Welcome to our community

Be a part of something great, join today!

Solving Radical Equation

  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,681
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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:
  • Thread starter
  • Admin
  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,681
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.:)