# How do you solve  x^2 + x + 1=0 using completing the square?

Jun 13, 2015

This quadratic has no real roots, but:

${x}^{2} + x + 1 = {\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4}$

giving complex solutions: $x = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$

#### Explanation:

${x}^{2} + x + 1$ is one of the factors of ${x}^{3} - 1$

${x}^{3} - 1 = \left(x - 1\right) \left({x}^{2} + x + 1\right) = \left(x - 1\right) \left(x - \omega\right) \left(x - {\omega}^{2}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i = \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right)$

is called the primitive complex root of unity.

Pretending we don't know that,

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

$= {\left(x + \frac{1}{2}\right)}^{2} + \frac{3}{4}$

So this is zero when ${\left(x + \frac{1}{2}\right)}^{2} = - \frac{3}{4}$

Hence $x + \frac{1}{2} = \pm \sqrt{- \frac{3}{4}} = \pm \frac{\sqrt{3}}{2} i$

So $x = - \frac{1}{2} \pm \frac{\sqrt{3}}{2} i$