The Root Of All Roots

Given: $\sqrt{x^2+ \sqrt{x^2+11}} + \sqrt{x^2- \sqrt{x^2+11}} =4$

Putting $\sqrt {x^2+11} = y$ , we get,

$\sqrt {y^2 + y -11} + \sqrt{y^2-y-11} =4 -------- \ (1)$

We also have the identity

$(y^2+y-11) - (y^2-y-11) =2y ------- \ (2)$

Dividing 2 by 1 , we get,

$\sqrt{y^2+y-11} - \sqrt{y^2-y-11} =\frac{y}{2} ------- \ (3)$

Adding 1 and 3 , we get,

$2( \sqrt{y^2+y-11}) =4 + \frac{y}{2}$

$\implies y^2+y-11=4+y+ \frac{y^2}{16}$

$\implies y=4$

$\implies x=\pm \sqrt{5}$

$\implies a^2 + 3ab + b^2 = \boxed{25}$

Let $n=\sqrt { { x }^{ 2 }+11 }$

So the equation becomes:

$\sqrt { { x }^{ 2 }+n } +\sqrt { { x }^{ 2 }-n } =4$

${ (x }^{ 2 }+n)+2\sqrt { ({ x }^{ 2 }+n)({ x }^{ 2 }-n) } +({ x }^{ 2 }-n)=16$

${ 2x }^{ 2 }=16-2\sqrt { ({ x }^{ 2 }+n)({ x }^{ 2 }-n) }$

${ x }^{ 2 }=8-\sqrt { ({ x }^{ 2 }+n)({ x }^{ 2 }-n) }$

${ (8-{ x }^{ 2 } })^{ 2 }=({ x }^{ 2 }+n)({ x }^{ 2 }-n)$

$64-16{ x }^{ 2 }+{ x }^{ 4 }={ x }^{ 4 }-{ n }^{ 2 }$

$64-16{ x }^{ 2 }=-({ x }^{ 2 }+11)$ (substitute back in the value of $n$ )

Solving the quadratic equations gives: $x=0\pm \sqrt { 5 } =a\pm \sqrt { b }$

So $a=0$ and $b=5$ . Substitute the values of $a$ and $b$ into the equation and obtaining $\boxed{25}$