How to Solve a Complex Radical Equation?

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SUMMARY

The equation $\sqrt{x+\sqrt{x+11}}+\sqrt{x+\sqrt{x+-11}}=4$ has no solution in the real number system. The functions defined as $f(x)= \sqrt{x+ \sqrt{x-11}}$ and $g(x)= \sqrt{x+ \sqrt{x+11}}$ are both monotonically increasing, with the domain of $f$ starting at $[11,\infty)$. At the lower endpoint $x=11$, the sum of $f(11)$ and $g(11)$ exceeds 4, confirming that no real solutions exist. Therefore, the problem requires consideration of complex numbers for potential solutions.

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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.
 
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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
$$ \sqrt{11}+ \sqrt{11+ \sqrt{22}} \ge 3+ \sqrt{11+ 4} \ge 3+ \sqrt{9} = 6.$$
 
Last edited:
Ackbach said:
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.:)
 

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