# How do you solve using the quadratic formula:  x^2+x+1=0?

Nov 20, 2015

See explanation...

#### Explanation:

${x}^{2} + x + 1$ is of the form $a {x}^{2} + b x + c$ with $a = b = c = 1$

This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- 1 \pm \sqrt{{1}^{2} - \left(4 \times 1 \times 1\right)}}{2 \times 1}$

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

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

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

is called the primitive Complex cube root of unity and is used extensively when solving cubic equations.

${\omega}^{2} = \overline{\omega} = - \frac{1}{2} - \frac{\sqrt{3}}{2} i$

${\omega}^{3} = 1$