To solve the equation

$|x^2 - 2x - 1| = 2,$

we need to address the absolute value, which splits the equation into two cases:

Case 1: $x^2 - 2x - 1 = 2$

1. Simplify: $x^2 - 2x - 1 - 2 = 0 \quad \Rightarrow \quad x^2 - 2x - 3 = 0.$

2. Factorize: $(x - 3)(x + 1) = 0.$

3. Solve: $x = 3 \quad \text{or} \quad x = -1.$

Case 2: $x^2 - 2x - 1 = -2$

1. Simplify: $x^2 - 2x - 1 + 2 = 0 \quad \Rightarrow \quad x^2 - 2x + 1 = 0.$

2. Factorize: $(x - 1)^2 = 0.$

3. Solve: $x = 1.$

Final Solution:

The solutions to the equation are: $x = 3, \, x = -1, \, x = 1.$

Let me know if you'd like further clarification or assistance! Here are some related questions to explore:

1. What happens when the right-hand side of the equation changes to a different number?
2. How do absolute value equations change if a quadratic term is inside the modulus?
3. Can this problem be solved graphically? If so, how?
4. What is the importance of checking for extraneous solutions in such equations?
5. How can we extend this technique to systems of equations involving absolute values?

Tip: Always split the equation based on the definition of absolute value to address all possible cases.