# How do you solve abs(x - 6)> 2?

Apr 6, 2015

If $| x - 6 | > 2$
it implies that ${\left(| x - 6 |\right)}^{2} > {2}^{2}$ is also true

$\implies {\left(x - 6\right)}^{2} > {2}^{2}$

$\implies {\left(x - 6\right)}^{2} - {2}^{2} > 0$

The left hand side is a difference of two squares that behaves like,
$\left(a - b\right) \left(a + b\right) = {a}^{2} - {b}^{2}$

$\implies \left(x - 6 - 2\right) \left(x - 6 + 2\right) > 0$

$\implies \left(x - 8\right) \left(x - 4\right) > 0$

The answer is $x < 4$ $\cup$ $x > 8$