Roots of a quadratic equation that is obtained from nested radicals Take any $ 0 \lt k \in \mathbb R$, and "algebraically massage" it into a nested radical:
$$k=\sqrt{k^2}=\sqrt{k^2+k-k}=\sqrt{k+(k-1)k}=\sqrt{k+(k-1)\sqrt{k^2}}= \ .\ .\ .\ = \sqrt{k+(k-1)\sqrt{k+(k-1)\sqrt{k+(k-1)\sqrt{.\ .\ .}}}}$$
Let's pretend we don't know what $k$ is, in the sense that we haven't seen the above derivation, and we're presented with the expression $\sqrt{k+(k-1)\sqrt{k+(k-1)\sqrt{.\ .\ .}}}$, and asked to find its value. 
We can set the expression equal to some $x$, and plug $x$ inside to obtain a quadratic equation:
$$x=\sqrt{k+(k-1)x} \quad \leftrightarrow \quad x^2-x(k-1)-k=0 $$
whose roots are  $\{k,-1\}$.
It does seem somewhat frivolous to say that $k$ is a root of this quadratic, but again we are pretending to not have known the value of this expression to begin with...
My question: is there any nontrivial reason for the appearance of $-1$?
We obviously can't do to $-1$ what we've done to $k$, so it would seem that the sequence $(a_n)_{n=1}^\infty=\underbrace{\sqrt{k+(k-1)\sqrt{k+(k-1)\sqrt{.\ .\ .}}}}_{\text{n times}}$ does not converge, since if we take $(a_n)_{n=1}^\infty$ as a subsequence of itself twice, we get 2 subsequences that converge to different limits, clearly an absurdity.
Perhaps I'm confused and this is nonsense, but I would appreciate both a clarification of what it is that I'm missing, and a reference to some books or articles which are concerned with the theory of nested, periodic, infinite etc. radicals. Thank you in advance.
 A: Having thought about this question some more, and having spoken to a few people with deeper knowledge and experience than mine, I've come to the following conclusions:


*

*I'm going to have to google my way to books and articles on nested and infinite radicals in general.

*There seem to be no deep facts behind the appearance of $-1$ here.
When one is presented with such an expression,$-1$ is a legitimate answer to the relation presented by our quadratic - and obviously, given that our expression converges - it cannot equal both $-1$ and $k$ at the same time, which means my rather inane mumble about divergence of the sequence was the result of pure confusion.
Now, having taken $0 \lt k$, it's quite obvious what the actual value of this expression is going to be, after applying the above strategy. One thing I didn't think of is numerical verification, and truncating such an expression - both before or after the $+$ sign -  does provide a good sense of the value that our expression will approach.
