Question

Solve the following inequation:
$\frac{x^2-5x+6}{x^2+x+1}<0$

Answer 1

Denominator $=x^2+x+1$

$\Rightarrow$Denominator $=x^2+x+1=(x+\frac{1}{2}{)}^2+\frac{3}{4}$

So we can say that the denominator is always greater than zero

Hence for the given inequality to hold, we must have

$\Rightarrow x^2-5x+6<0$

$\Rightarrow x^2-3x-2x+6<0$

$\Rightarrow x(x-3)-2(x-3)<0$

$\Rightarrow (x-3)(x-2)<0$

Thus we can say that $\Rightarrow 2<x<3$