deriving $y=\sqrt{x+\sqrt{x+\sqrt{x}}\cdots} $ How to derive $y=\sqrt{x+\sqrt{x+\sqrt{x}}\cdots}$ at $x=6$ ?
 A: $y^2=x+y\implies y^2-y-x=0$ For $x=6$, $y^2-y-6=0\implies y=3$ or $y=-2$.
$y$ can't be negative, hence $y=3$.
A: Hint: First you want to rewrite $y=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$ in a friendly form. To do this note that $y=\sqrt{x+y}$.
A: We have:

$$\begin{equation}
\begin{split}
y=\sqrt{6+\sqrt{6+\sqrt{6}}\cdots}\Leftrightarrow 
y^2&=6+\sqrt{6+\sqrt{6+\sqrt{6}}\cdots}, \\&=6+y.
\end{split}
\end{equation}$$

Hence:

Find  $y$ in $$y^2-y-6=0.$$

Use the following: $y^2-y-6=(y-3)(y+2).$
Hence: $y=3$ or $y=-2$.
Since, $\sqrt{6+\sqrt{6+\sqrt{6}}\cdots}\geqslant0$ then $\sqrt{6+\sqrt{6+\sqrt{6}}\cdots}=3.$
A: Notice that if you square both sides:
$$y^2 = x + y$$
If x = 6:
$$y^2 = 6 + y$$
$$y^2 - y -6 = 0$$
$$(y-3)(y+2) = 0$$
$$y = 3, -2$$
Since y cannot be negative, y = 3.
A: Assuming that
$$
y = \sqrt{ x + \sqrt{ x + \sqrt{ x + \sqrt{ x + \cdots } } } },
$$
means to "repeat up to infinity".

So we write
$$
\begin{eqnarray}
y_0 &=& 0,\\
y_{n+1} &=& \sqrt{ x + y_n },\\
\end{eqnarray}
$$
and we define
$$
y = \lim_{n \rightarrow \infty} y_n.
$$
We must first show that $y$ exists.


If $y$ exists we obtain
$$
y = \lim_{n \rightarrow \infty} y_n = \lim_{n \rightarrow \infty} y_{n+1}
\Rightarrow y = \sqrt{x+y} \Rightarrow y^2 - y - x = 0 $$
At this point the quadratic formula tells us that $$y = \frac{1 \pm \sqrt{1 + 4x}}{2}$$
and we take the positive solution.
The case $x=6$ yields $y = \tfrac{1}{2} + \tfrac{1}{2} \times 5 = 3$.
