Solving inequality with modulus How do I go about simplifying 
$\frac{x^2-2x+2^{|a|}}{x^2-a^2}>0$
I have a pretty decent idea about solving general inequalities but I'm stuck on this one. I tried taking $2^{|a|}$ as $t$ and then using log both sides but that got me nowhere. Someone please help.
Where $|a|$ is any Real number and I have to solve the inequality for the values of $x$.
 A: Case $1$: if $a=0$: then $x^2-2x+2^{|a|}=x^2-2x+1=(x-1)^2$
The problem reduces to $\frac{(x-1)^2}{x^2}>0$
Hence $x \neq 0$ and $x \neq 1$.
Case $2$: $a \neq 0$
$x \neq \pm a$,
Since 
$$x^2-2x+2^{|a|}=(x-1)^2+(2^{|a|}-1)>0$$
We just have to make sure that the denominator is positive
$$(x-a)(x+a) >0$$
$$x > |a| \text{ or } x < -|a|$$
A: The numerator doesn't do a whole lot to affect where this function is positive. What happens is that it acts like $\frac{1}{x}$ around points where the denominator goes to zero, and thus on one side of each vertical asymtote the function goes to $+\infty$ while on the other side of each asymtote the function goes to $-\infty$. Accordingly, since $x^2-a^2 = (x+a)(x-a)$ we conclude that the poles occur when $x=\pm a$, and since $\lim_{x \to \infty} f(x) = \infty$ we conclude that the function must be negative for $-a < x < a$ and thus $f$ is positive for $(-\infty,a) \cup (a,\infty)$   
Note: $a=0$ is an exceptional case since we no longer have an interval, and so $x$ is positive everywhere
