A square, of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle 30° with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is:

Let the square have vertices $$O, A, B, C$$ in that order, with $$O$$ at the origin. The length of every side is given to be $$2$$.

The side $$OA$$ passes through the origin and makes an angle of $$30^{\circ}$$ with the positive $$x$$-axis. Hence the direction ratios of $$OA$$ are $$\bigl(\cos 30^{\circ},\; \sin 30^{\circ}\bigr).$$ Using the length $$2,$$ the exact coordinates of $$A$$ are obtained by simple trigonometry: $$\begin{aligned} x_A &= 2\cos 30^{\circ}=2\left(\frac{\sqrt3}{2}\right)=\sqrt3,\\[2pt] y_A &= 2\sin 30^{\circ}=2\left(\frac12\right)=1. \end{aligned}$$ Thus $$A(\sqrt3,\;1).$$

Write the vector $$\overrightarrow{OA}$$ in component form: $$\overrightarrow{OA}=(\sqrt3,\,1).$$ To obtain the adjacent side of the square we need a vector of the same length that is perpendicular to $$\overrightarrow{OA}.$$ Rotating a vector $$(x,\,y)$$ through $$90^{\circ}$$ counter-clockwise produces $$(-y,\,x).$$ Therefore $$\overrightarrow{w}=(-1,\;\sqrt3)$$ is perpendicular to $$\overrightarrow{OA}$$ and has the same magnitude because $$\bigl|(-1,\sqrt3)\bigr|=\sqrt{(-1)^2+(\sqrt3)^2}=\sqrt{1+3}=2.$$

If instead we had rotated clockwise, we would get $$(1,-\sqrt3),$$ whose $$y$$-component is negative; that would force two vertices below the $$x$$-axis, contradicting the condition that the whole square lies above it. Hence we retain $$\overrightarrow{w}=(-1,\;\sqrt3).$$

The remaining two vertices are obtained by adding this vector once to $$A$$ and once to $$O$$:

For $$B:$$ $$\begin{aligned} B &= A+\overrightarrow{w}\\ &=\bigl(\sqrt3,\,1\bigr)+\bigl(-1,\,\sqrt3\bigr)\\ &=\bigl(\sqrt3-1,\,1+\sqrt3\bigr). \end{aligned}$$

For $$C:$$ $$\begin{aligned} C &= O+\overrightarrow{w}\\ &=\bigl(0,\,0\bigr)+\bigl(-1,\,\sqrt3\bigr)\\ &=\bigl(-1,\,\sqrt3\bigr). \end{aligned}$$

Collecting all four vertices with their $$x$$-coordinates:

$$\begin{aligned} O &: (0,\,0) &\Rightarrow&\ x_O=0,\\ A &: (\sqrt3,\,1) &\Rightarrow&\ x_A=\sqrt3,\\ B &: (\sqrt3-1,\,1+\sqrt3) &\Rightarrow&\ x_B=\sqrt3-1,\\ C &: (-1,\,\sqrt3) &\Rightarrow&\ x_C=-1. \end{aligned}$$

Now add these four $$x$$-coordinates explicitly:

$$\begin{aligned} x_O+x_A+x_B+x_C&=0+\sqrt3+(\sqrt3-1)+(-1)\\ &=\sqrt3+\sqrt3-1-1\\ &=2\sqrt3-2. \end{aligned}$$

Thus the required sum of the $$x$$-coordinates is $$2\sqrt3-2.$

Hence, the correct answer is Option A.