Question

Solve the following inequality:
$\frac{x^4+x^2+1}{x^2-4x-5}>0$

Answer 1

$\frac{x^4+x^2+1}{x^2-4x-5}>0$

As $x^4+x^2+1>0\forall xϵR$

$\frac{1}{x^2-4x-5}>0$

$\frac{1}{(x-5)(x+1)}>0$

$xϵ(-\infty ,-1)\cup (5,\infty )$