How to solve $x + \sqrt{x + \sqrt{11+x}}=11$ algebraically? How can I solve the following equation algebraically?
$x + \sqrt{x + \sqrt{11+x}}=11$
 A: Using the same steps as Gio67 in his/her answer, we arrive to the following quartic equation
$$x^4-46 x^3+771 x^2-5567 x+14630=0$$ But, a (tedious) inspection shows that $x_1=14$ is a solution. However, this solution does not satisfy the original equation (it was introduced by the multiple squaring processes).
So, what remains is the cubic $$x^3-32 x^2+323 x-1045=0$$ the discriminant of which being $\Delta=5073$ which implies  three distinct real roots. Using the trigonometric method, the three roots are given by 
$$x_2=\frac{2\sqrt{55}}{3}  \cos (\theta )+\frac{32}{3}\approx 15.5500 $$
$$x_3=\sqrt{\frac{55}{3}} \sin (\theta )-\frac{\sqrt{55}}{3}  \cos (\theta
   )+\frac{32}{3}\approx 8.89430$$
$$x_4=-\sqrt{\frac{55}{3}} \sin (\theta )-\frac{\sqrt{55}}{3}  \cos (\theta
   )+\frac{32}{3}\approx 7.55568$$ where $$\theta=\frac{1}{3} \tan ^{-1}\left(\frac{9 \sqrt{1691}}{727}\right)\approx 0.156960$$ But, again, $x_2$ and $x_3$ do not satisfy the original equation.
So, the only root is $x_4\approx 7.55568$ which can be confirmed by a plot of the original function.
For sure, the plot showing a solution close to $x=7$, we could use Newton method using $$f(x)=x+\sqrt{x+\sqrt{x+11}}-11$$ 
$$f'(x)=\frac{\frac{1}{2 \sqrt{x+11}}+1}{2 \sqrt{x+\sqrt{x+11}}}+1$$ and, starting with $x_0=7$, the successive iterates will be
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 7.000000000 \\
 1 & 7.554554415 \\
 2 & 7.555683721 \\
 3 & 7.555683726
\end{array}
\right)$$ which is the solution for ten significant figures.
A: $x-11=-\sqrt{x+\sqrt{11+x}}$. Square both sides to get $(x-11)^2=x+\sqrt{11+x}$.
Then $(x-11)^2-x=\sqrt{11+x}$ and you square again
$((x-11)^2-x)^2=11+x$. Unfortunately you get a 4th order equation. Try this quartic equation
