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

I square the both sides and get $(5+2\sqrt{6})^{x}+(5-2\sqrt{6})^{x}=98$. But I don't know how to carry on. Please help. Thank you.

Let $t_1=(\sqrt{5+2\sqrt6})^x$, and $t_2=(\sqrt{5-2\sqrt6})^x$. 
Now the given equation is:
$$\tag 1 t_1+t_2=10$$ 
But
$$\displaylines{
  {t_1}\cdot{t_2}&=& {\left( {\sqrt {5 + 2\sqrt 6 } } \right)^x}{\left( {\sqrt {5 - 2\sqrt 6 } } \right)^x} \cr 
   &=& {\left[ {\sqrt {\left( {5 + 2\sqrt 6 } \right)\left( {5 - 2\sqrt 6 } \right)} } \right]^x} \cr 
   &=& {\left[ {\sqrt {{5^2} - {{\left( {2\sqrt 6 } \right)}^2}} } \right]^x} \cr 
   &=& {\left( {\sqrt {25 - 24} } \right)^x} \cr &=& {1^x} = 1 \cr} $$
Thus
$$\begin{cases} t_1+t_2=10 \\t_1\cdot t_2=1\end{cases}$$
$\Rightarrow$ $t^2-10t+1=0$
Since $$\frac{1}{{{t_1}}} = {t_2}$$
we
have $${t_1} + \frac{1}{{{t_1}}} = 10$$ or
$$t_1^2 - 10{t_1} + 1 = 0$$
Note the system is symmetric on the unknowns.
For this quadratic equation we have:
$t_{1,2}=\frac{10\pm\sqrt{100-4}}{2}$
$t_{1,2}=\frac{10\pm 4\sqrt{6}}{2}$
$t_1=5+2\sqrt 6$, $t_2=5-2\sqrt 6$
Now return the inital substition:
$t_1=(\sqrt{5+2\sqrt6})^x$
$5+2\sqrt 6=(\sqrt{5+2\sqrt6})^x$ 
$(\sqrt{5+2\sqrt 6})^2=(\sqrt{5+2\sqrt6})^x$ $\Rightarrow$ $x=2$, and 
$(\sqrt{5+2\sqrt6})^x=5-2\sqrt 6$
$(\sqrt{5+2\sqrt6})^x=(5+2\sqrt 6)^{-2}$ $\Rightarrow$ $x=-2$
Definitly $x=2$, and $x=-2$ is solve.
