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

Question

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! — John Doe, Jan 11 '18 at 00:09
@JohnDoe: Solving an equation requires proving rigorously what exactly are the solutions. Just finding one is nowhere near enough, but sometimes finding one can be an important part of the solution. For example one can use real analysis to solve $x^2 = 2^x$ where $x$ is a positive real. — user21820, Jan 11 '18 at 05:22
@user21820 I agree that that is true, but what can be done is to find solutions and then prove no more solutions exist (as done by Alex's answer). I didn't spot the method of Dylan's answer myself, however. — John Doe, Jan 11 '18 at 21:34
@JohnDoe: That's what I said in my comment, so I don't understand your earlier comment in response to the asker, who obviously doesn't just want the answer but wants a proof. — user21820, Jan 12 '18 at 02:55
Does this answer your question? How to solve $(5-2\sqrt6)^n + (5+2\sqrt6)^n = 98$ — Anne Bauval, Sep 22 '23 at 21:45

score 8 · Accepted Answer · answered Jan 11 '18 at 00:18

8

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 $x = \pm 2$

answered Jan 11 '18 at 00:18

Dylan

16,575

Okay thanks I've got it – arsene stein Jan 11 '18 at 00:19
If you're done, make sure to "tick" your favorite answer to close the question – Dylan Jan 11 '18 at 00:21
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 $x$ – Will Jagy Jan 11 '18 at 01:08

Alex · Answer 2 · 2018-01-11T15:38:21.227

2

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 ${\mathbb R}^-$.

edited Jan 11 '18 at 15:38

answered Jan 11 '18 at 00:12

Alex

4,873

OK but isn't there a way of arriving at x=2? – arsene stein Jan 11 '18 at 00:16
This can be solved without using calculus – Dylan Jan 11 '18 at 00:21
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 '18 at 01:22
@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 '18 at 02:10
Thanks @dxiv. :) – Alex Jan 11 '18 at 02:15
Or just notice that $a^x+a^{-x}=2\cosh\xi$, where $\xi=x\log a$, and use properties of the $\cosh$ function. – Alex Jan 11 '18 at 02:48

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

2 Answers2

Linked