If product of roots of the equation `mx^(2)+6x+(2m-1)=0` is -1 then m equals

Options

A. -1
B. 1
C. `1//3`
D. `-1//3`

Answer

To solve the problem, we will follow these steps:

1. **Identify the quadratic equation**: The given quadratic equation is \( mx^2 + 6x + (2m - 1) = 0 \).

2. **Use the product of roots formula**: For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the product of the roots (\( r_1 \cdot r_2 \)) is given by \( \frac{c}{a} \). Here, \( a = m \), \( b = 6 \), and \( c = 2m - 1 \).

3. **Set up the equation for the product of roots**: According to the problem, the product of the roots is -1. Therefore, we can write:
\[ \frac{2m - 1}{m} = -1 \]

4. **Cross-multiply to eliminate the fraction**: This gives us:
\[ 2m - 1 = -m \]

5. **Rearrange the equation**: Add \( m \) to both sides:
\[ 2m + m - 1 = 0 \]
Simplifying this gives:
\[ 3m - 1 = 0 \]

6. **Solve for \( m \)**: Add 1 to both sides:
\[ 3m = 1 \]
Now, divide by 3:
\[ m = \frac{1}{3} \]

Thus, the value of \( m \) is \( \frac{1}{3} \).