If the two roots of (a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2=0 are real and distinct, then the se…

Question

If the two roots of $(a-1)(x^4+x^2+1)+(a+1)(x^2+x+1)^2=0$ are real and distinct, then the set of all values of 'a' is:

Accepted Answer

Factoring $x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, the equation becomes

$$(x^2+x+1)[(a-1)(x^2-x+1)+(a+1)(x^2+x+1)]=0 \implies (x^2+x+1)(ax^2+x+a)=0.$$

For $ax^2+x+a=0$ to have real, distinct roots: $a\ne0$ and $1-4a^2>0 \implies a\in\left(-\dfrac12,0\right)\cup\left(0,\dfrac12\right).$

Solutions · 1