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

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

• Isn't it quite clear what the answer should be? Try some small natural numbers in the initial equation you are given. This is a good lesson in inspecting the problem before you run into a bunch of algebra! Jan 11, 2018 at 0:09
• I know it's 2 but that's not what I want thanks Jan 11, 2018 at 0:11
• I don't understand - what do you want then? Jan 11, 2018 at 0:13
• Get a formal way of arriving at the answer Jan 11, 2018 at 0:14
• @arsenestein I know it's 2 $x=-2$ is a solution as well.
– dxiv
Jan 11, 2018 at 0:18

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 $x = \pm 2$

• Okay thanks I've got it Jan 11, 2018 at 0:19
• If you're done, make sure to "tick" your favorite answer to close the question Jan 11, 2018 at 0:21
• your $\lambda^2 - 10 \lambda + 1 = 0$ also appears as a degree two linear recurrence for the values, once for even $x,$ once for odd $x$ Jan 11, 2018 at 1:08

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 ${\mathbb R}^-$.

• OK but isn't there a way of arriving at x=2? Jan 11, 2018 at 0:16
• This can be solved without using calculus Jan 11, 2018 at 0:21
• left-hand side is an increasing function on R+ That could be elaborated some more, since it's not trivially obvious. The LHS is the sum of two terms with opposite monotonicity, $(5+2 \sqrt{6})^x$ is increasing while $(5-2 \sqrt{6})^x$ is decreasing.
– dxiv
Jan 11, 2018 at 1:22
• @Alex Using calculus, you could write $f(x) = a^x + a^{-x}$ then argue that $f'(x) = (a^x-a^{-x}) \log(a)$ is strictly monotonic and has a zero at $x=0$, so it must keep the same sign on $\mathbb{R}^+$ and $\mathbb{R}^-$ respectively.
– dxiv
Jan 11, 2018 at 2:10
• Thanks @dxiv. :)
– Alex
Jan 11, 2018 at 2:15