The integral value of m for which the quadratic equation `(2m-3)x^2-4x+2m-3=0` has both the roots negative is given by

Options

A. 1
B. 2
C. 3
D. 5

Answer

To find the integral value of \( m \) for which the quadratic equation \( (2m-3)x^2 - 4x + (2m-3) = 0 \) has both roots negative, we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation can be expressed in the standard form \( ax^2 + bx + c = 0 \) where:
- \( a = 2m - 3 \)
- \( b = -4 \)
- \( c = 2m - 3 \)

### Step 2: Conditions for negative roots
For the quadratic equation to have both roots negative, we need to satisfy the following conditions:
1. The discriminant \( D \) must be greater than 0: \( D > 0 \)
2. The sum of the roots \( \alpha + \beta < 0 \)
3. The product of the roots \( \alpha \beta > 0 \)

### Step 3: Calculate the discriminant
The discriminant \( D \) is given by:
\[ D = b^2 - 4ac = (-4)^2 - 4(2m - 3)(2m - 3) \]
\[ D = 16 - 4(2m - 3)^2 \]
For \( D > 0 \):
\[ 16 - 4(2m - 3)^2 > 0 \]
Dividing through by 4:
\[ 4 - (2m - 3)^2 > 0 \]
This can be rewritten as:
\[ (2m - 3)^2 < 4 \]
Taking the square root:
\[ -2 < 2m - 3 < 2 \]
Adding 3 to all parts:
\[ 1 < 2m < 5 \]
Dividing by 2:
\[ \frac{1}{2} < m < \frac{5}{2} \]

### Step 4: Sum of the roots
The sum of the roots \( \alpha + \beta \) is given by:
\[ \alpha + \beta = -\frac{b}{a} = \frac{4}{2m - 3} \]
For \( \alpha + \beta < 0 \):
\[ \frac{4}{2m - 3} < 0 \]
This implies:
\[ 2m - 3 < 0 \implies m < \frac{3}{2} \]

### Step 5: Product of the roots
The product of the roots \( \alpha \beta \) is given by:
\[ \alpha \beta = \frac{c}{a} = \frac{2m - 3}{2m - 3} \]
Since \( \alpha \beta > 0 \), and \( 2m - 3 \) must be positive:
\[ 2m - 3 > 0 \implies m > \frac{3}{2} \]

### Step 6: Combine conditions
Now we have three conditions:
1. \( \frac{1}{2} < m < \frac{5}{2} \)
2. \( m < \frac{3}{2} \)
3. \( m > \frac{3}{2} \)

The only feasible condition is:
\[ \frac{1}{2} < m < \frac{3}{2} \]

### Step 7: Integral values of \( m \)
The integral values of \( m \) in the interval \( \left(\frac{1}{2}, \frac{3}{2}\right) \) is only:
\[ m = 1 \]

### Conclusion
Thus, the integral value of \( m \) for which the quadratic equation has both roots negative is:
\[ \boxed{1} \]