<s>[INST] 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?
 [/INST] 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:
$-\sqrt {10} < x < \sqrt {10}$
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