If `x=2` and `x=3` are roots of the equation `3x^(2)-2mx+2n=0`, find the values of m and n.

Answer

To find the values of \( m \) and \( n \) given that \( x = 2 \) and \( x = 3 \) are roots of the equation \( 3x^2 - 2mx + 2n = 0 \), we can substitute these values into the equation and form two equations.

### Step 1: Substitute \( x = 2 \) into the equation
Substituting \( x = 2 \):
\[ 3(2)^2 - 2m(2) + 2n = 0 \]
Calculating this gives:
\[ 3 \cdot 4 - 4m + 2n = 0 \]
\[ 12 - 4m + 2n = 0 \]
Rearranging this equation:
\[ -4m + 2n = -12 \]
Dividing the entire equation by 2:
\[ -2m + n = -6 \]
Rearranging gives us:
\[ 2m - n = 6 \quad \text{(Equation 1)} \]

### Step 2: Substitute \( x = 3 \) into the equation
Substituting \( x = 3 \):
\[ 3(3)^2 - 2m(3) + 2n = 0 \]
Calculating this gives:
\[ 3 \cdot 9 - 6m + 2n = 0 \]
\[ 27 - 6m + 2n = 0 \]
Rearranging this equation:
\[ -6m + 2n = -27 \]
Dividing the entire equation by 2:
\[ -3m + n = -\frac{27}{2} \]
Rearranging gives us:
\[ 3m - n = \frac{27}{2} \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of equations:
1. \( 2m - n = 6 \) (Equation 1)
2. \( 3m - n = \frac{27}{2} \) (Equation 2)

We can subtract Equation 1 from Equation 2:
\[ (3m - n) - (2m - n) = \frac{27}{2} - 6 \]
This simplifies to:
\[ 3m - 2m = \frac{27}{2} - \frac{12}{2} \]
\[ m = \frac{15}{2} \]

### Step 4: Substitute \( m \) back to find \( n \)
Now, substitute \( m = \frac{15}{2} \) back into Equation 1:
\[ 2\left(\frac{15}{2}\right) - n = 6 \]
This simplifies to:
\[ 15 - n = 6 \]
Rearranging gives:
\[ n = 15 - 6 = 9 \]

### Final Values
Thus, the values are:
\[ m = \frac{15}{2}, \quad n = 9 \]