# What is the discriminant of m^2 + m + 1 = 0 and what does that mean?

Jul 21, 2015

The discriminant $\Delta$ of ${m}^{2} + m + 1 = 0$ is $- 3$.

So ${m}^{2} + m + 1 = 0$ has no real solutions. It has a conjugate pair of complex solutions.

#### Explanation:

${m}^{2} + m + 1 = 0$ is of the form $a {m}^{2} + b m + c = 0$, with $a = 1$, $b = 1$, $c = 1$.

This has discriminant $\Delta$ given by the formula:

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

We can conclude that ${m}^{2} + m + 1 = 0$ has no real roots.

The roots of ${m}^{2} + m + 1 = 0$ are given by the quadratic formula:

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

Notice that the discriminant is the part inside the square root. So if $\Delta > 0$ then the quadratic equation has two distinct real roots. If $\Delta = 0$ then it has one repeated real root. If $\Delta < 0$ then it has a pair of distinct complex roots.

In our case:

$m = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 1 \pm \sqrt{- 3}}{2} = \frac{- 1 \pm i \sqrt{3}}{2}$

The number $\frac{- 1 + i \sqrt{3}}{2}$ is often denoted by the Greek letter $\omega$.

It is the primitive cube root of $1$ and is important when finding all roots of a general cubic equation.

Notice that $\left(m - 1\right) \left({m}^{2} + m + 1\right) = {m}^{3} - 1$

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

Jul 21, 2015

The discriminant of $\left({m}^{2} + m + 1 = 0\right)$ is $\left(- 3\right)$ which tells us that there are no Real solutions to the equation (a graph of the equation does not cross the m-axis).

#### Explanation:

Given a quadratic equation (using $m$ as the variable) in the form:
$\textcolor{w h i t e}{\text{XXXX}}$$a {m}^{2} + b m + c = 0$

The solution (in terms of $m$) is given by the quadratic formula:
$\textcolor{w h i t e}{\text{XXXX}}$$m = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The discriminant is the portion:
$\textcolor{w h i t e}{\text{XXXX}}$${b}^{2} - 4 a c$

If the discriminant is negative
$\textcolor{w h i t e}{\text{XXXX}}$there can be no real solutions
$\textcolor{w h i t e}{\text{XXXX}}$(since there is no real value which is the square root of a negative number).

For the given example
$\textcolor{w h i t e}{\text{XXXX}}$${m}^{2} + m + 1 = 0$
the discriminant, $\Delta$ is
$\textcolor{w h i t e}{\text{XXXX}}$${\left(1\right)}^{2} - 4 \left(1\right) \left(1\right) = - 3$
and therefore
$\textcolor{w h i t e}{\text{XXXX}}$there are no Real solutions to this quadratic.