Question

The sum of the squares of the roots of $ |x - 2|^2 + |x - 2| - 2 = 0 $ and the squares of the roots of $ x^2 |x - 3| - 5 = 0 $, is:

• A. \(24\)
• B. \(26\)
• C. \(36\)
• D. \(30\)

Hint:

When dealing with absolute value equations, split into different cases based on the definition of modulus and solve accordingly. Combine the results carefully for summation problems.

Answer 1

The Correct Option is C

Part 1: Solving \( |x - 2|^2 + |x - 2| - 2 = 0 \)

1. Substitution: Let \( y = |x - 2| \). The equation becomes \( y^2 + y - 2 = 0 \).
2. Factoring: \( (y + 2)(y - 1) = 0 \). So, \( y = -2 \) or \( y = 1 \).
3. Since \( y = |x - 2| \), \( y \) must be non-negative: Therefore, \( y = -2 \) is not a valid solution.
4. Solve \( |x - 2| = 1 \):
   • For \( x - 2 = 1 \), we get \( x = 3 \).
   • For \( x - 2 = -1 \), we get \( x = 1 \).
5. Squares of the roots: \( 3^2 + 1^2 = 9 + 1 = 10 \).

Part 2: Solving \( x^2 - 2|x - 3| - 5 = 0 \)

We need to consider two cases for the absolute value:

• Case 1: \( x \geq 3 \): Then \( |x - 3| = x - 3 \).
  • The equation becomes \( x^2 - 2(x - 3) - 5 = 0 \).
  • Simplifying: \( x^2 - 2x + 6 - 5 = 0 \) which gives \( x^2 - 2x + 1 = 0 \).
  • Factoring: \( (x - 1)^2 = 0 \), so \( x = 1 \).
  • But this contradicts the condition \( x \geq 3 \), so \( x = 1 \) is not a solution.
• Case 2: \( x < 3 \): Then \( |x - 3| = -(x - 3) = 3 - x \).
  • The equation becomes \( x^2 - 2(3 - x) - 5 = 0 \).
  • Simplifying: \( x^2 - 6 + 2x - 5 = 0 \) which gives \( x^2 + 2x - 11 = 0 \).
  • Using the quadratic formula: \[ x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-11)}}{2 \times 1} = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2} = \frac{-2 \pm 4\sqrt{3}}{2} \]
  • Thus, \( x = -1 \pm 2\sqrt{3} \).

Check if these solutions satisfy \( x < 3 \):

• For \( x = -1 + 2\sqrt{3} \), we get: \[ x \approx -1 + 2 \times 1.732 \approx -1 + 3.464 \approx 2.464 \] This satisfies \( x < 3 \).
• For \( x = -1 - 2\sqrt{3} \), we get: \[ x \approx -1 - 2 \times 1.732 \approx -1 - 3.464 \approx -4.464 \] This satisfies \( x < 3 \).

Squares of the roots: \[ (-1 + 2\sqrt{3})^2 + (-1 - 2\sqrt{3})^2 = (1 - 4\sqrt{3} + 12) + (1 + 4\sqrt{3} + 12) = 13 - 4\sqrt{3} + 13 + 4\sqrt{3} = 26 \]

Final Calculation: The sum of the squares of the roots of the first equation is 10, and the sum of the squares of the roots of the second equation is 26. Total sum = 10 + 26 = 36.

Answer: The answer is 36 (Option 3).