Question

For what values of $a\in R$, the quadratic equation $(a^2+1)x^2-(a+1)x+(a^2-a-2)=0$ will have roots of opposite sign.

Answer 1

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

The above equation has roots of opposite sign, then product of roots $=\frac{a^2-a-2}{a^2+1}<0$

$a^2-a-2<0$ ($a^2+1$ is a always positive)

$(a+1)(a-2)<0$

$a\in (-1,2)$ for that equation to have roots of opposite signs.