Inequality: $x > \pm\sqrt{2}$ $x^2=2$
and $x=+\sqrt2$ and $x=-\sqrt2$.
That's ok.
But when $x^2>2$, (It's my fault. $x^2<2$ (inequality fault) => $x^2>2$ (ok))
$x > \pm\sqrt{2}$ ?
The answer is $x > \sqrt{2}$, and $x<-\sqrt{2}$.
When I was young, maybe I studied $x$ bigger than positive number, and smaller than negative number.
 A: From 
$$x^2\gt 2$$
we can conclude that
$$\sqrt{x^2}\gt \sqrt{2}.$$
However, the important point to remember is that $\sqrt{x^2}$ is not equal to $x$, it is equal to $|x|$, the absolute value of $x$. That is, we have
$$|x|\gt \sqrt{2}.$$
And, by the definition of the absolute value,
$|x|\gt\sqrt{2}$ if and only if $x\gt 0$ and $x\gt \sqrt{2}$, or $x\lt 0$ and $-x\gt\sqrt{2}$, which is equivalent to $x\lt-\sqrt{2}$; so
$$|x|\gt\sqrt{2}\text{ is equivalent to }x\gt\sqrt{2}\text{ or }x\lt-\sqrt{2}.$$
You either write it as two inequalities, with an "or" connective, or as the single inequality using the absolute value. 
A: I think you have a typo: $x^2<2$ implies $-\sqrt2<x<\sqrt2$ which is fine, but the inequality $x^2>2$ means that either $x>\sqrt2$ or $x<-\sqrt2$, not both simultaneously (which is impossible). See for yourself that the  graph of $y=x^2$ is above that of $y=2$ only outside the interval $[-\sqrt2,\sqrt2]$:
$\hskip 1.8in$ 
A: If $x > \sqrt{2}$, then $x² > 2$. You seem to be having trouble with the difference between "smaller" and "more negative". A number's relative "size" is its absolute value, or its distance from zero. 
Since the problem asks for $x² < 2$, the "less than" sign is unaffected and your desired $x$ is $x < \pm\sqrt{2} $.
If $x<-\sqrt{2} \Rightarrow x² < (-\sqrt{2})² =2.$
If $x < +\sqrt{2} \Rightarrow x² < (+\sqrt{2}) = 2$. 
Good luck.
A: This is how I always viewed this. Not sure if totally correct.
$x^2 > 2$ is similar to $|x|^2 > 2$, so $|x| > \sqrt{2}$, so $x < -\sqrt{2}$ or $x > \sqrt{2}$
