If sum and product of the zeroes of `ky^(2)+2y+3k ` are equal , find k .

Answer

To solve the problem, we need to find the value of \( k \) such that the sum and product of the zeroes of the quadratic equation \( ky^2 + 2y + 3k \) are equal.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation is \( ky^2 + 2y + 3k \). Here, we can identify the coefficients:
- \( a = k \)
- \( b = 2 \)
- \( c = 3k \)

2. **Use the formulas for sum and product of the zeroes**:
The sum of the zeroes \( \alpha + \beta \) of a quadratic equation \( ay^2 + by + c = 0 \) is given by:
\[ \text{Sum of zeroes} = -\frac{b}{a} \]
The product of the zeroes \( \alpha \beta \) is given by:
\[ \text{Product of zeroes} = \frac{c}{a} \]

3. **Substitute the values of \( a \), \( b \), and \( c \)**:
Now substituting the values we identified:
\[ \text{Sum of zeroes} = -\frac{2}{k} \]
\[ \text{Product of zeroes} = \frac{3k}{k} = 3 \]

4. **Set the sum equal to the product**:
According to the problem, the sum of the zeroes is equal to the product of the zeroes:
\[ -\frac{2}{k} = 3 \]

5. **Solve for \( k \)**:
To solve for \( k \), we can cross-multiply:
\[ -2 = 3k \]
Now, divide both sides by 3:
\[ k = -\frac{2}{3} \]

### Final Answer:
The value of \( k \) is \( -\frac{2}{3} \).