# Solutions Using the Discriminant

## Key Questions

For $a {x}^{2} + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:

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

The discriminate is the portion of the quadratic equation within the radical: ${\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}$

If the discriminate is:
- Positive, you will get two real solutions
- Zero you get just ONE solution
- Negative you get complex solutions

To determine how many roots are there in a quadratic equation.

#### Explanation:

There are 4 natures

${b}^{2} - 4 a c > 0$ and is a perfect square -># 2 rational roots

${b}^{2} - 4 a c > 0$ and is not a perfect square $\to$ 2 irrational root
${b}^{2} - 4 a c = 0 \to$ 1 root
${b}^{2} - 4 a c < 0$ No root

$\Delta = {b}^{2} - 4 a c$ for a quadratic $a {x}^{2} + b x + c = 0$

#### Explanation:

The discriminant indicated normally by $\Delta$, is a part of the quadratic formula used to solve second degree equations.
Given a second degree equation in the general form:
$a {x}^{2} + b x + c = 0$
the discriminant is:
$\Delta = {b}^{2} - 4 a c$

The discriminant can be used to characterize the solutions of the equation as:
1) $\Delta > 0$ two separate real solutions;
2) $\Delta = 0$ two coincident real solutions (or one repeated root);
3) $\Delta < 0$ no real solutions.

For example:
${x}^{2} - x - 2 = 0$
Where: $a = 1$, $b = - 1$ and $c = - 2$
So:
$\Delta = {b}^{2} - 4 a c = 1 + 4 \cdot 2 = 9 > 0$, giving $2$ real distinct solutions.

The discriminant can also come in handy when attempting to factorize quadratics. If $\Delta$ is a square number, then the quadratic will factorize, (since the square root in the quadratic formula will be rational). If it is not a square number, then the quadratic will not factorize. This can save you spending ages trying to factorize when it won't work. Instead, solve by completing the square or using the formula.

I hope that helps!