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

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Answer 1 · 2016-10-24T01:16:45

$x>=2, x<=0$

Fist we must find where,

$x^2-2x=0$

$x(x-2)=0$

$x=0 , x=2$

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.

$(-1)^2-2(-1)= 3$ which is greater than $0$ so $x<0$ is a solution

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

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

for $x < 2$ well sub in 3

$(3)^2-2(3)= 3$ which is greater than 0 so $x < 2$ is a solution.

leaving us with,

$x>=2, x<=0$

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.

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