Solving the inequality ${x^2+2x+2^{|a|}\over x^2-a^2} > 0$ How do I solve the inequality ${x^2+2x+2^{|a|}\over x^2-a^2} > 0$?  
My idea is that for $a\ne 1$, the numerator will always be positive. So the inequality reduces to $
{1 \over x^2-a^2 } > 0
$
My doubt is in this part. If we factorise the denominator, we get
$
{1 \over (x+a)(x-a)}
$
which, by using the wavy curve method, gives me,
$
x \epsilon (-\infty,-a) \  \ \ \cup \ \ \ (a,\infty).
$
But according to my textbook, the answer with $a\ne1$ is 
$
x \epsilon (-\infty,-|a|) \  \ \ \cup \ \ \ (|a|,\infty).
$
Why?

The problem is that in your solution you can have the $2$ intervals ovelapping, i.e. $-a > a$ which holds for all $a<0$
Hence in order for the intervals to not overlap, you have to use the modulus
