Sign system in inequalities

Question

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 .

Answer 1

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 $u\ge 0$.

Answer 2

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 $$ x \gt a \\ {\large OR}\\ x \lt -a $$

Answer 3

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 $x>1$.