Question

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

Answer 1

Suppose $y=\frac{1+x^2}{x^2-5x+6}<0$

Numerator $\Rightarrow 1+x^2=0$

$\Rightarrow D=0-4<0$

Here,$a>0$, so $1+(x{)}^2>0\forall x\in R$

Denominator, $x^2=5x+6$

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

but $x\ne 2,3$ as denominator becomes zero

y will be less than zero if numerator & denominator have opposite sign as Numerator is positive $\forall x\in R$, Denominator must be negative.

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

$\Rightarrow x\in (2,3)$