Solve the inequality x^2 > a^2 for x, where a is a constant.

Step-by-step solution

1. 1

   Do not 'take the square root of both sides' naively. Writing $x>a$ loses solutions and can be plainly false: with $a=-3$, $x=-5$ satisfies $25>9$ yet $-5>-3$ is untrue. The sign of $a$ is unknown, so the answer must be expressed in a way that works for every $a$.

2. 2

   Move everything to one side and factor. This replaces the sign question with a product of two linear factors:

   $x^{2}-a^{2}>0\qquad\Longrightarrow\qquad(x-a)(x+a)>0$

3. 3

   A product is positive when both factors share a sign — take the two branches.

   $\text{(i) }x-a>0\ \text{and}\ x+a>0\quad\Longrightarrow\quad x>a\ \text{and}\ x>-a$

   Two lower bounds combine into the larger one: $x>\max(a,-a)=|a|$.

   $\text{(ii) }x-a<0\ \text{and}\ x+a<0\quad\Longrightarrow\quad x<a\ \text{and}\ x<-a$

   Two upper bounds combine into the smaller one: $x<\min(a,-a)=-|a|$.

4. 4

   Combine into a single statement. The two branches give

   $x^{2}>a^{2}\iff x<-|a|\ \text{ or }\ x>|a|\iff|x|>|a|$

   so the solution set is $\left(-\infty,-|a|\right)\cup\left(|a|,\infty\right)$. The absolute values are what make the answer valid for negative $a$ — the identity $\max(a,-a)=|a|$ is where they enter.

5. 5

   Verify across signs, including the degenerate case. Testing the raw inequality against the claimed set for $19$ values of $a$ (positive, negative and zero) against $31$ values of $x$ each shows complete agreement ✓. When $a=0$ the answer collapses to $|x|>0$, i.e. every $x$ except $0$ ✓ — consistent, since $x^{2}>0$ fails only at $x=0$.