Solving Radical Equation

[anemone]

Problem:
Solve for x in the following equation:
$\sqrt{x+\sqrt{x+11}}+\sqrt{x+\sqrt{x+-11}}=4$

I have not attempted this particular problem simply because I haven't the faintest idea how to even start it...

Could anyone please give me some hints on how to approach it, please?

Many thanks in advance.

 

[Ackbach]

If you're not allowed in the complex world, there is no solution. Proof: let
$$f(x)= \sqrt{x+ \sqrt{x-11}} \quad \text{and} \quad g(x)= \sqrt{x+ \sqrt{x+11}}.$$
It is fairly easy to see that both functions are monotonically increasing. The domain of $f$ is $[11,\infty)$, but at the lower endpoint, $f(11)= \sqrt{11}$ and $g(11)= \sqrt{11+ \sqrt{22}}$. The sum of $f$ and $g$ at $11$ is definitely greater than $4$, since
$$ \sqrt{11}+ \sqrt{11+ \sqrt{22}} \ge 3+ \sqrt{11+ 4} \ge 3+ \sqrt{9} = 6.$$

 

[anemone]

If you're not allowed in the complex world, there is no solution. Proof: let
$$f(x)= \sqrt{x+ \sqrt{x-11}} \quad \text{and} \quad g(x)= \sqrt{x+ \sqrt{x+11}}.$$
It is fairly easy to see that both functions are monotonically increasing. The domain of $f$ is $[11,\infty)$, but at the lower endpoint, $f(11)= \sqrt{11}$ and $g(11)= \sqrt{11+ \sqrt{22}}$. The sum of $f$ and $g$ at $11$ is definitely greater than $4$, since
$$ \sqrt{11}+ \sqrt{11+ \sqrt{22}} \ge 3+ \sqrt{11+ 4} \ge 3+ \sqrt{9} = 6.$$

Hi Ackbach, thanks for pointing this out to me.