also solves for x=-2:(-8)+16-8=0.

Since this implies that (x+2)(x^2-2x-4)=0, the roots are: {-2, 1-5^(1/2), 1+5^(1/2)}

But in all combinations the stipulated summations total -8, not 8.

Posted by broll on 2012-12-24 04:23:35