# How to use the discriminant to find out what type of solutions the equation has for x^2 + 6x + 5 = 0?

May 18, 2015

${x}^{2} + 6 x + 5$ is of the form $a {x}^{2} + b x + c$ with $a = 1$, $b = 6$ and $c = 5$.

The discriminant is given by the formula:

$\Delta = {b}^{2} - 4 a c = {6}^{2} - \left(4 \times 1 \times 5\right) = 36 - 20 = 16 = {4}^{2}$

Since $\Delta > 0$, the quadratic equation ${x}^{2} + 6 x + 5 = 0$ has two distinct real roots.

Further, since $\Delta = {4}^{2}$ is a perfect square the roots are rational.

(In fact the solutions are given by the formula:

$x = \frac{- b \pm \sqrt{\Delta}}{2 a} = \frac{- 6 \pm 4}{2} = - 3 \pm - 2$

That is $x = - 1$ or $x = - 5$.)