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 find the value of \( k \) given that \(\alpha\) and \(\frac{1}{\alpha}\) are the zeroes of the quadratic polynomial \(2x^2 - x + 8k\), follow these steps:

1. **Identify the coefficients**:
The given quadratic polynomial is \(2x^2 - x + 8k\). Here, the coefficients are:
\[ a = 2, \quad b = -1, \quad c = 8k \]

2. **Use the properties of zeroes**:
For a quadratic polynomial \(ax^2 + bx + c\) with zeroes \(\alpha\) and \(\beta\), the sum and product of the zeroes are given by:
\[ \alpha + \beta = -\frac{b}{a} \quad \text{and} \quad \alpha \beta = \frac{c}{a} \]
Given the zeroes are \(\alpha\) and \(\frac{1}{\alpha}\), we can use these properties.

3. **Sum of the zeroes**:
\[ \alpha + \frac{1}{\alpha} = -\frac{b}{a} = -\frac{-1}{2} = \frac{1}{2} \]

4. **Product of the zeroes**:
\[ \alpha \cdot \frac{1}{\alpha} = \frac{c}{a} = \frac{8k}{2} = 4k \]
Since \(\alpha \cdot \frac{1}{\alpha} = 1\), we have:
\[ 1 = 4k \]

5. **Solve for \(k\)**:
\[ 4k = 1 \implies k = \frac{1}{4} \]

Thus, the value of \( k \) is \(\frac{1}{4}\).