The given equation is:

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

Steps to solve:

1. Isolate one of the square roots:

   Let's isolate $\sqrt{4 - x}$: $\sqrt{4 - x} = 2 - \sqrt{x - 2}$

2. Square both sides:

   Squaring both sides to eliminate the square root: $(\sqrt{4 - x})^2 = (2 - \sqrt{x - 2})^2$ Simplifying both sides: $4 - x = 4 - 4\sqrt{x - 2} + (x - 2)$ Which simplifies to: $4 - x = 2 + x - 4\sqrt{x - 2}$

3. Rearrange and isolate the square root:

   Now, isolate the square root term: $4 - x - 2 - x = -4\sqrt{x - 2}$ Simplifying: $2 - 2x = -4\sqrt{x - 2}$ Divide by -4: $\frac{2 - 2x}{-4} = \sqrt{x - 2}$ Simplify: $\frac{x - 1}{2} = \sqrt{x - 2}$

4. Square both sides again:

   Square both sides again to eliminate the square root: $\left( \frac{x - 1}{2} \right)^2 = (x - 2)$ Expanding both sides: $\frac{(x - 1)^2}{4} = x - 2$ Multiply by 4 to clear the denominator: $(x - 1)^2 = 4(x - 2)$ Expand both sides: $x^2 - 2x + 1 = 4x - 8$ Rearrange into a quadratic equation: $x^2 - 6x + 9 = 0$

5. Solve the quadratic equation:

   Factorize the quadratic: $(x - 3)^2 = 0$ Therefore: $x = 3$

6. Check the solution:

   Substituting $x = 3$ into the original equation: $\sqrt{4 - 3} + \sqrt{3 - 2} = \sqrt{1} + \sqrt{1} = 1 + 1 = 2$ This satisfies the equation.

Final Answer:

$x = 3$

Would you like further details on any part of this solution or have any additional questions?

Here are 5 related questions for further exploration:

1. What happens if we substitute values other than 3 into the equation?
2. How do you solve equations with three square root terms?
3. Can square root equations have more than one solution?
4. What are the potential pitfalls of squaring both sides in an equation?
5. How can quadratic equations be solved when they are not easily factorable?

Tip: Always check for extraneous solutions after solving square root equations, as squaring both sides can introduce invalid results.