1. **State the problem:** Solve the equation $$\sqrt{x^2 - 4x + 4} = 2x - 5$$ in the set of real numbers. 2. **Understand the equation:** The expression inside the square root must be non-negative, and the right side must be non-negative because the square root always gives a non-negative result. 3. **Rewrite the equation:** Note that $$x^2 - 4x + 4 = (x - 2)^2$$, so the equation becomes: $$\sqrt{(x - 2)^2} = 2x - 5$$ 4. **Simplify the square root:** Since $$\sqrt{(x - 2)^2} = |x - 2|$$, the equation is: $$|x - 2| = 2x - 5$$ 5. **Consider cases for the absolute value:** - Case 1: $$x - 2 = 2x - 5$$ - Case 2: $$-(x - 2) = 2x - 5$$ 6. **Solve Case 1:** $$x - 2 = 2x - 5$$ Subtract $$x$$ from both sides: $$\cancel{x} - 2 = \cancel{x} + x - 5$$ $$-2 = x - 5$$ Add 5 to both sides: $$-2 + 5 = x$$ $$3 = x$$ 7. **Check domain for Case 1:** Right side $$2x - 5 = 2(3) - 5 = 6 - 5 = 1 \geq 0$$, valid. 8. **Solve Case 2:** $$-(x - 2) = 2x - 5$$ $$-x + 2 = 2x - 5$$ Add $$x$$ to both sides: $$2 = 3x - 5$$ Add 5 to both sides: $$7 = 3x$$ Divide both sides by 3: $$x = \frac{7}{3}$$ 9. **Check domain for Case 2:** Right side $$2x - 5 = 2 \times \frac{7}{3} - 5 = \frac{14}{3} - 5 = \frac{14}{3} - \frac{15}{3} = -\frac{1}{3} < 0$$, invalid because right side must be non-negative. 10. **Final solution:** Only $$x = 3$$ satisfies the equation and domain restrictions. **Answer:** $$x = 3$$