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

Feb 27, 2016

This would be a very roundabout way in this case.

#### Explanation:

$\to {x}^{2} - x = 0 \to x \left(x - 1\right) = 0 \to x = 0 \mathmr{and} x = 1$

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:
${x}^{2} + 2 \cdot \left(- \frac{1}{2}\right) \cdot x + {\left(- \frac{1}{2}\right)}^{2} = {\left(x - \frac{1}{2}\right)}^{2}$, is the square we look for.

We have to add the ${\left(- \frac{1}{2}\right)}^{2}$ also to other side.

So: ${\left(x - \frac{1}{2}\right)}^{2} = {\left(- \frac{1}{2}\right)}^{2}$

$\to x - \frac{1}{2} = \pm \frac{1}{2} \to x = 0 \mathmr{and} x = 1$

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.