How do I assume correct values for x in absolute value function?

Question

Given, $|\frac{1}{x}-2|<4$, I can solve this via the theorem approach $|x-a|<b\Rightarrow-b<x-a<b$..... but in the above question, there comes a possibility in $-4<\frac{1}{x}-2$ where the solution for it is less than $-\frac{1}{2}$ but it can change if I assume $\frac{1}{x}$ to be a negative value. How do I decide in these question of structure $x-a$ about value of $x$?

Answer 1

First, note that $x \neq 0$. You can avoid casework by squaring both sides, then solving the resulting quadratic. \begin{align*} \left|\frac{1}{x} - 2\right| & < 4\\ \left|\frac{1 - 2x}{x}\right| & < 4\\ \left(\frac{1 - 2x}{x}\right)^2 & < 16\\ \frac{1 - 4x + 4x^2}{x^2} & < 16\\ 1 - 4x + 4x^2 & < 16x^2\\ 0 & < 12x^2 + 4x - 1\\ 0 & < (6x - 1)(2x + 1) \end{align*} The inequality is true when both factors have the same sign. Both factors are positive if $x > 1/6$. Both factors are negative if $x < -1/2$. Hence, the solution set is $$\left(-\infty, -\frac{1}{2}\right) \cup \left(\frac{1}{6}, \infty\right)$$

Answer 2

You have $-2<{1\over x} <6$ thus:

If $x>0$ we have $-2x<1<6x$ so $x>1/6$.

If $x<0$ we have $-2x>1>6x$ so $x<-1/2$.

So $x\in (-\infty, -{1\over 2})\cup ({1\over 6},\infty)$.