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 quadratic equation \(2x^2 - 6x + k = 0\) given that one of the roots is \(\frac{\alpha + 5i}{2}\), where \(\alpha\) and \(k\) are real numbers, we proceed as follows:

### Step 1: Identify the roots
Since one root is given as \(\frac{\alpha + 5i}{2}\), the other root (due to the property of complex conjugates) will be \(\frac{\alpha - 5i}{2}\).

### Step 2: Use the sum of the roots
The sum of the roots of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula:
\[ \text{Sum of roots} = -\frac{b}{a} \]
For our equation \(2x^2 - 6x + k = 0\), we have \(a = 2\) and \(b = -6\). Therefore, the sum of the roots is:
\[ \text{Sum of roots} = -\frac{-6}{2} = \frac{6}{2} = 3 \]
Now, substituting the roots we have:
\[ \frac{\alpha + 5i}{2} + \frac{\alpha - 5i}{2} = \alpha \]
Setting this equal to the sum we found:
\[ \alpha = 3 \]

### Step 3: Use the product of the roots
The product of the roots of the quadratic equation is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
In our case, this means:
\[ \text{Product of roots} = \frac{k}{2} \]
Now, we calculate the product of the roots:
\[ \frac{\alpha + 5i}{2} \cdot \frac{\alpha - 5i}{2} = \frac{(\alpha + 5i)(\alpha - 5i)}{4} = \frac{\alpha^2 - (5i)^2}{4} = \frac{\alpha^2 + 25}{4} \]
Substituting \(\alpha = 3\):
\[ \frac{3^2 + 25}{4} = \frac{9 + 25}{4} = \frac{34}{4} = \frac{17}{2} \]
Setting this equal to the product of the roots:
\[ \frac{17}{2} = \frac{k}{2} \]
Multiplying both sides by 2 gives:
\[ k = 17 \]

### Conclusion
From our calculations, we have found:
- The value of \(\alpha\) is \(3\).
- The value of \(k\) is \(17\).