If one of the roots of the equation `2x^(2)-6x+k=0` is `(alpha + 5i)/(2)` (where `alpha, k in R`) then
Statement 1 : the value of `alpha` is 3
because
Statement 2 : the value of k is 17.

Options

A. Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
B. Statement - 1 is True, Statement - 2 is True, Statement - 2 is NOT a correct explanation for statement - 1
C. Statement - 1 is True, Statement - 2 is False
D. Statement - 1 is False, Statement - 2 is True

Answer

To solve the problem, we need to find the values of \( \alpha \) and \( k \) given that one of the roots of the quadratic equation \( 2x^2 - 6x + k = 0 \) is \( \frac{\alpha + 5i}{2} \).

### Step-by-Step Solution:

1. **Identify the given root**:
We know one of the roots is \( \frac{\alpha + 5i}{2} \). Since the coefficients of the quadratic equation are real, the complex roots must occur in conjugate pairs. Therefore, the other root must be \( \frac{\alpha - 5i}{2} \).

2. **Use Vieta's Formulas**:
According to Vieta's formulas, the sum of the roots \( r_1 + r_2 \) is equal to \( -\frac{b}{a} \) and the product of the roots \( r_1 \cdot r_2 \) is equal to \( \frac{c}{a} \).
- Here, \( a = 2 \), \( b = -6 \), and \( c = k \).
- The sum of the roots is:
\[ \frac{\alpha + 5i}{2} + \frac{\alpha - 5i}{2} = \frac{2\alpha}{2} = \alpha \]
- According to Vieta's, this should equal \( \frac{6}{2} = 3 \):
\[ \alpha = 3 \]

3. **Calculate the product of the roots**:
- The product of the roots is:
\[ \left(\frac{\alpha + 5i}{2}\right) \left(\frac{\alpha - 5i}{2}\right) = \frac{(\alpha)^2 - (5i)^2}{4} = \frac{\alpha^2 + 25}{4} \]
- According to Vieta's, this should equal \( \frac{k}{2} \):
\[ \frac{\alpha^2 + 25}{4} = \frac{k}{2} \]
- Substituting \( \alpha = 3 \):
\[ \frac{3^2 + 25}{4} = \frac{k}{2} \]
\[ \frac{9 + 25}{4} = \frac{k}{2} \]
\[ \frac{34}{4} = \frac{k}{2} \]
\[ \frac{17}{2} = \frac{k}{2} \]
- Multiplying both sides by 2 gives:
\[ k = 17 \]

### Conclusion:
- From the calculations, we have found that \( \alpha = 3 \) and \( k = 17 \). Therefore, both statements are true.