# How do you solve x^2+x=1 by completing the square?

May 5, 2015

${x}^{2} + x = 1$

In order for the left side to be a perfect square, it must be the square of some $x + a$.

Since ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$,

we see that we must have $2 a = 1$, so $a = \frac{1}{2}$.

This make ${a}^{2} = \frac{1}{4}$ which I need on the left to make a perfect square. We'll add 1/4 to bot side to get:

${x}^{2} + x + \frac{1}{4} = 1 + \frac{1}{4}$

Factoring the left and simplifying the right, we get:

${\left(x + \frac{1}{2}\right)}^{2} = \frac{5}{4}$.

Now solve by the square root method:

$x + \frac{1}{2} = \pm \sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2}$

So $x = - \frac{1}{2} \pm \frac{\sqrt{5}}{2}$, which may also be written:

$x = \frac{- 1 \pm \sqrt{5}}{2}$