# How do you solve x^2+9<=0?

May 13, 2018

$x \in \phi$, the null set.

#### Explanation:

We can rewrite our inequality as

${x}^{2} \le - 9$

However, we know that ${x}^{2}$ is always bigger or equal to $0$, for all $x \in \mathbb{R}$.

This means there is no real $x$ for which our inequality is true.

$x \in \phi$

You can also check it by the graphic of the polynomial:
graph{x^2+9 [-36.57, 36.48, -5.37, 31.17]}