The value of m for which the equation `x^3-mx^2+3x-2=0` has two roots equal rea magnitude but opposite in sign, is

Options

A. `4//5`
B. `3//4`
C. `2//3`
D. `1//2`

Answer

To find the value of \( m \) for which the equation \( x^3 - mx^2 + 3x - 2 = 0 \) has two roots that are equal in magnitude but opposite in sign, we can follow these steps:

### Step 1: Define the Roots
Let the roots of the cubic equation be \( \alpha, \beta, \) and \( \gamma \). According to the problem, we can assume that two roots are equal in magnitude but opposite in sign. Thus, we can set:
\[ \alpha = a \quad \text{and} \quad \beta = -a \]
where \( a \) is some real number.

### Step 2: Use Vieta's Formulas
From Vieta's formulas, we know that the sum of the roots \( \alpha + \beta + \gamma \) is equal to \( m \) (the coefficient of \( x^2 \) with a negative sign):
\[ \alpha + \beta + \gamma = m \]
Substituting the values of \( \alpha \) and \( \beta \):
\[ a + (-a) + \gamma = m \implies \gamma = m \]

### Step 3: Substitute the Roots into the Polynomial
Now, we substitute \( \gamma = m \) into the original polynomial equation:
\[ x^3 - mx^2 + 3x - 2 = 0 \]
We can substitute \( x = m \):
\[ m^3 - m(m^2) + 3m - 2 = 0 \]
This simplifies to:
\[ m^3 - m^3 + 3m - 2 = 0 \implies 3m - 2 = 0 \]

### Step 4: Solve for \( m \)
Now, we can solve for \( m \):
\[ 3m - 2 = 0 \implies 3m = 2 \implies m = \frac{2}{3} \]

### Conclusion
Thus, the value of \( m \) for which the equation \( x^3 - mx^2 + 3x - 2 = 0 \) has two roots equal in magnitude but opposite in sign is:
\[ \boxed{\frac{2}{3}} \]