# How do you know if a polynomial of degree 2 has an x-intercept?

Mar 6, 2016

Once you have it in standard $a {x}^{2} + b x + c$ form, check the discriminant $\Delta = {b}^{2} - 4 a c$ is non-negative.

#### Explanation:

Any polynomial of degree $2$ in $x$ can be expressed in the form:

$a {x}^{2} + b x + c$

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

$\Delta = {b}^{2} - 4 a c$

If $\Delta \ge 0$ then $a {x}^{2} + b x + c = 0$ has at least one Real root.

Roots of $a {x}^{2} + b x + c = 0$ are $x$ intercepts.

The roots of $a {x}^{2} + b x + c = 0$ are given by the formula:

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

Notice that the discriminant is the expression under the square root, so the square root takes Real values if and only if $\Delta \ge 0$.