Real solutions of $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$

Question

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?

After squaring three times, I guess that you have obtained an $8$th degree equation. Can you write it? — ajotatxe, Feb 20 '17 at 18:45
@ajotatxe: I can write it but is very hard to try to factorize it. — Arnaldo, Feb 20 '17 at 18:50

score 0 · Answer 1 · answered Feb 20 '17 at 18:51

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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?

answered Feb 20 '17 at 18:51

Dr. Sonnhard Graubner

95,283

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Is $x=2$ a solution? – Arnaldo Feb 20 '17 at 18:52
Both of those cubics also give you three (*2) more solutions ... but they are quite messy ... cube roots of root threes. – Donald Splutterwit Feb 20 '17 at 18:54
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No, $x \neq 2$, we get $2 = \sqrt 2$ when we insert $x=2$ in the original equation. – rookie Feb 20 '17 at 18:54
@stud_iisc ... I think you can choose to take the negative root on the second square rooting. – Donald Splutterwit Feb 20 '17 at 18:56
i have squared the equation, so you must make a control if the solutions fulfill also your equation – Dr. Sonnhard Graubner Feb 20 '17 at 18:56
Yes, $x=2$ is a solution if you take $\sqrt4 = -2$ – jonask Feb 20 '17 at 18:57
@DonaldSplutterwit Yes. Now makes sense. Thanks! – rookie Feb 20 '17 at 18:58

Real solutions of $x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$

1 Answers1