How do you solve x^2-2x>=0?

Oct 24, 2016

$x \ge 2 , x \le 0$

Explanation:

Fist we must find where,

${x}^{2} - 2 x = 0$

$x \left(x - 2\right) = 0$

$x = 0 , x = 2$

That split's our graph up into three sections.

$x < 0 , 0 < x < 2$ and $x < 2$

so for $x < 0$ we sub in some value (-1) that meets the requirment.

${\left(- 1\right)}^{2} - 2 \left(- 1\right) = 3$ which is greater than $0$ so $x < 0$ is a solution

for $0 < x < 2$ i'll sub in 1

${\left(1\right)}^{2} - 2 \left(1\right) = - 1$ which is less than 0 so this isn't a solution.

for $x < 2$ well sub in 3

${\left(3\right)}^{2} - 2 \left(3\right) = 3$ which is greater than 0 so $x < 2$ is a solution.

leaving us with,

$x \ge 2 , x \le 0$

This can also be done visually by graphing the function,

graph{x^2-2x [-3.182, 4.613, -1.682, 2.215]}

It is true when x is less or equal to 0 or greater or equal to than 2.