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

Solution 1:

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

Solution 2:

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 $t=\left(\sqrt{5+2\sqrt{6}}\right)^x$......

Solution 3:

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.