If one of the values of x of the equation `2x^(2)-6x + k= 0 " be " (1)/(2) (a+5i)`, find the values of a and k.

Answer

To solve the problem, we need to find the values of \( a \) and \( k \) given that one of the roots of the quadratic equation \( 2x^2 - 6x + k = 0 \) is \( \frac{1}{2}(a + 5i) \).

### Step 1: Identify the roots
Given that one root is \( \frac{1}{2}(a + 5i) \), the other root, by the property of complex conjugates, will be \( \frac{1}{2}(a - 5i) \).

### Step 2: Calculate 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 \( 2x^2 - 6x + k = 0 \):
- \( a = 2 \)
- \( b = -6 \)

Thus, the sum of the roots is:
\[ \text{Sum of roots} = -\frac{-6}{2} = 3 \]

### Step 3: Set up the equation for the sum of the roots
Now, we can express the sum of the roots using the roots we found:
\[ \frac{1}{2}(a + 5i) + \frac{1}{2}(a - 5i) = 3 \]
Simplifying this gives:
\[ \frac{1}{2}(a + 5i + a - 5i) = \frac{1}{2}(2a) = a \]
So, we have:
\[ a = 3 \]

### Step 4: Calculate the product of the roots
The product of the roots of the quadratic equation is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
For our equation:
- \( c = k \)
- \( a = 2 \)

Thus, the product of the roots is:
\[ \text{Product of roots} = \frac{k}{2} \]

### Step 5: Set up the equation for the product of the roots
Now we can express the product of the roots using the roots we found:
\[ \left(\frac{1}{2}(a + 5i)\right) \left(\frac{1}{2}(a - 5i)\right) = \frac{1}{4}(a^2 + 25) \]
Substituting \( a = 3 \):
\[ \frac{1}{4}(3^2 + 25) = \frac{1}{4}(9 + 25) = \frac{1}{4}(34) = \frac{34}{4} = \frac{17}{2} \]

### Step 6: Set the product equal to \( \frac{k}{2} \)
Now we equate this to the product of the roots:
\[ \frac{17}{2} = \frac{k}{2} \]
Multiplying both sides by 2 gives:
\[ k = 17 \]

### Final Values
Thus, the values of \( a \) and \( k \) are:
\[ a = 3, \quad k = 17 \]