Let `alpha,beta and gamma` be the roots of the equation `x^(3)+6x^(2)-px-42=0`. If `alpha, beta and gamma` are in arithmetic progression then `|alpha|+|beta|+|gamma|=`

Options

A. 10
B. 11
C. 12
D. 13

Answer

To solve the problem, we need to find the values of the roots \( \alpha, \beta, \gamma \) of the polynomial equation \( x^3 + 6x^2 - px - 42 = 0 \) given that these roots are in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Identify the relationship between roots and coefficients**:
According to Vieta's formulas, for the polynomial \( x^3 + 6x^2 - px - 42 = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} = -6 \)
- The product of the roots \( \alpha \beta \gamma = -\frac{d}{a} = 42 \)

2. **Express roots in terms of \( \beta \)**:
Since \( \alpha, \beta, \gamma \) are in AP, we can express them as:
- Let \( \beta = b \)
- Then \( \alpha = b - d \) and \( \gamma = b + d \) for some common difference \( d \).

3. **Set up the equations**:
From the sum of the roots:
\[ (b - d) + b + (b + d) = -6 \]
Simplifying this gives:
\[ 3b = -6 \implies b = -2 \]
Thus, \( \beta = -2 \).

4. **Find \( \alpha \) and \( \gamma \)**:
Using \( \beta = -2 \):
- \( \alpha = -2 - d \)
- \( \gamma = -2 + d \)

5. **Use the product of the roots**:
From the product of the roots:
\[ (b - d) \cdot b \cdot (b + d) = 42 \]
Substituting \( b = -2 \):
\[ (-2 - d)(-2)(-2 + d) = 42 \]
This simplifies to:
\[ -2(-2^2 - d^2) = 42 \implies -2(4 - d^2) = 42 \implies 4 - d^2 = -21 \implies d^2 = 25 \implies d = 5 \text{ or } d = -5 \]

6. **Calculate the roots**:
- If \( d = 5 \):
- \( \alpha = -2 - 5 = -7 \)
- \( \beta = -2 \)
- \( \gamma = -2 + 5 = 3 \)
- If \( d = -5 \):
- \( \alpha = -2 + 5 = 3 \)
- \( \beta = -2 \)
- \( \gamma = -7 \)

In both cases, the roots are \( -7, -2, 3 \).

7. **Calculate the absolute values**:
Now we find \( |\alpha| + |\beta| + |\gamma| \):
\[ |-7| + |-2| + |3| = 7 + 2 + 3 = 12 \]

### Final Answer:
\[ |\alpha| + |\beta| + |\gamma| = 12 \]