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


$LHS \geq 0$, so: $|x^2-a|<b \text{ when } b > 0 \tag{0}$

$-b<x^2-a<b \text{ when } b > 0 \tag{1}$

$a-b<x^2<a+b \text{ when } b > 0 \tag{2}$

$a-b<x^2 \text{ when } b > 0 \tag{2.1}$ $x^2<a+b \text{ when } b > 0 \tag{2.2}$

$0 \leq a-b<x^2 \text{ when } b > 0 \tag{2.1.1}$

$a-b < x^2 \text{ when } 0 < b \leq a \tag{2.1.2}$

$\sqrt{a-b} < |x| \text{ when } 0 <b \leq a \tag{2.1.3}$

$\sqrt{a-b} < x \text{, } x < -\sqrt{a-b} \text{, when } 0 < b \leq a \tag{2.1.4}$

$|x|<\sqrt{a+b} \text{ when } a \geq -b, b > 0 \tag{2.2.1}$

$-\sqrt{a+b}<x<\sqrt{a+b} \text{ when } a \geq -b, b > 0 \tag{2.2.2}$


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

$$-\sqrt{a+b}<x<-\sqrt{a-b}, \sqrt{a-b}<x<\sqrt{a+b} \text{, when } b\leq a$$ $$-\sqrt{a+b}<x<\sqrt{a+b} \text{ when } b > a$$ $$\text{all when } a > -b, b > 0$$


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?

  • $\begingroup$ You have to consider that all expressions that you introduce make sense. For example, you can use $\sqrt{a+b}$ iff $a+b \geq 0$. $\endgroup$ – Jean Marie Dec 14 '16 at 0:21
  • $\begingroup$ @JeanMarie -- Oh, yes. I didn't notice that. If I put that into consideration, my requirement that $b > 0$ will be covered automagically, correct? $\endgroup$ – Fine Man Dec 14 '16 at 0:24
  • $\begingroup$ If $b \le 0$ there are no solutions since $|x| \ge 0$ for $\forall x$ so $|x^2-a| \lt b \le 0$ is impossible. $\endgroup$ – dxiv Dec 14 '16 at 0:27
  • $\begingroup$ @dxiv -- So, this is a proof has no relation to what JeanMarie mentioned, correct? $\endgroup$ – Fine Man Dec 14 '16 at 0:31
  • $\begingroup$ It is a necessary condition for any solutions to exist, and it is independent of the other conditions on $a$. $\endgroup$ – dxiv Dec 14 '16 at 0:33


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 $(3)+(5)$: $x \in (-\sqrt{a+b}, -\sqrt{a-b}) \cup (\sqrt{a-b}, \sqrt{a+b})$.

  • $\begingroup$ Thank! This is a fairly elegant and complete solution. Every time I try solving this type of problem, I end up with an incomplete mess. Any suggestions on keeping solutions complete and concise? $\endgroup$ – Fine Man Dec 14 '16 at 4:51
  • $\begingroup$ @SirJony Thanks. That's the kind of messy problem which invites messy solutions (as often happens in real world problems, unfortunately). Don't know there is a universal recipe, but it generally helps to eliminate the most obvious cases, first, so as to leave fewer subcases to consider in full detail. And, of course, keep good tabs on the progress and assumptions at each step. $\endgroup$ – dxiv Dec 14 '16 at 4:56
  • 1
    $\begingroup$ I meant "thanks", not you to thank me. Oops. Sorry. :) I guess I was hoping for a universal recipe (as if Wolfram|Alpha were to solve it), but I'll have to stay satisfied with your tips and tricks. Again, thanks! $\endgroup$ – Fine Man Dec 14 '16 at 5:03

I advise you to have a graphical view of the situation as depicted below with $f(x)=|x^2-a|$ (green curve) and $g(x)=b$ (black horizontal line), looking for values of $x$ such that

$$|x^2-a|<b \ \ \ \ \iff \ \ \ \ f(x)<g(x).$$

(graphics obtained with Geogebra, with sliders for values of $a$ and $b$).

This figure allows to consider the different cases according to resp. values of $a$ and $b$, by "sweeping" the horizontal line along the curve, and looking for cases where the line is above the curve.

For example, for the displayed case ($b=3 \leq a=4$), the line is above the curve for $-\sqrt{7}<x<-1$ and for $1<x<\sqrt{7}.$ (case of two disjoint validity intervals), with $\sqrt{7}=\sqrt{a+b}$ and $1=\sqrt{a-b}$.

Were $b>a$, it is visible that there would be a single validity interval.

And of course, if $b<0$, no solution exist.

enter image description here

  • $\begingroup$ Is my graphical presentation understandable ? $\endgroup$ – Jean Marie Dec 14 '16 at 7:41

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