Nice and Simple (edited)

$\sqrt{(2 + \sqrt3)^x} +\sqrt{(2 - \sqrt3)^x} = 4$ Observe that if $\Rightarrow y = \sqrt{(2 + \sqrt3)^x} \quad ...\color{#3D99F6}{1}$ $\Rightarrow \dfrac{1}{y} = \sqrt{(2 - \sqrt3)^x} \quad ...\color{#D61F06}{2}$ $\Rightarrow y + \dfrac{1}{y} = 4$ $\Rightarrow y^2 - 4y + 1 = 0$ Using Quadratic formula,
$\Rightarrow y = 2 \pm \sqrt3 \quad\quad ..\color{#20A900}{3}$ Comparing values of y, $\Rightarrow (2 + \sqrt3) = \sqrt{(2 + \sqrt3)^x}$ $\Rightarrow (2 - \sqrt3)^{-1} = \sqrt{(2 - \sqrt3)^x}$ Bases are same, equating power $\Rightarrow \dfrac{x}{2} = 1 \ \ ; \ \ \dfrac{x}{2} = -1$ $\boxed{x = \pm 2 }$

Observe that $(2+\sqrt{3}) = \frac{1}{(2-\sqrt{3})}$ .

So, $\sqrt{(2+\sqrt{3})^{-x}} = \sqrt{(2-\sqrt{3})^{x}}$ .

And if $y$ is a solution, so is $-y$ . Hence the sum of solutions is $0$ .