How to solve the inequality $x^2>10$ using square roots? Solve the inequality: $$x^2>10$$
How am I supposed to do this? It doesn't make sense when I take into account that if $x^2=10$ then $x=+\sqrt{10}$ and $x=-\sqrt{10}$
But how am I supposed to apply this to an inequality, I would get $x>\sqrt{10}$ and $x>-\sqrt{10}$
But for some reason this just doesn't make sense to me.  Can someone explain it to me mathematically, instead of just having to memorize these kinds of things?

Here is how I think about problems like this:
If $x^2 > 10$, then
$x^2 - 10 > 0$
$(x + \sqrt {10})(x - \sqrt {10}) > 0 $
$(x + \sqrt {10})$ and $(x - \sqrt {10})$ are either both positive or both negative.
Thus, when they are both positive and we divide both sides by one of the terms:
$x + \sqrt {10} > 0$ and $x - \sqrt {10} > 0$
This simplifies to:
$x > \sqrt {10}$
When they are both negative and we divide both sides by one of the terms:
$x + \sqrt {10} < 0$ and $x - \sqrt {10} < 0$
This simplifies to:
$x < -\sqrt {10}$
The final solution is:
$x < -\sqrt {10}$ or $x > \sqrt {10}$

Conversely,
if $x^2 < 10$, then
$x^2 - 10 < 0$
$(x - \sqrt{10})(x + \sqrt{10}) < 0$
We know that one of the terms, $(x - \sqrt{10})$ or $(x + \sqrt{10})$, is negative.
Since $(x - \sqrt{10})$ is always smaller, we know that it is the negative term.
Thus, when we divide both sides by $(x - \sqrt{10})$ we get:
$x + \sqrt{10} > 0$
This simplifies to:
$x > -\sqrt{10}$
when we divide both sides by $(x + \sqrt{10})$ we get:
$x - \sqrt{10} < 0$
This simplifies to:
$x < \sqrt{10}$
The final solution is:
$-\sqrt {10} < x < \sqrt {10}$
