algebra precalculus - exponential equation

$$\sqrt{(5+2\sqrt6)^x}+\sqrt{(5-2\sqrt6)^x}=10$$

So I have squared both sides and got:

$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2\sqrt{1^x}=100$$

$$(5-2\sqrt6)^x+(5+2\sqrt6)^x+2=100$$

I don't know what to do now

Answer

You don't have to square the equation in the first place.

Let $y = \sqrt{(5+2\sqrt{6})^x}$, then $\frac{1}{y} = \sqrt{(5-2\sqrt{6})^x}$. Hence you have $y + \frac{1}{y} = 10$ i.e. $y^2 + 1 = 10y$ i.e. $y^2-10y+1 = 0$.

Hence, $(y-5)^2 =24 \Rightarrow y = 5 \pm 2 \sqrt{6}$.

Hence, $$\sqrt{(5+2\sqrt{6})^x} = 5 \pm 2\sqrt{6} \Rightarrow x = \pm 2$$

(If you plug in $x = \pm 2$, you will get $5+2\sqrt{6} + 5-2\sqrt{6} $ which is nothing but $10$)