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#### anemone

##### MHB POTW Director
Staff member
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?

#### Ackbach

##### Indicium Physicus
Staff member
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:

#### anemone

##### MHB POTW Director
Staff member
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. 