# What kind of solutions does m^2 + m + 1 = 0 have?

Jul 8, 2015

${m}^{2} + m + 1 = 0$
has two imaginary solutions

#### Explanation:

If expressed in a standard quadratic form
$\textcolor{w h i t e}{\text{XXXX}}$$a {m}^{2} + b m + c = 0$

The discriminant $\Delta = {b}^{2} - 4 a c$
indicates the number of roots
$\Delta = \left\{\begin{matrix}> 0 \Rightarrow \text{2 Real roots" \\ =0 rArr "1 Real root" \\ <0 rArr "2 Imaginary roots}\end{matrix}\right.$

${b}^{2} - 4 a c = {1}^{2} - 4 \left(1\right) \left(1\right) = - 3 < 0$

Jul 8, 2015

The solutions include an imaginary number, $\sqrt{- 3} = \sqrt{3} i$.

#### Explanation:

${m}^{2} + m + 1 = 0$ is in the form of a quadratic equation $a {x}^{2} + b x + c = 0$, where $a = 1 ,$ $b = 1 ,$ $c = 1$.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substitute the values for $a$, $b$, and $c$ into the quadratic formula.

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$ =

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

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

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

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

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

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