Solve $(5+2\sqrt{6})^{\frac{x}{2}} + ( 5-2\sqrt{6})^{\frac{x}{2}} = 10$

Question

I wish to solve the equation $$(5+2\sqrt{6})^{\frac{x}{2}} + ( 5-2\sqrt{6})^{\frac{x}{2}} = 10$$

I tried factorizing until I reached $(\sqrt{2}+\sqrt{3})^x + (\sqrt{2}-\sqrt{3})^x = 10$ But from there I don't know what to do any help would be welcome Thanks in advance

Answer 1

Since $\sqrt{3} - \sqrt{2} > 0$, your equation should simplify to

$$ (\sqrt{3} + \sqrt{2})^x + (\sqrt{3} - \sqrt{2})^x = 10 $$

Also note that $$ \sqrt{3} - \sqrt{2} = \frac{1}{\sqrt{3}+\sqrt{2}} $$

Let $t = (\sqrt{3}+\sqrt{2})^x$, then

$$ t + \frac{1}{t} = 10 $$ $$ t^2 - 10t + 1 = 0 $$

which gives $t = 5 \pm 2\sqrt{6}$

Therefore $x = \pm 2$

Answer 2

There are two solutions: $x=2$ and $x=-2$.

We easily see that $x=2$ is a solution. There are no other solutions $x>0$ because the left-hand side is an increasing function on ${\mathbb R}^+$. Indeed, noticing that $1/(5+2\sqrt{6})=5-2\sqrt{6}$, we then find that $$ f(x) = (5+2√6)^{\frac{x}{2}} + ( 5-2√6)^{\frac{x}{2}} $$ is an even function $(f(x)=a^x+a^{-x}=2\cosh(x\log a))$. So we have the second solution $x=-2$, and no other solutions for $x<0$ because $f(x)$ is decreasing on ${\mathbb R}^-$.