Problem 1

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

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

Square both sides ,

$\displaystyle{2{ x }^{ 2 }+2\sqrt { { x }^{ 4 }-{ A }^{ 2 } } =16\\ { (8-{ x }^{ 2 }) }^{ 2 }={ x }^{ 4 }-{ A }^{ 2 }\\ { A }^{ 2 }={ x }^{ 4 }-{ (8-{ x }^{ 2 }) }^{ 2 }\\ { x }^{ 2 }+11=(8)(2{ x }^{ 2 }-8)\\ 15{ x }^{ 2 }=75\\ \boxed { x=\pm \sqrt { 5 } } }$

So we have,

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

On squaring both sides,we are left with,

$2x^{2} + 2\sqrt{ x^{4} - x^{2} - 11 } = 16$

$x^{2} + \sqrt{ x^{4} - x^{2} - 11 } = 8$

$8 - x^{2} = \sqrt{ x^{4} - x^{2} - 11 }$

Again squaring both sides.we get,

$64 + x^{4} - 16x^{2} = x^{4} - x^{2} - 11$

$15x^{2} = 75$

$x = +\sqrt{5} or - \sqrt{5}$

Therefore,

Product of roots $= \boxed{ -5 }$

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

$B=\sqrt{x^2-\sqrt{x^2+11}}$

Given $A+B=4$ .............. $\boxed{1^{st}}$

Then clearly $A^2-B^2=(A+B)(A-B)=2\sqrt{x^2+11}$

$\Rightarrow$ $A-B=\dfrac{\sqrt{x^2+11}}{2}$ .............. $\boxed{2^{nd}}$

$\Rightarrow$ Adding (or one can go for subtracting too ) $1^{st}$ and $2^{nd}$ equation we get

$2A=\dfrac{8+\sqrt{x^2+11}}{2}$

$\Rightarrow$ $4A=8+\sqrt{x^2+11}$

Squaring $L.H.S$ and $R.H.S$ we get,

$16A^2=64+(x^2+11)+16\sqrt{x^2+11}$

Using $A=\sqrt{x^2+\sqrt{x^2+11}}$ we get,

$16(x^2+\sqrt{x^2+11})=64+(x^2+11)+16\sqrt{x^2+11}$

$\Rightarrow$ $16x^2=64+x^2+11$

$\Rightarrow$ $15x^2=75$

$\Rightarrow$ $\boxed{x=\sqrt{5}}$ or $\boxed{x=-\sqrt{5}}$

Since both values of $x$ are irrational therefore required $answer$ is $\boxed{-5}$ .

$\sqrt { { x }^{ 2 }+\sqrt { { x }^{ 2 }+11 } } +\sqrt { { x }^{ 2 }-\sqrt { { x }^{ 2 }+11 } } =4\quad ........(1)\\ Put\quad \sqrt { { x }^{ 2 }+11 } =t\quad ,so\quad that\quad the\quad equation\quad becomes\\ \sqrt { { t }^{ 2 }+t-11 } +\sqrt { { t }^{ 2 }-t-11 } =4\quad ......(2)\\ We\quad also\quad have\quad an\quad identity\quad \\ ({ t }^{ 2 }+t-11)-({ t }^{ 2 }-t-11)=2t\quad .........(3)\\ Dividing\quad (2)\quad by\quad (3)\\ \sqrt { { t }^{ 2 }+t-11 } -\sqrt { { t }^{ 2 }-t-11 } =\frac { t }{ 2 } \quad .......(4)\\ Adding\quad (2)\quad and\quad (4)\\ 2\left( \sqrt { { t }^{ 2 }+t-11 } \right) =\quad 4+\frac { t }{ 2 } \\ \Rightarrow \quad { t }^{ 2 }+t-11=4+t+\frac { { t }^{ 2 } }{ 16 } \\ \Rightarrow t=4\quad \Rightarrow \quad x=\boxed { \pm \sqrt { 5 } } \\$