Range of $x$ given $|x^2-a|Exercise:


Using signs of inequality alone (not using signs of absolute value) specify the values of $x$ which satisfy the following relations. Discuss all cases.
  $$|x^2-a|<b$$


Solution:
$LHS \geq 0$, so:
$|x^2-a|<b \text{ when } b > 0 \tag{0}$
$-b<x^2-a<b \text{ when } b > 0 \tag{1}$
$a-b<x^2<a+b \text{ when } b > 0 \tag{2}$
$a-b<x^2 \text{ when } b > 0 \tag{2.1}$ $x^2<a+b \text{ when } b > 0 \tag{2.2}$
$0 \leq a-b<x^2 \text{ when } b > 0 \tag{2.1.1}$
$a-b < x^2 \text{ when } 0 < b \leq a \tag{2.1.2}$
$\sqrt{a-b} < |x| \text{ when } 0 <b \leq a \tag{2.1.3}$
$\sqrt{a-b} < x \text{, } x < -\sqrt{a-b} \text{, when } 0 < b \leq a \tag{2.1.4}$
$|x|<\sqrt{a+b} \text{ when } a \geq -b, b > 0 \tag{2.2.1}$
$-\sqrt{a+b}<x<\sqrt{a+b} \text{ when } a \geq -b, b > 0 \tag{2.2.2}$

Answer:
Formulation 1
$$\sqrt{a-b} < x \text{, } x < -\sqrt{a-b} \text{, when } b \leq a$$
$$\text{and}$$
$$-\sqrt{a+b}<x<\sqrt{a+b}$$
$$\text{all when } a > -b, b > 0$$
Formulation 2
$$-\sqrt{a+b}<x<-\sqrt{a-b}, \sqrt{a-b}<x<\sqrt{a+b} \text{, when } b\leq a$$
$$-\sqrt{a+b}<x<\sqrt{a+b} \text{ when } b > a$$
$$\text{all when } a > -b, b > 0$$

Request:
Is my answer correct? If not, where in my solution did I go wrong?
Update: I just noticed (graphically) that $b > 0$ for there to be a range of $x$. Where do I prove (algebraically ) that this is so?
Another Update: Ignore the previous update. Thanks to @JeanMarie comment, I realized that this isn't necessarily so.
Yet Another Update: Ignore the previous update, and pay attention to the first. Thanks to @dxiv comment, I realized that this is necessarily so.
An Update Once Again: I've updated my attempt with the input given my @dxiv and @JeanMarie. How is it now?


$$|x^2-a|<b$$

The LHS side is non-negative, so there are no solutions if $\,b \le 0 \;\;\;(1)\,$.
For $b \gt 0$ the inequality can be rewritten as:
$$
-b \lt x^2 - a \lt b \quad \iff 
\quad \begin{cases}
x^2 \lt a+b \\
x^2 \gt a-b
\end{cases}
$$


*

*$(x^2 \lt a+b)\,$  Since $x^2 \ge 0$ there are no solutions if $a+b \le 0 \;\;\;(2)\,$. 
Otherwise for $a+b \gt 0$ the inequality is equivalent to $-\sqrt{a+b} \lt x \lt \sqrt{a+b}\;\;\;(3)$.

*$(x^2 \gt a-b)$  If $a-b \lt 0$ then the inequality holds for $\forall x\;\;\;(4)\,$.
Otherwise for $a-b \ge 0$ the inequality is equivalent to $x \lt -\sqrt{a-b}\,$ or $\,x \gt \sqrt{a-b}\;\;\;(5)\,$.
To combine $(1)\cdots(5)$ into one final answer, note that $\sqrt{a+b} \ge \sqrt{a-b} \ge 0$ when $a \ge b \ge 0$.


*

*From $(1)+(2)\,$: if $b \le 0$ or $a \le -b$ then there are no solutions i.e. $x \in \emptyset$.

*Otherwise ($b \gt 0$ and $a \gt -b$) if $a \lt b$ from $(3)+(4)$: $x \in (-\sqrt{a+b}, \sqrt{a+b})$.

*Otherwise ($a \ge b \gt 0$) from $(3)+(5)$: $x \in (-\sqrt{a+b}, -\sqrt{a-b}) \cup (\sqrt{a-b}, \sqrt{a+b})$.
