Range of $x$ given $|x^2-a|<b$

Question

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

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Solution:

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

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

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

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

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

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

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

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

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

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

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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

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

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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?

Answer 1

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

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

Answer 2

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.