Question

If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:

• A. 6
• B. 2
• C. -2
• D. -6

Hint:

For complex roots, use the quadratic formula to find the roots and compute products directly.

Answer 1

The Correct Option is B

Step 1: Write the given quadratic equation

The given quadratic equation is:
\( x^2 - (3 - 2i)x - (2i - 2) = 0 \)

Step 2: Apply the quadratic formula

Using the quadratic formula:
\[ x = \frac{(3 - 2i) \pm \sqrt{(3 - 2i)^2 - 4(1)(-(2i - 2))}}{2(1)} \]

Step 3: Simplify the equation

Expanding the terms inside the square root:
\[ x = \frac{(3 - 2i) \pm \sqrt{9 - 4i^2 - 4(1)(-2i + 2)}}{2} \] \[ = \frac{3 - 2i \pm \sqrt{9 - 4(-1) - 12i + 8i - 8}}{2} \] \[ = \frac{3 - 2i \pm \sqrt{-3 - 4i}}{2} \]

Step 4: Further simplify the root term

Breaking the square root term into a solvable form:
\[ = 3 - 2i \pm \sqrt{(1)^2 + (2i)^2 - 2(1)(2i)} \] \[ = 3 - 2i \pm (1)^{2} + (2i)^{2} - 2(1)(2i) \]

Step 5: Final solutions

The final roots are:
\( x = 2 - 2i \quad \text{or} \quad x = 1 + 0i \)

Step 6: Find the product of the roots

From the roots obtained, we have:
\( \alpha \beta = 2(1) \cdot (-2)(0) = 2 \)