How do you solve using the completing the square method #x^2=x#?

1 Answer
Feb 27, 2016

Answer:

This would be a very roundabout way in this case.

Explanation:

#->x^2-x=0->x(x-1)=0->x=0orx=1#

Completing the square (if you really must):
#x^2-x=0#, so the number with #x# is #-1#
Take half of that and square it:
#x^2+2*(-1/2)*x+(-1/2)^2=(x-1/2)^2#, is the square we look for.

We have to add the #(-1/2)^2# also to other side.

So: #(x-1/2)^2=(-1/2)^2#

#->x-1/2=+-1/2->x=0orx=1#

I prefer the first way, to be honest, but if you have to use the other method, you can still use the first to check your answer.