Solving quadratic inequality $x^2 > 0$ This particular inequality has me puzzled:
$$ x^2 > 0 $$
Graphically it is simple to solve, as you can see the curve only touches $0$ but stretches off in both negative and positive directions as shown here:

However I cannot express in a calculation how I can come up with $$x > 0 \lor x < 0. $$
For instance if you attempt to solve the inequality by finding the square root on either side, you end up with:
$$ x > \pm\sqrt{0} $$
Since zero is neither positive or negative, this doesn't really make much sense to me. Or is there a different way to interpret this?
Thanks
 A: We have that

*

*for $x=0 \implies x^2=0$


*for $x \neq 0 \implies x^2>0$
and the proof is complete by exhaustion.
Following your idea, using that $\sqrt{x^2}= |x|$, we can take the square root both sides to obtain
$$x^2>0 \iff \sqrt{x^2}>\sqrt 0 \iff |x|>0$$
which is always true for $x\neq 0$.
A: Caution,
$$a^2>b$$ does not imply $$a>\pm\sqrt b.$$
But $$\pm a>\sqrt b$$ is correct. (With a somewhat sloppy notation.)

More rigorously
$$a^2>b\\\iff (a-\sqrt b)(a+\sqrt b)>0\\\iff (a>\sqrt b\land a>-\sqrt b)\lor (a<\sqrt b\land a<-\sqrt b)\\\iff a>\sqrt b\lor a<-\sqrt b.$$
A: When you applied the square root on each side you shouldn't have "cancelled out" the square root with the square power.
I think you are used to do something like this: $x^2=a$ then $x=\pm\sqrt a$, which is fine, but doesn't work with inequalities. Instead you should be thinking it more like this: if $x^2=a$ then $|x|=+\sqrt a$, which also works with inequalities.
For example: if $x^2=4$ usually we say straightforward $x=\pm2$, but you could do a middle step, which is $|x|=2$, and then think that the absolute value just gets rid of the minus sign, so your solutions can be $x=\pm2$.
This is the way you should do the inequality: if $x^2>0$ then $|x|>0$, so you can have either $x>0$ or $x<0$, since the absolute value will get rid of the minus sign.
A: $$x^2=0\iff x=0.$$ Hence
$$x^2>0\iff x\ne0.$$
A: As a general rule, I have discovered the following:
$$\mathrm{For\:}u^n\:>\:0\mathrm{,\:if\:}n\:\mathrm{is\:even}\mathrm{\:then\:}u\:<\:0\quad \mathrm{or}\quad \:u\:>\:0$$
Courtesy of Symbolab. https://www.symbolab.com/solver/inequalities-calculator/x%5E%7B2%7D%3E0
