MHB How to Solve a Complex Radical Equation?

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The equation $\sqrt{x+\sqrt{x+11}}+\sqrt{x+\sqrt{x-11}}=4$ poses a challenge in finding solutions. It is noted that both functions involved, $f(x)$ and $g(x)$, are monotonically increasing, with $f$ defined for $x \geq 11$. At the lower boundary of this domain, the sum of the functions exceeds 4, indicating that no real solutions exist. The discussion concludes that if complex solutions are not permitted, the equation has no solution. Understanding the behavior of these functions is crucial for tackling similar radical equations.
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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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