The real values of a for which the quadratic equation 3x² + 2(a² + 1)x + (a² – 3a + 2) = 0 possesses roots of opposite signs lie in

The quadratic equation $3x^2+2\left(a^2+1\right)x+a^2-3a+2=0$ will have two roots of opposite signs if it has real roots and the product of the roots is negative, that is , if $4{\left(a^2+1\right)}^2-12\left(a^2-3a+2\right)\ge 0$ and $\frac{a^2-3a+2}{3}<0$ Both of these conditions are met if $a^2-3a+2<0$ i.e. if $(a-1)(a-2)<0$ or 1 < a < 2.