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Assuming convergence of the following series, find the value of $\sqrt{6+\sqrt{6+\sqrt{6+...}}}$

I was advised to proceed with this problem through substitution but that does not seem to help unless I am substituting the wrong parts. If i substitute the $6$, well then i am just stuck with above.

Any ideas on how to proceed. Also, what is the purpose of stating that it is convergent.

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marked as duplicate by MJD, Claude Leibovici, Jyrki Lahtonen Oct 7 '15 at 6:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ @GAVD I think this question quite is different from that one since it involves no unknown, just a simple "evaluate this expression...". $\endgroup$ – BigbearZzz Oct 7 '15 at 3:42
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Let $x=\sqrt{6+\sqrt{6+\sqrt{6+...}}}$, then observe that $x=\sqrt{6+x}$. Squaring both sides yields $$ x^2=x+6 $$ , which is a quadratic formula. Solve it normally and choose the wise answer out of the 2 roots.

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  • $\begingroup$ I see, so I would get -2 and 3. I would have to eliminate -2 since we cant have a negative number within the square root. $\endgroup$ – Caddy Heron Oct 7 '15 at 3:43
  • $\begingroup$ Exactly. I have a question that you might be interested in: What if BOTH roots are positive? Which one will you choose? :) $\endgroup$ – BigbearZzz Oct 7 '15 at 3:51
  • $\begingroup$ Interesting, I would say pick none but since its convergent, i would have to pick one? So i'm not sure. $\endgroup$ – Caddy Heron Oct 7 '15 at 3:54

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