Question

The sum of the squares of all the roots of the equation \( x^2 + [2x - 3] - 4 = 0 \) is:

• A. \(3(2 - \sqrt{2})\)
• B. \(6(2 - \sqrt{2})\)
• C. \(3(3 - \sqrt{2})\)
• D. \(6(2 - \sqrt{2})\)

Hint:

Always remember to check both roots in quadratic equations for full solutions.

Answer 1

The Correct Option is D

We are given the equation: \[ x^2 + |2x - 3| - 4 = 0 \]

Case 1: \( x \geq \frac{3}{2} \)
The absolute value simplifies to: \[ x^2 + 2x - 3 - 4 = 0 \] Simplified equation: \[ x^2 + 2x - 7 = 0 \] Solution: \[ x = 2\sqrt{2} - 1 \]

Case 2: \( x < \frac{3}{2} \)
The absolute value becomes: \[ x^2 + 3 - 2x - 4 = 0 \] Simplified equation: \[ x^2 - 2x - 1 = 0 \] Solution: \[ x = 1 - \sqrt{2} \]

Sum of Squares Calculation:
\[ \left( 2\sqrt{2} - 1 \right)^2 + \left( 1 - \sqrt{2} \right)^2 \] Expansion: \[ = 8 - 4\sqrt{2} + 1 + 1 - 2\sqrt{2} + 2 \] Simplification: \[ = 12 - 6\sqrt{2} = 6(2 - \sqrt{2}) \]

Final Answer:
The sum of squares of the solutions is \(6(2 - \sqrt{2})\).