Mathematics - Quadratic Equation Question with Solution | TestHub

We know that $1+x^2+x^4=(1+x^2{)}^2-x^2=(1+x^2-x)(1+x^2+x)$.

Let $y=1+x+x^2$

The equation becomes $(a-1)y^2=(a+1)(1-x+x^2)y$.

Since $1+x+x^2=0$ has imaginary roots, we can divide by $y$.

$(a-1)(1+x+x^2)=(a+1)(1-x+x^2)$.

$((a+1)-(a-1))x^2+(-(a+1)-(a-1))x+((a+1)-(a-1))=0$.

$x^2-ax+1=0$.

For all roots to be imaginary, the discriminant must be negative.

$D=(-a{)}^2-4(1)(1)<0$.

$a^2<4$.

$-2<a<2$.

So, $a\in (-2,2)$.