# How to use the discriminant to find out how many real number roots an equation has for x^2+4x+5?

May 17, 2015

${x}^{2} + 4 x + 5$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 4$ and $c = 5$.

The discriminant is given by the formula:

$\Delta = {b}^{2} - 4 a c = {4}^{2} - \left(4 \times 1 \times 5\right) = 16 - 20 = - 4$

Since this is negative, ${x}^{2} + 4 x + 5 = 0$ has no real roots. It has two distinct complex roots.

The various possible cases are:
$\Delta > 0$ : The quadratic has two distinct real roots.
$\Delta = 0$ : The quadratic has one repeated real root.
$\Delta < 0$ : The quadratic has no real roots. It has two distinct complex roots.

In addition, if the original coefficients are integers (or rational numbers) and $\Delta \ge 0$ is a perfect square, then the roots are rational numbers.