# How to use the discriminant to find out what type of solutions the equation has for m^2 + m + 1 = 0?

May 17, 2015

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

The discriminant is given by the formula:

$\Delta = {b}^{2} - 4 a c = {1}^{2} - \left(4 \times 1 \times 1\right) = 1 - 4 = - 3$

$\Delta < 0$ so there are no real roots. There are two distinct complex roots.

As a matter of interest, the two roots are $\omega$ and ${\omega}^{2}$, where

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

is called the primitive cube root of unity.

Another way of expressing it is:

$\omega = \cos \left(\frac{2 \pi}{3}\right) + i \sin \left(\frac{2 \pi}{3}\right) = {e}^{\frac{2 \pi}{3} i}$