Sign system in inequalities If it is the case that:
$$x^2-a^2 > 0$$
then $x < -a$ or $x > a$.
But if we were to solve the following inequality:
$$x^2 > a^2$$
then we should get $x > +a$ or $x > -a$, right?
How is that possible? Please correct me if I am wrong .
 A: I don't think so since the inequalities $x^2>a^2$ and $x^2-a^2>0$ are actually the same. Also $$x^2>a^2\iff |x|^2>|a|^2\iff |x|>|a|\iff x>a\text{ or }x<-a$$since the function $f(u)=u^2$ is strictly increasing for $u\ge 0$.
A: Considering $x, a \in \mathbb R$ and if it is given that $$x^2 - a^2 \gt 0$$ than we can add $-a^2$ to both sides and get
$$x^2 \gt a^2$$ Since, $f : x \mapsto \sqrt x$ is a monotonically increasing function, therefore taking square root on both sides of the inequality won’t invert the inequality sign, so we get
$$
|x| \gt |a|
$$
The reason we have got absolute values is that taking square root of both sides is like applying a function to both sides, and a square root function always gives positive output hence absolute sign is necessary.
So, writing things out will give us
$$ x \gt a \\
{\large OR}\\
x \lt -a 
$$
A: Solving the inequality $x^2>a^{2}$ gives  $x>a$ or $x<-a$ (not $x>-a$). For example assuming $a>0$, if $x=\frac{-a}{2}>-a$ then $x^{2}=(\frac{-a}{2})^{2}=\frac{a^{2}}{4}<a^{2}$.
We have $x^2-a^2=(x-a)(x+a)$ and visually for say $a=1$: 
Then clearly $x^2>1$ when $x<-1$ or $x>1$.
