$x^2 > k \implies |x| > \sqrt{k}$?

Question

Is the following true?

If $k \geq 0$...

$x^2 > k \implies$

$|x| > \sqrt k \implies$

$x > \sqrt k$ or $x < -\sqrt k$

If $k < 0$...

$x^2 > k \implies$

$x^2 \geq 0$

Answer 1

Just distinguish the cases

If $k$ is nonnegative then $x^2>k$ is true if and only if $x>\sqrt{k}$ or $x<-\sqrt{k}$

If $k$ is negative, then $x^2>k$ is true for all $x\in \mathbb R$

In your case we have the additional restriction $x\ge 0$. Then, we have :

If $k$ is nonnegative, then $x^2>k$ is true if and only if $x>\sqrt{k}$

If $k$ is negative, then $x^2>k$ is true if and only if $x\ge 0$

Answer 2

The second half: if $k < 0$ then $x^2 > k$ then $x^2 \ge 0$ is trivial and irrelevant as $x^2 \ge 0 > k$ is always true.

So a better statement is: for $k \ge 0$ is it true $x^2 > k \implies |x| > \sqrt{k}$.

The answer is yes. If $|x| \le \sqrt{k}$ while $|x| \ge 0$ then $x^2 = |x||x| \le |x|\sqrt{k} \le \sqrt{k}\sqrt{k} = k$, which is a contradiction.