\hspace{-16}$Solve for $\mathbf{x}$\\\\ $\mathbf{x+\sqrt{11+\sqrt{x}}=11}$
71If I try to solve it blindly, just by squaring and all that (I set x = z2 ; since x ≥ 0), It would end up in a fourth degree Polynomial like this :
z4-22z2-z+110 = 0.
I could only analyse from here that it has 2 positive and 2 negative roots which are equal.
So I tried a graph and it reveals the answer to be around ~ 7.3.
But Certainly, that is not a good way to solve. Kindly suggest.
21a similar question
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=442616&p2494058#p2494058
for this one, after what vivke has done see this http://www.artofproblemsolving.com/Forum/viewtopic.php?t=442616&p2494058#p2494058
11Hint:
Take 11+\sqrt{x}=y
\Rightarrow y-\sqrt{x}=11
From original equation we get x+√y=11
Subtract the equations to get the solution.
11Like this-
y-√x-x-√y=0
=>y-x=√x+√y
=>√y-√x=1
=>√y=√x+1
=>y=x+1+2√x
=>11+√x=x+1+2√x
=>0=x+√x-10
Now, this a quadratic in √x. On solving we get √x=(-1+√41)/2
So, x=(21-√41)/2
1708Sambit thanks for Nice solution
\hspace{-16}$Here $\mathbf{x+\sqrt{11+\sqrt{x}}=11\Leftrightarrow \sqrt{11+\sqrt{x}}=11-x}$\\\\ Now for Domain $\mathbf{x\geq 0}$ and $\mathbf{11-x\geq 0\Leftrightarrow x\leq 11}$\\\\ So $\mathbf{x\in \left[0\;,11\right]}$\\\\ Now Let $\mathbf{11=a\;, a>0}$\\\\ So $\mathbf{x+\sqrt{a+\sqrt{x}}=a\Leftrightarrow \sqrt{a+\sqrt{x}}=a-x}$\\\\ $\mathbf{a+\sqrt{x}=(a-x)^2\Leftrightarrow a^2+x+2a.\sqrt{x}=a^2+x^2-2ax}$\\\\ $\mathbf{a^2-(2x+1).a+x^2-\sqrt{x}=0}$\\\\ So $\mathbf{a=\frac{(2x+1)\pm (2\sqrt{x}+1)}{2}}$\\\\ so $\mathbf{a=x+\sqrt{x}+1}$ and $\mathbf{a=x-\sqrt{x}}$\\\\ Now Put value of $\mathbf{a}\;,$ we Get \\\\ so $\mathbf{x+\sqrt{x}-10=0}$ and $\mathbf{x-\sqrt{x}-11=0}$\\\\
