Continued fraction of the golden ratio It is known, that the continued fraction of $\phi = \frac{1+\sqrt{5}}{2}$ is $[\bar{1}]$. This can be shown via the equation $x^2-x-1=0$:
$$ x^2-x-1=0 \Rightarrow x = 1+\frac{1}{x} = 1+ \frac{1}{1+\frac{1}{x}} = \cdots $$
As far as I can see, the only thing that has been used here is that $\phi$ is a root of the polynomial $x^2-x-1$. My question:
This polynomial has 2 roots. Why do we get the continued fraction of $\phi$ and not those of the other root? What has to be done to get the continued fraction of the other root with this method?

The golden ratio
$$
x : 1 = 1 + x : x
$$
leads to the equation
$$
x^2 - x - 1 = 0 \quad (\#)
$$
It can be transformed to two different equations of the form
$$
x = F(x)
$$
which then can be used to substitute the $x$ on the right hand side by $F(x)$
$$
x = F(F(x)) = F(F(F(x))) = \cdots
$$
Your transformed version of $(\#)$ was this equation:
$$
x = 1 + \frac{1}{x} \quad (*)
$$ 
The repeated substitution of the RHS $x$ in $(*)$ with the RHS term 
$$
x \mapsto 1 + \frac{1}{x}
$$
results in the expansion
\begin{align}
x 
&= 1 + \frac{1}{x} \quad (\mbox{EX}1.1) \\
&= 1 + \frac{1}{1 + \frac{1}{x}} \quad (\mbox{EX}1.2) \\
&= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{x}}} \quad (\mbox{EX}1.3) \\
&= \cdots
\end{align}
and leads to the continued fraction 
$$
x = 1 + \frac{1\rvert}{\lvert 1} + \frac{1\rvert}{\lvert1} + \cdots \quad (\mbox{CF}1)
$$
This continued fraction is positive (consisting only of additions and divisions of positive numbers) and will converge to the positive solution of $(\#)$.
Note that each of the steps $(\mbox{EX}1.n)$ is an equation equivalent
to equation $(\#)$, having two solutions for $x$, while $(\mbox{CF}1)$
converges only to the positive root.
This is because the continued fraction is the limit of these finite fractions:
\begin{align}
c_0 &= 1 \\
c_1 &= 1 + \frac{1}{1} = 2 \\
c_2 &= 1 + \frac{1}{1 + \frac{1}{1}} = 1.5 \\
c_3 &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}} = 1.\overline{6} \\
c_4 &= 1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1}}}} = 1.6 \\
\end{align}
Note that compared to the expansions $(\mbox{EX}1.n)$ the right hand side $1/x$ is dropped, which makes the difference.
To get the other root, the golden ratio equation $(\#)$
needs to be transformed into
$$
x = \frac{1}{-1 + x} \quad (**)
$$
Repeated substitution of the RHS $x$ in $(**)$ with the RHS term
$$
x \mapsto \frac{1}{-1 + x}
$$
results in the expansion
\begin{align}
x 
&= 0 + \frac{1}{-1 + x} \quad (\mbox{EX}2.1) \\
&= 0 + \frac{1}{-1 + \frac{1}{-1 + x}} \quad (\mbox{EX}2.2) \\
&= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1 + x}}} \quad (\mbox{EX}2.3) \\
&= \cdots
\end{align}
leads to the continued fraction
$$
x = 0 + \frac{1\rvert}{\lvert-1} + \frac{1\rvert}{\lvert-1} + \cdots \quad (\mbox{CF}2)
$$
for the negative root of $(\#)$.
Note that all equations $(\mbox{EX}2.m)$ are equivalent to equation $(\#)$ and the equations $(\mbox{EX}1.n)$, thus have two solutions for $x$.
However the derived continued fraction $(\mbox{CF}2)$ is the limit of these finite fractions:
\begin{align}
d_0 &= 0 \\
d_1 &= 0 + \frac{1}{-1} = -1 \\
d_2 &= 0 + \frac{1}{-1 + \frac{1}{-1}} = -0.5 \\
d_3 &= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1}}} = -0.\overline{6} \\
d_4 &= 0 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1 + \frac{1}{-1}}}} = -0.6\\
\end{align}
They result from dropping the right hand side $1/(-1+x)$ sub term. 
Note that the negative root is nicely approached from above by the even numbered terms and from below by the odd numbered terms.
