Key Concepts and Formulas

• Absolute Value Definition: $|a| = a$ if $a \geq 0$ and $|a| = -a$ if $a < 0$.
• Sum and Product of Roots: For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$.
• Sum of Squares of Roots: If $\alpha$ and $\beta$ are roots of a quadratic, then $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

Step-by-Step Solution

• Step 1: Solve the first equation: $|x-2|^2 + |x-2| - 2 = 0$

  • What and Why: We are solving the first equation which contains an absolute value term. To simplify, we'll use substitution to transform it into a quadratic equation.
  • Let $y = |x-2|$. Then the equation becomes $y^2 + y - 2 = 0$.
  • Reasoning: This substitution makes the equation easier to solve by temporarily removing the absolute value.

• Step 2: Solve the quadratic equation for $y$

  • What and Why: We are solving the quadratic to find possible values for $y$, which represents the absolute value term.
  • Factoring the quadratic, we get $(y+2)(y-1) = 0$.
  • Therefore, $y = -2$ or $y = 1$.
  • Reasoning: Factoring allows us to easily find the roots of the quadratic equation.

• Step 3: Back-substitute and solve for $x$

  • What and Why: We substitute back $|x-2|$ for $y$ and solve for $x$. Remember that absolute value cannot be negative.
  • If $y = -2$, then $|x-2| = -2$. This is impossible since absolute value is always non-negative. Thus, there are no solutions in this case.
  • If $y = 1$, then $|x-2| = 1$. This means $x-2 = 1$ or $x-2 = -1$.
  • Solving for $x$:
    • $x - 2 = 1 \implies x = 3$
    • $x - 2 = -1 \implies x = 1$
  • Reasoning: We consider both positive and negative possibilities for the expression inside the absolute value.

• Step 4: Calculate the sum of squares of the roots of the first equation

  • What and Why: We calculate the sum of the squares of the roots we found in the previous step.
  • The roots are $x = 1$ and $x = 3$.
  • The sum of their squares is $1^2 + 3^2 = 1 + 9 = 10$.
  • Reasoning: This is a straightforward calculation using the roots we found.

• Step 5: Solve the second equation: $x^2 - 2|x-3| - 5 = 0$

  • What and Why: We solve the second equation, which also contains an absolute value term, using casework.
  • Case 1: $x \geq 3$. Then $|x-3| = x-3$, and the equation becomes $x^2 - 2(x-3) - 5 = 0$.
  • Simplifying: $x^2 - 2x + 6 - 5 = 0 \implies x^2 - 2x + 1 = 0 \implies (x-1)^2 = 0 \implies x = 1$.
  • However, we assumed $x \geq 3$, and $x=1$ does not satisfy this condition. Therefore, there are no solutions in this case.
  • Case 2: $x < 3$. Then $|x-3| = -(x-3) = 3-x$, and the equation becomes $x^2 - 2(3-x) - 5 = 0$.
  • Simplifying: $x^2 - 6 + 2x - 5 = 0 \implies x^2 + 2x - 11 = 0$.
  • Reasoning: We consider both cases for the absolute value and check if the solutions satisfy the initial assumption for each case.

• Step 6: Solve the quadratic equation for the second case

  • What and Why: We find the roots of the quadratic equation from the second case.
  • Using the quadratic formula, $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-11)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}$.
  • We need to check if both roots satisfy the condition $x < 3$.
  • $x_1 = -1 + 2\sqrt{3} \approx -1 + 2(1.732) = 2.464 < 3$.
  • $x_2 = -1 - 2\sqrt{3} \approx -1 - 2(1.732) = -4.464 < 3$.
  • Both roots satisfy the condition $x < 3$.
  • Reasoning: We use the quadratic formula to find the roots and verify that they satisfy the initial condition of the case.

• Step 7: Calculate the sum of squares of the roots of the second equation

  • What and Why: We calculate the sum of the squares of the roots we found in the previous step for the second equation.
  • For the quadratic $x^2 + 2x - 11 = 0$, the sum of the roots is $-2$ and the product of the roots is $-11$.
  • The sum of squares of the roots is $(-2)^2 - 2(-11) = 4 + 22 = 26$.
  • Reasoning: Using Vieta's formulas, we can easily find the sum and product of the roots, and then calculate the sum of their squares.

• Step 8: Combine the results

  • What and Why: We add the sum of squares of the roots from both equations to get the final answer.
  • The sum of squares of the roots from the first equation is $10$.
  • The sum of squares of the roots from the second equation is $26$.
  • The total sum of squares is $10 + 26 = 36$.
  • Reasoning: This is the final step, combining the results from the previous calculations.

Common Mistakes & Tips

• Always check if the solutions obtained in each case of an absolute value problem satisfy the initial conditions for that case.
• Remember that $|x|^2 = x^2$ can be useful for simplification, but be mindful of the original absolute value when back-substituting.
• Vieta's formulas are helpful for finding the sum and product of roots without explicitly calculating them.

Summary

We solved two equations, each involving absolute values. For the first equation, we used substitution to simplify it into a quadratic and found the sum of the squares of its roots. For the second equation, we used casework based on the definition of absolute value and again found the sum of the squares of its roots. Finally, we added the results from both equations to get the final answer.

The final answer is \boxed{36}, which corresponds to option (C).