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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 $x$.

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    $\begingroup$ what is $a$? I assume $a \in \mathbb{R}$, but in that case by symmetry we can just let $a \in [0,\infty)$ and drop the absolute value sign. Also, what is the domain on $x$? $\endgroup$ – Brevan Ellefsen Jul 28 '17 at 6:41
  • $\begingroup$ Sorry for not being crystal clear, I have edited the question now. Please give it a shot now. $\endgroup$ – Tanuj Jul 28 '17 at 6:43
  • $\begingroup$ @MONNET the sign would depend on $-2x$ how do I know if it's positive or negative? $\endgroup$ – Tanuj Jul 28 '17 at 6:46
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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 $$(x-a)(x+a) >0$$

$$x > |a| \text{ or } x < -|a|$$

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  • $\begingroup$ How did you simplify the term in the second case? Sorry if it's too easy and I should get it already but I'm not, can you please simplify so I can understand? $\endgroup$ – Tanuj Jul 28 '17 at 6:52
  • $\begingroup$ which part are you referring to? $\endgroup$ – Siong Thye Goh Jul 28 '17 at 6:52
  • $\begingroup$ Like in case 2 where a is not equal to 0, how did you simplify the expression $\endgroup$ – Tanuj Jul 28 '17 at 6:54
  • $\begingroup$ The numerator is positive, I can always divide them away and focus and the denominator. $\endgroup$ – Siong Thye Goh Jul 28 '17 at 6:56
  • $\begingroup$ Very true, that's where my concern lies. Why is the numerator always positive? Won't $-2x$ affect the sign of the numerator? $\endgroup$ – Tanuj Jul 28 '17 at 6:57
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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

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