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.
 A: Nice solution, just for clarification
$\left(5+2\sqrt{6}\right)^{\frac{x}{2}}\cdot \left(5-2\sqrt{6}\right)^{\frac{x}{2}}=\left[\left(5+2\sqrt{6}\right)\cdot \left(5-2\sqrt{6}\right)\right]^{\frac{x}{2}}=\left[5^2-\left(2\sqrt{6}\right)^2\right]^{\frac{x}{2}}=\left(25-24\right)^{\frac{x}{2}}=1$
and so for $t=\left(\sqrt{5+2\sqrt{6}}\right)^x$......
A: Hint $\ $ Put $\rm\ b = 5 + 2\sqrt{6},\,\ a = b^{\,x/2}\ $ in
$$\rm a+ a^{-1} =\, b + b^{-1}\, \Rightarrow\ \{a,a^{-1}\} = \{b,b^{-1}\}\ \ since\ \ (x-a)(x-a^{-1})\, =\, (x-b)(x-b^{-1})$$
Note $\, $ Generally every pair of numbers in a ring is uniquely determined by their sum and product iff the ring is a domain. Above is the special case of this uniqueness result where the product $= 1.\:$  As I often emphasize, uniqueness theorems provide powerful tools for proving equalities.
A: Let $t=( \sqrt{5 + 2\sqrt{6}})^{x}\implies (\sqrt{5-2\sqrt{6}})^{x}=\frac{1}{t} $
Thus the equation becomes,
$ t+\frac{1}{t}= 10\implies t^2-10t+1=0$ which is a quadratic equation and have roots $t=5+2\sqrt 6,5-2\sqrt 6$
If $t=5+2\sqrt 6\implies ( \sqrt{5 + 2\sqrt{6}})^{x}=5+2\sqrt 6\implies (5+2\sqrt 6)^{1-x/2}=1\implies 1-x/2=0\implies x=2$
If $t=5-2\sqrt 6 \implies (\sqrt{5+2\sqrt{6}})^{x}=5-2\sqrt 6=\frac{1}{5+2\sqrt 6}\implies (5+2\sqrt 6)^{1+x/2}=1\implies x=-2$
Thus, $x=2,-2$ are the two solutions.
Verification:
Putting $x=2$ in the original equation gives L.H.S=$5+2\sqrt 6+5-2\sqrt 6=10=$R.H.S
Similarly, putting $x=-2$ also gives L.H.S=$5-2\sqrt 6+5+2\sqrt 6=10=$R.H.S
A: 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.
