To find the value of \( m \) such that the product of roots of the quadratic equation \( mx^2 + 6x + (2m-1) = 0 \) is \(-1\), we start by using the formula for the product of roots of a quadratic equation \( ax^2 + bx + c = 0 \), which is given by:
\(\text{Product of roots} = \frac{c}{a}\)
Here, \( a = m \), \( b = 6 \), and \( c = 2m - 1 \).
According to the problem statement, the product of roots is \(-1\). Thus, we have:
\(\frac{2m - 1}{m} = -1\)
We can solve this equation for \( m \) by cross-multiplying:
\(2m - 1 = -m\)
Bringing the terms involving \( m \) to one side gives:
\(2m + m = 1\)
\(3m = 1\)
Solving for \( m \) gives:
\(m = \frac{1}{3}\)
Upon reviewing the options, we notice that the answer provided as correct is \(m = -\frac{1}{3}\). Thus, let's confirm and re-calculate:
Starting from the equation:
\(\frac{2m - 1}{m} = -1\)
Clearing denominators:
\(2m - 1 = -m\)
Add \( m \) to both sides:
\(2m + m = 1\)
This leads to:
\(3m = 1\)
Thus, solving yields:
\(m = \frac{1}{3}\)
The correct solution according to this calculation does not match with the given correct answer, suggesting there may be an error in the provided correct answer.
The computed value \(m = \frac{1}{3}\) correctly satisfies the condition that the product of roots is \(-1\).