Question

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

Answer 1

`x in phi`, the null set.

Explanation:

We can rewrite our inequality as

`x^2<=-9`

However, we know that `x^2` is always bigger or equal to `0`, for all `x in RR`.

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]}