Basic question $|x^2| < 9$

Question

I have a rather basic question.

Let's assume that $|x^2| < 9$, where $x\in \mathbb{R}$. Then everyone knows that $x \in$ (-3,3). However, I have trouble arriving at the answer based on basic operations.

$$|x^2| < 9$$

$$ -9 < x^2 < 9$$

Since $x^2$ cannot be negative:

$$ 0 \leq x^2 < 9$$

$ 0 \leq x < 3$ when x is positive or zero, OR $-3 < x \leq 0$ when x is negative or zero.

It means $-3 < x < 3$.

Am my reasoning correct?

Answer 1

What you did is correct, but here is another way of solving it:

(1) $$|x^2| = x^2$$

(2) If $$x^2 < A^2$$ (of course, under the assumption that $A > 0$), then $$x^2 - A = (x + A)(x - A) < 0.$$ It follows that $$\left(x + A > 0\right) \land \left(x - A < 0\right),$$ or $$\left(x + A < 0\right) \land \left(x - A > 0\right).$$ Since $A^2 > x^2 \geq 0$, $x + A < 0$ is false. Consequently, $$\left(x > -A\right) \land \left(x < A\right)$$ which can be abbreviated as $$-A < x < A.$$