Solving the inequality ${x^2+2x+2^{|a|}\over x^2-a^2} > 0$

Question

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?

Answer 1

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

Answer 2

Your textbook is right.

For instance, take $a=-1$. Your solution would be $(-\infty,1)\cup(-1,\infty)=\mathbb R$, which is wrong.

In fact, the solution must be in terms of $|a|$, as the sign of $a$ plays no role.