Simple question on treatment of negative case in $|x^{2} - 1| < \frac{1}{2}$

Question

Question: Indicate the set of all $x$ satisfying the following condition:

$|x^{2} - 1| < \frac{1}{2}$

I'm having an issue with how to handle a particular case of this question formally. I know from experience and doing these questions in the past that the two sets are: $\bigg(-\sqrt{\frac{3}{2}}, -\sqrt{\frac{1}{2}}\bigg) \cup \bigg(\sqrt{\frac{1}{2}}, \sqrt{\frac{3}{2}}\bigg)$. My issue is with deriving the negative set.

$$|x^{2} - 1| < \frac{1}{2} \\ \Rightarrow \ \frac{1}{2} < x^{2} < \frac{3}{2}$$

Taking the square root I will be left with positive and negative cases for my inequality and this is where I'm having issue. How should I treat the negatives? I was interpreting it like this with regards to one part of the inequality:

$$\frac{1}{2} < x^{2} \\ \pm \sqrt{\frac{1}{2}} < x$$ and concerning myself with the negative case I get:

$$-\sqrt{\frac{1}{2}} < x$$

Should I invert that last line the moment I take the square root? Thus it would be:

$$-\sqrt{\frac{1}{2}} > -x$$

But as you see I just added another negative sign to the right side. Is that what is happening implicitly but we don't express it? Because in this form combined with doing the same thing for the other part of the inequality I would get what I want, but I'm not sure if it is the right procedure to arrive at the conclusion.

Answer 1

Remember that $\sqrt{x^2}=|x|$.

So once you get to $$ \frac{1}{2} \lt x^2 \lt \frac{3}{2}$$ if you take square roots, then you really get is $$\sqrt{\frac{1}{2}}\lt |x| \lt \sqrt{\frac{3}{2}}.$$

So now you have to deal with the absolute value; here the easiest thing is just to consider the two cases of whether $x\geq 0$ or $x\lt 0$.

If $x\geq 0$, then $|x|=x$, so this becomes $$\sqrt{\frac{1}{2}}\lt x \lt \sqrt{\frac{3}{2}}.$$ As this is compatible with $x\geq 0$, this gives you one interval.

If $x\lt 0$, then $|x|=-x$, so this becomes $$\sqrt{\frac{1}{2}}\lt -x \lt \sqrt{\frac{3}{2}},$$ and multiplying through by $-1$ you get $$-\sqrt{\frac{1}{2}}\gt x \gt -\sqrt{\frac{3}{2}}.$$ As this is compatible with $x\lt 0$, this gives you the other interval.

Answer 2

$0 \le M < x^2 < N$ means

$\sqrt M < |x| < \sqrt N$. Which means either $\sqrt M < x < \sqrt N$ or $-\sqrt N < x < -\sqrt N$.

That's all.

$|x^2 - 1| < \frac 12$

$-\frac 12 < x^2 - 1< \frac 12$

$\frac 12 < x^2 < \frac 32$

$\sqrt \frac 12 < |x| < \sqrt {\frac 32}$ so

$-\sqrt{\frac 32} < x < -\sqrt{\frac 12}$ or $\sqrt{\frac 12} < x < \sqrt{\frac 32}$.

Answer 3

For notational convenience, we introduce $a,b>0$.$$a^2<x^2<b^2\iff a<|x|<b.$$

If $x>0$, this is $a<x<b$.

If $x<0$, this is $a<-x<b$, or $-b<x<-a$.

Answer 4

Squaring both sides of the original inequality (since they are both nonnegative) gives $$(x^2-1)^2< 1/4,$$ or $$(x^2-1)^2-(1/2)^2<0.$$ Factorising then gives $$(x^2-1-1/2)(x^2-1+1/2)<0,$$ or $$(x^2-3/2)(x^2-1/2)<0,$$ which, being quadratic in $x^2,$ tells us that $$1/2<x^2<3/2,$$ and all sides being positive, taking square roots gives $$\frac{1}{\sqrt 2}<|x|<\frac{\sqrt 3}{\sqrt 2}.$$

If you want to interpret this further, you may now considering the cases, $x>0$ or $x\le 0,$ and take the union.