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

Options

A. `+- 1`
B. `+- 2`
C. `+- 3`
D. `+- 4`

Answer

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

### Step-by-Step Solution:

1. **Assume the Roots**:
Let the roots of the equation be \( a - d, a, a + d \), where \( a \) is the middle root and \( d \) is the common difference.

2. **Sum of the Roots**:
According to Vieta's formulas, the sum of the roots of the polynomial \( ax^3 + bx^2 + cx + d = 0 \) is given by:
\[ -\frac{b}{a} \]
For our equation, \( a = 1 \) and \( b = -12 \). Therefore, the sum of the roots is:
\[ (a - d) + a + (a + d) = 3a = -\frac{-12}{1} = 12 \]
From this, we can derive:
\[ 3a = 12 \implies a = \frac{12}{3} = 4 \]

3. **Product of the Roots**:
The product of the roots is given by:
\[ (a - d) \cdot a \cdot (a + d) = -\frac{d}{a} \]
For our equation, the product of the roots is:
\[ (a - d) \cdot a \cdot (a + d) = -\frac{-28}{1} = 28 \]
Substituting \( a = 4 \):
\[ (4 - d) \cdot 4 \cdot (4 + d) = 28 \]

4. **Simplifying the Product**:
Expanding the left side:
\[ 4 \cdot (4^2 - d^2) = 28 \]
This simplifies to:
\[ 4 \cdot (16 - d^2) = 28 \]
Dividing both sides by 4:
\[ 16 - d^2 = 7 \]
Rearranging gives:
\[ d^2 = 16 - 7 = 9 \]

5. **Finding the Common Difference**:
Taking the square root of both sides:
\[ d = \pm \sqrt{9} = \pm 3 \]

Thus, the common difference \( d \) is \( 3 \) or \( -3 \).

### Final Answer:
The common difference is \( d = \pm 3 \).

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