1
$\begingroup$

Find all real solutions of the equation:

$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$

My approach:

One idea was square $3$ times untill get a equation and try factoring it. Another was try to get a system, calling, for example, $y=\sqrt{2+x}$, but I didn't have luck with any of them.

Any idea?

  • 0
    After squaring three times, I guess that you have obtained an $8$th degree equation. Can you write it? – 2017-02-20
  • 0
    $x \in [-2, 2]$, actually $x \in [\sqrt 2, 2]$ – 2017-02-20
  • 0
    @stud_iisc Right – 2017-02-20
  • 0
    And $x>\sqrt 2$ – 2017-02-20
  • 0
    @ajotatxe: I can write it but is very hard to try to factorize it. – 2017-02-20

1 Answers 1

0

after squaring three times and factorizing we get $$- \left( x-2 \right) \left( x+1 \right) \left( {x}^{3}-3\,x+1 \right) \left( {x}^{3}+{x}^{2}-2\,x-1 \right) =0$$ can you go further from here?

  • 1
    Is $x=2$ a solution? – 2017-02-20
  • 0
    Both of those cubics also give you three (*2) more solutions ... but they are quite messy ... cube roots of root threes. – 2017-02-20
  • 1
    No, $x \neq 2$, we get $2 = \sqrt 2$ when we insert $x=2$ in the original equation. – 2017-02-20
  • 0
    @stud_iisc ... I think you can choose to take the negative root on the second square rooting. – 2017-02-20
  • 0
    i have squared the equation, so you must make a control if the solutions fulfill also your equation – 2017-02-20
  • 0
    Yes, $x=2$ is a solution if you take $\sqrt4 = -2$ – 2017-02-20
  • 0
    @DonaldSplutterwit Yes. Now makes sense. Thanks! – 2017-02-20