Need help in finding bounds on inequality

Question

Determine the condition on $|x-2|$ that will assure that $|x^2-4|<1/2$ will work

So $|x^2-4|=|x+2||x-2$| Do I assume that $|x-2|<1$ or $|x-2|<0.5$ I would like to see full details. Can someone show all the steps thanks

Answer 1

$$|x^2-4|\le\frac12\iff\frac72\le x^2\le\frac92$$ $$\iff -\frac3{\sqrt2}\le x\le-\frac{\sqrt7}{\sqrt2}\textbf{ or } \frac{\sqrt7}{\sqrt2}\le x\le\frac{3} {\sqrt2}$$ $$\iff -\frac3{\sqrt2}-1\le x-1\le-\frac{\sqrt7}{\sqrt2}-1\textbf{ or } \frac{\sqrt7}{\sqrt2}-1\le x-1\le\frac{3} {\sqrt2}-1$$

Answer 2

The factors $|x+2|$ and $|x-2|$ hints at case boundaries $-2$ and $2$.

$$\begin{array}{c|c|c|c|c|c} x&(-\infty,-2)&-2&(-2,2)&2&(2,\infty)\\\hline x+2&-&0&+&+&+\\\hline x-2&-&-&-&0&+\\\hline |x+2||x-2|&(-x-2)(2-x)&0&(x+2)(2-x)&0&\cdots \end{array}$$

For each of the cases of $x$, solve the inequality.

E.g. for $x\in(-\infty,-2)$,

$$\begin{align*} |x+2||x-2|&<\frac12\\ (-x-2)(2-x) &<\frac12\\ x^2 - 4 &< \frac12\\ x &\in \left(-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) \end{align*}$$

Combining with the case condition that $x\in(-\infty,-2)$, these $x$ satisfies the inequality:

$$x\in(-\infty,-2)\cap \left(-\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) = \left(-\frac{3}{\sqrt{2}}, -2\right)$$

The remaining 2 cases are similar.