Question

Solve the following inequalities:
$|x^2-2x-8|>2x.$

Answer 1

$|x^2-2x-8|>2x$

$x^2-2x-8<-2x$ or $x^2-2x-8>2x$

$x^2-8<0$ $x^2-4x-8>0$

$(x-2\sqrt{2})(x+2\sqrt{2})<0$ $(x-(2+\sqrt{3}\left)\right(x-(2+\sqrt{3})>0$

$xϵ(-2\sqrt{2},2\sqrt{2})$ $xϵ(-\infty ,2-\sqrt{3})\cup (2+\sqrt{3},\infty )$

$xϵ(-2\sqrt{2},2\sqrt{2})\cup (-\infty ,2-\sqrt{3})\cup (2+\sqrt{3},\infty )$