Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $(7+4\sqrt 3)^{x^2-8}+(7-4\sqrt 3)^{x^2-8}=14,\quad x=?$

How do I solve this algebra question?

share|cite|improve this question
up vote 4 down vote accepted

Let $a=(7+4\sqrt{3})^{x^2-8}$ then $\dfrac{1}{a}=(7-4\sqrt{3})^{x^2-8}$

So the given equation will be $a+\dfrac{1}{a}=14\implies a^2-14a+1=0$

Solving quadratic for $a=7+4\sqrt{3}$ or $a=7-4\sqrt{3}\implies x^2-8=1$ or $-1$

therefore possible values of $x=3,-3,\sqrt{7},-\sqrt{7}$

share|cite|improve this answer

Hint: if $a$ be the first term on LHS, then the second term is $\frac{1}{a}$.

share|cite|improve this answer

Note that $$(7+4 \sqrt{3}) \cdot (7-4\sqrt{3}) = 49-48 =1$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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