How do I go about simplifying


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 $x$.

  • 1
    $\begingroup$ 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

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 $$(x-a)(x+a) >0$$

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

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.