How do you solve $x^{2} + 9 \leq 0$ $x^2+9<=0$ ?

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Answer 1 · 2018-05-13T11:22:20

$x \in ϕ$ $x in phi$ , the null set.

We can rewrite our inequality as

$x^{2} \leq - 9$ $x^2<=-9$

However, we know that $x^{2}$ $x^2$ is always bigger or equal to $0$ $0$ , 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]}

How do you solve x^2+9<=0x2+9≤0x^2+9<=0?