Sum of absolute values of solutions of $(2 + \sqrt 3)^x + (2 - \sqrt 3)^x=4$

Question

What is the sum of the absolute values of all the $x$-s that satisfy the equation:$$(2 + \sqrt 3)^x + (2 - \sqrt 3)^x=4$$

Answer 1

Hint: $2-\sqrt{3}= \frac{1}{2+\sqrt{3}}$.

Answer 2

Let $f(x)=(2 + \sqrt 3)^x + (2 - \sqrt 3)^x-4$. Now show that $f$ decreases for negative $x$ and increases for positive $x$.

Obviously $1$ is a zero of $f$. Then from the symmetry of $f$, namely $f(-x)=f(x)$, it follows that $-1$ is the only other zero of $f$, so their absolute values sum up to $2$.

Answer 3

Let $(2+\sqrt3)^x=t$.

Hence, we need $t+\frac{1}{t}=4$, which gives $t=2+\sqrt3$ or $t=2-\sqrt3$ and we get the answer: $$\{1,-1\}.$$