# How do you solve x^2-4x>=0 using a sign chart?

Oct 15, 2017

We first consider where ${x}^{2} - 4 x = 0$

i.e. when $x \left(x - 4\right) = 0$

Hence, $x = 0 \mathmr{and} x = 4$

We then need to consider three seperate regions.

i.e. $x < 0$, $0 < x < 4$ and $x > 4$

For $x < 0$, we have that ${x}^{2} - 4 x > 0$

For $0 < x < 4$, we have that ${x}^{2} - 4 x < 0$

and for $x > 4$ we have that ${x}^{2} - 4 x > 0$

Hence, for ${x}^{2} - 4 x \ge 0$ we have two regions.

i.e. $x \le 0$ or $x \ge 4$

:)>