If `alpha` and `1/alpha` are the zeroes of the quadratic polynomial `2x^2-x+8k`, then k is:

Options

A. 4
B. `1/4`
C. `(-1)/4`
D. 2

Answer

To solve the problem, we need to find the value of \( k \) given that \( \alpha \) and \( \frac{1}{\alpha} \) are the zeros of the quadratic polynomial \( 2x^2 - x + 8k \).

### Step-by-Step Solution:

1. **Identify the Polynomial and Zeros**:
The polynomial is given as:
\[ p(x) = 2x^2 - x + 8k \]
The zeros of the polynomial are \( \alpha \) and \( \frac{1}{\alpha} \).

2. **Use the Product of Zeros**:
The product of the zeros of a quadratic polynomial \( ax^2 + bx + c \) is given by:
\[ \text{Product of zeros} = \frac{c}{a} \]
Here, \( a = 2 \), \( b = -1 \), and \( c = 8k \).

3. **Calculate the Product of Zeros**:
For our polynomial:
\[ \alpha \cdot \frac{1}{\alpha} = 1 \]
According to the formula:
\[ \text{Product of zeros} = \frac{8k}{2} \]
Thus, we have:
\[ 1 = \frac{8k}{2} \]

4. **Simplify the Equation**:
Simplifying the right side:
\[ 1 = 4k \]

5. **Solve for \( k \)**:
To find \( k \), we rearrange the equation:
\[ k = \frac{1}{4} \]

### Conclusion:
The value of \( k \) is:
\[ \boxed{\frac{1}{4}} \]