**Introduction and Problem Statement**

Given a regular hexagon ABCDEF with each side measuring 2 units, we extend side AB to point X such that AX 3AB. The task is to find the length of segment FX, expressed in simplest radical form.

Steps to Solve the Problem

We approach this problem using both geometric and coordinate geometry methods.

Coordinate Geometry Method

The vertices of the hexagon can be placed in the coordinate system as follows:

A(2, 0) B(1, sqrt{3}) C(-1, sqrt{3}) D(-2, 0) E(-1, -sqrt{3}) F(1, -sqrt{3})

Let's go through the steps to find the coordinates of point X and subsequently determine the length of segment FX.

Step 1: Find the coordinates of point X

Given that AX 3AB and AB 2, we find AX 3 times 2 6.

The vector vec{AB} is calculated as follows:

vec{AB} B - A (1 - 2, sqrt{3} - 0) (-1, sqrt{3}).

The unit vector in the direction of vec{AB} is:

vec{u} frac{vec{AB}}{| vec{AB} |} frac{(-1, sqrt{3})}{sqrt{1 3}} frac{(-1, sqrt{3})}{2}.

Hence, the coordinates of point X are:

X A 6 cdot vec{u} (2, 0) 6 cdot left( frac{-1, sqrt{3}}{2} right) (-1, 3sqrt{3}).

Step 2: Find the length of segment FX

The coordinates of F and X are:

F(1, -sqrt{3}) and X(-1, 3sqrt{3}).

We use the distance formula to find:

FX sqrt{(-1 - 1)^2 (3sqrt{3} - (-sqrt{3}))^2} sqrt{(-2)^2 (4sqrt{3})^2} sqrt{4 48} sqrt{52} 2sqrt{13}.

Cosine Rule Method

Using the cosine rule in triangle AFX:

We know AF 2, AX 6, and angle FAB 120^circ (since it’s a regular hexagon).

Applying the cosine rule:

FX^2 AF^2 AX^2 - 2 cdot AF cdot AX cdot cos(120^circ).

Solving for FX:

FX^2 2^2 6^2 - 2 cdot 2 cdot 6 cdot (-frac{1}{2}) 4 36 12 52.

Hence, FX 2sqrt{13}.

Alternative Geometry Method

Another approach involves constructing a right-angled triangle:

Hypotenuse AF 2 units, angle angle FAB 120^circ is split into 60° and 60°.

Given AX 3AB 6 units, a perpendicular FG is drawn to the extended line of BA.

Thus, in the right-angled triangle AGF with FAG 60^circ, AG frac{AF}{2} 1 unit.

Hence, GX 7 units, and GF sqrt{2^2 - 1^2} sqrt{3} units.

Using the Pythagorean theorem, the length of FX:

FX sqrt{GF^2 GX^2} sqrt{(2sqrt{3})^2 7^2} sqrt{12 49} 7.21 units.

Conclusion

The length of segment FX is 2sqrt{13} units.