Rules of inequality and modulus inequality. I am very confused whenever I encounter an inequality .
Like suppose 
$ \frac{| \sqrt {1 -4x^{2}}|}{|x|} \le 1 $ 
Now I am confused whether to square or not .
 I am very well aware of the rule at we do not square if the sign of $x$ is not known . But what if the range of $x$ is given as 
$ \frac {-1}{2} \le x \le \frac 12 $ .
Now here I am confused whether to square or not as $ x $ can take either negative value. 
If someone could give me a link to a set of rules of inequalities under such conditions , it would be appreciated .


Note:  All of what follows is based on the axiom that if $a < b$ (regardless of the signs of either) and $c > 0$ then $ac < bc$ and likewise if $d < 0$ then $ad > bd$.  
I'm assuming that is known and familiar.

You can do whatever you want. You can take them out to dinner for all I care.  You just have to know what you are doing.
If you have $x < 3$ and $x$ is negative, say, then you can not say $x^2 < 9$ because:  If we multiple both sides of the equation $x < 3$ by $x$ we flip the inequality to get $x^2 > 3x$ and if we multiply both sides of $x < 3$ by $3$ we get $3x < 9$.  So we get $x^2 > 3x$ and $9> 3x$ but we have no way of comparing $x^2$ to $9$.
That's not wrong.  It's just not what we'd expect if we were stupid and just monkeying around with rules we didn't understand and "squaring both sides" without thinking.
Anyway.... If we do know the signs say $0 < x < 3$ then we can square both sides because $x< 3$ and $x > 0$ means $x^2 < 3x$ and $x < 3$ and $3>0$ so $3x < 9$.  So $0 < x^2 < 9$.
We can actually do if $-a < x < b$ where $-a < 0$ and $b < 0$ we can get.
If $x < 0$ then $-a < x \implies -ax > x^2; (-a)^2 = a^2 > -ax$ so $x^2 < a^2$.  But if $x > 0$ we have $-ax < x^2$ and $(-a)^2 =a^2 > -ax$ and we have no way of compare $a^2$ to $x^2$.
And if $x < 0$ we get $x < b$ means $x^2 > bx$ and $bx < b^2$ but we have no way to compare $x^2$ to $b^2$.  But if $x > 0$ we have $x^2 < b^2$.
So we have either $x^2 < a^2$ or $x^2 < b^2$ so we do have $x^2 < \max(a^2,b^2)$.
And note.  If $a = b$ and you have $-a < x < a$ we have $0 \le x^2 < a^2$.
.....
But to your problem.....
Well, yours is simple.  $|K| \ge 0$ so you have
$0 \le \frac {|1-4x^2|}{|x|} \le 1$
So we know the signs and we can square both sides.
$0 \le \frac {1-4x^2}{x^2} \le 1$ so
$0 \le 1-4x^2 \le x^2$ so
$4x^2 \le 1 \le 5x^2$ so
$0 \le 4x^2 \le 1 \le 5x^2$ so
$0 \le 2|x| \le 1 \le \sqrt 5|x|$.  So $|x| \le \frac 12$ and $\frac 1{\sqrt 5} \le |x|$ so $\frac 1{\sqrt {5} }\le |x| \le \frac 12$.
We don't know the sign of $x$ so
Either $-\frac 12 \le x \le -\frac 1{\sqrt 5}$ or $\frac 1{\sqrt {5} }\le x \le \frac 12$
