Where did the negative answer come from in the continued fraction $1+\frac{1}{1+1/(1+\dots)}$? In this question we increased solution domain by squaring both sides of equation but what about this one ? 
Here the question is to evaluate $1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots  }}}}}$
$$x=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots}}}}}$$
$$x-1=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots  }}}}}$$
$$\cfrac{1}{x-1}=1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{\ddots }}}}}$$
$$\cfrac{1}{x-1}=x$$
$$x^2-x-1 = 0$$
$$x=\frac{1\pm \sqrt{5}}{2}$$
Now, it is obvious that answer can't be negative so : 
$$x=\frac{1+\sqrt{5}}{2}$$

If we set $x = \dfrac{1}{-1 +\underbrace{\dfrac{1}{-1+\dfrac{1}{-1+\dfrac{1}{-1+\cdots}}}}_{x}}$, then we have $x = \frac{1}{-1 + x}$, which gives also $x^2 - x - 1=0$
That's a similar presentation of the negative answer.
