3
$\begingroup$

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

$\endgroup$
8
  • 1
    $\begingroup$ 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! $\endgroup$
    – John Doe
    Jan 11, 2018 at 0:09
  • $\begingroup$ I know it's 2 but that's not what I want thanks $\endgroup$ Jan 11, 2018 at 0:11
  • $\begingroup$ I don't understand - what do you want then? $\endgroup$
    – John Doe
    Jan 11, 2018 at 0:13
  • $\begingroup$ Get a formal way of arriving at the answer $\endgroup$ Jan 11, 2018 at 0:14
  • 1
    $\begingroup$ @arsenestein I know it's 2 $x=-2$ is a solution as well. $\endgroup$
    – dxiv
    Jan 11, 2018 at 0:18

2 Answers 2

8
$\begingroup$

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 $x = \pm 2$

$\endgroup$
3
  • $\begingroup$ Okay thanks I've got it $\endgroup$ Jan 11, 2018 at 0:19
  • $\begingroup$ If you're done, make sure to "tick" your favorite answer to close the question $\endgroup$
    – Dylan
    Jan 11, 2018 at 0:21
  • $\begingroup$ 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 $x$ $\endgroup$
    – Will Jagy
    Jan 11, 2018 at 1:08
2
$\begingroup$

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 ${\mathbb R}^-$.

$\endgroup$
6
  • $\begingroup$ OK but isn't there a way of arriving at x=2? $\endgroup$ Jan 11, 2018 at 0:16
  • $\begingroup$ This can be solved without using calculus $\endgroup$
    – Dylan
    Jan 11, 2018 at 0:21
  • $\begingroup$ 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. $\endgroup$
    – dxiv
    Jan 11, 2018 at 1:22
  • $\begingroup$ @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. $\endgroup$
    – dxiv
    Jan 11, 2018 at 2:10
  • $\begingroup$ Thanks @dxiv. :) $\endgroup$
    – Alex
    Jan 11, 2018 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.