Let `alpha, beta and gamma` are the roots of the equation `2x^(2)+9x^(2)-27x-54=0`. If `alpha, beta, gamma` are in geometric progression, then the value of `|alpha|+|beta|+|gamma|=`

Options

A. `(19)/(2)`
B. `(21)/(2)`
C. 13
D. 11

Answer

To solve the problem, we need to find the roots of the cubic equation given and then determine the values of \(|\alpha| + |\beta| + |\gamma|\) where \(\alpha\), \(\beta\), and \(\gamma\) are in geometric progression.

### Step 1: Write down the equation
The given equation is:
\[ 2x^3 + 9x^2 - 27x - 54 = 0 \]

### Step 2: Identify coefficients
From the equation, we identify:
- \(a = 2\)
- \(b = 9\)
- \(c = -27\)
- \(d = -54\)

### Step 3: Use Vieta's formulas
According to Vieta's formulas for a cubic equation \(ax^3 + bx^2 + cx + d = 0\):
- The sum of the roots \(\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{9}{2}\)
- The sum of the product of the roots taken two at a time \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = -\frac{27}{2}\)
- The product of the roots \(\alpha\beta\gamma = -\frac{d}{a} = \frac{54}{2} = 27\)

### Step 4: Assume roots in GP
Since \(\alpha\), \(\beta\), and \(\gamma\) are in geometric progression, we can express them as:
- \(\alpha = \frac{a}{r}\)
- \(\beta = a\)
- \(\gamma = ar\)

### Step 5: Substitute into Vieta's formulas
1. From the sum of the roots:
\[ \frac{a}{r} + a + ar = -\frac{9}{2} \]
Multiplying through by \(r\):
\[ a + ar + a r^2 = -\frac{9r}{2} \]

2. From the product of the roots:
\[ \frac{a}{r} \cdot a \cdot ar = 27 \implies a^3 = 27 \implies a = 3 \]

### Step 6: Substitute \(a\) back into the equations
Substituting \(a = 3\):
1. The sum of the roots becomes:
\[ \frac{3}{r} + 3 + 3r = -\frac{9}{2} \]
Multiplying through by \(r\):
\[ 3 + 3r + 3r^2 = -\frac{9r}{2} \]
Rearranging gives:
\[ 6r^2 + 9r + 6 = 0 \quad \text{(after multiplying by 2)} \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-9 \pm \sqrt{81 - 48}}{12} = \frac{-9 \pm \sqrt{33}}{12} \]

### Step 8: Calculate roots
We can find the roots \(\alpha\), \(\beta\), and \(\gamma\) using the values of \(a\) and \(r\):
1. For \(r = \frac{-9 + \sqrt{33}}{12}\):
- \(\alpha = \frac{3}{r}\)
- \(\beta = 3\)
- \(\gamma = 3r\)

2. For \(r = \frac{-9 - \sqrt{33}}{12}\):
- \(\alpha = \frac{3}{r}\)
- \(\beta = 3\)
- \(\gamma = 3r\)

### Step 9: Calculate \(|\alpha| + |\beta| + |\gamma|\)
In both cases, we can find the absolute values and sum them:
\[ |\alpha| + |\beta| + |\gamma| = |3| + |3| + |3| = 3 + 3 + 3 = 9 \]

### Final Answer
Thus, the value of \(|\alpha| + |\beta| + |\gamma| = 9\).