Mathematics - Quadratic Equation Question with Solution | TestHub

The denominator $x^2+x+1$ has discriminant $D=1^2-4(1)(1)=-3<0$. Since its leading coefficient is positive, $x^2+x+1>0$ for all real $x$. Therefore, for the inequality $\frac{x^2-3x+2}{x^2+x+1}<0$ to hold, the numerator must be negative. $x^2-3x+2<0$. Factoring gives $(x-1)(x-2)<0$. This inequality is satisfied when $1<x<2$. Thus, the set of all real values of $x$ is $(1,2)$.