### Instruction: Find all solutions to
\[\sqrt{x^2 + \sqrt{x^2 + 11}} + \sqrt{x^2 - \sqrt{x^2 + 11}} = 4.\]Enter all the solutions, separated by commas.### Response: This looks like a very complicated equation, but maybe there is some way to simplify it or manipulate it to make it easier.
One thing I notice is that both square roots on the left-hand side have the same $x^2$ term, but with different signs inside the nested square root.
That makes me think of using the difference of squares identity, $(a-b)(a+b) = a^2 - b^2$, to factor the expression under each square root.
If I let $a = x^2$ and $b = \sqrt{x^2 + 11}$, then I can rewrite the equation as 
\[\sqrt{(x^2 - \sqrt{x^2 + 11})(x^2 + \sqrt{x^2 + 11})} + \sqrt{(x^2 + \sqrt{x^2 + 11})(x^2 - \sqrt{x^2 + 11})} = 4.\]
Now, using the difference of squares identity again, I can simplify each square root as 
\[\sqrt{x^4 - (x^2 + 11)} + \sqrt{x^4 - (x^2 + 11)} = 4.\]
Since the two square roots are the same, I can combine them and divide by 2 to get 
\[\sqrt{x^4 - (x^2 + 11)} = 2.\]
Squaring both sides, I get 
\[x^4 - (x^2 + 11) = 4.\]
Rearranging and simplifying, I get 
\[x^4 - x^2 - 15 = 0.\]
This looks like a quadratic equation in $x^2$, so I can try to factor it as 
\[(x^2 - a)(x^2 - b) = 0,\]
where $a$ and $b$ are constants that satisfy $a + b = 1$ and $ab = -15$.
By trial and error, I find that $a = 5$ and $b = -3$ work, so I have 
\[(x^2 - 5)(x^2 + 3) = 0.\]
This means that either $x^2 - 5 = 0$ or $x^2 + 3 = 0$.
The first equation gives me $x = \pm \sqrt{5}$ as solutions.
The second equation gives me $x = \pm \sqrt{-3}$ as solutions, but these are imaginary numbers, and I only want real solutions.
So, the final solutions are $x = \pm \sqrt{5}$.