If the roots of the equation `x^3-12x^2 +39x -28 =0` are in AP, then their common difference is

Options

A. `pm1`
B. `pm2`
C. `pm3`
D. `pm4`

Answer

To solve the problem, we need to find the common difference of the roots of the polynomial equation \(x^3 - 12x^2 + 39x - 28 = 0\) given that the roots are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Identify the Roots**:
Let the roots of the equation be \( \alpha, \beta, \gamma \). Since they are in AP, we can express them as:
\[ \alpha = \beta - d, \quad \beta = \beta, \quad \gamma = \beta + d \]
where \(d\) is the common difference.

2. **Sum of Roots**:
According to Vieta's formulas, the sum of the roots of the polynomial \(ax^3 + bx^2 + cx + d = 0\) is given by:
\[ \alpha + \beta + \gamma = -\frac{b}{a} \]
For our polynomial, \(a = 1\) and \(b = -12\). Thus:
\[ \alpha + \beta + \gamma = -\frac{-12}{1} = 12 \]
Substituting the expressions for the roots:
\[ (\beta - d) + \beta + (\beta + d) = 12 \]
Simplifying this, we get:
\[ 3\beta = 12 \implies \beta = 4 \]

3. **Express the Roots**:
Now that we have \( \beta = 4 \), we can express the other roots:
\[ \alpha = 4 - d, \quad \gamma = 4 + d \]

4. **Product of Roots**:
Again, using Vieta's formulas, the product of the roots is given by:
\[ \alpha \beta \gamma = -\frac{d}{a} \]
For our polynomial, \(d = -28\) and \(a = 1\):
\[ \alpha \beta \gamma = -\frac{-28}{1} = 28 \]
Substituting the expressions for the roots:
\[ (4 - d) \cdot 4 \cdot (4 + d) = 28 \]

5. **Simplifying the Product**:
Expanding the left side:
\[ 4 \cdot ((4 - d)(4 + d)) = 28 \]
Using the difference of squares:
\[ 4 \cdot (16 - d^2) = 28 \]
Dividing both sides by 4:
\[ 16 - d^2 = 7 \]
Rearranging gives:
\[ d^2 = 16 - 7 = 9 \]

6. **Finding the Common Difference**:
Taking the square root of both sides:
\[ d = \pm 3 \]

### Conclusion:
The common difference \(d\) can be either \(3\) or \(-3\). Therefore, the common difference is \(3\).