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.