If `alpha, beta and gamma` are the roots of the equation `x^(3)-13x^(2)+15x+189=0` and one root exceeds the other by 2, then the value of `|alpha|+|beta|+|gamma|` is equal to

Options

A. 23
B. 17
C. 13
D. 19

Answer

To solve the cubic equation \( x^3 - 13x^2 + 15x + 189 = 0 \) with roots \( \alpha, \beta, \gamma \), where one root exceeds another by 2, we can follow these steps:

### Step 1: Identify the relationship between the roots
Let’s assume \( \alpha = x \), \( \beta = x + 2 \), and \( \gamma \) is the third root. According to Vieta's formulas, we know:
- \( \alpha + \beta + \gamma = 13 \)
- \( \alpha \beta + \beta \gamma + \gamma \alpha = 15 \)
- \( \alpha \beta \gamma = -189 \)

### Step 2: Substitute the values of \( \alpha \) and \( \beta \)
Substituting \( \alpha \) and \( \beta \) into the first equation:
\[ x + (x + 2) + \gamma = 13 \]
This simplifies to:
\[ 2x + 2 + \gamma = 13 \]
Thus:
\[ \gamma = 11 - 2x \]

### Step 3: Substitute into the second Vieta's formula
Now, substituting \( \alpha \), \( \beta \), and \( \gamma \) into the second equation:
\[ x(x + 2) + (x + 2)(11 - 2x) + x(11 - 2x) = 15 \]
Expanding this:
\[ x^2 + 2x + (11x + 22 - 2x^2 - 2x) + (11x - 2x^2) = 15 \]
Combining like terms:
\[ -x^2 + 15x + 22 = 15 \]
Rearranging gives:
\[ -x^2 + 15x + 7 = 0 \]
Multiplying through by -1:
\[ x^2 - 15x - 7 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{15 \pm \sqrt{225 + 28}}{2} \]
\[ x = \frac{15 \pm \sqrt{253}}{2} \]

### Step 5: Calculate the roots
Now we have:
- \( \alpha = \frac{15 + \sqrt{253}}{2} \)
- \( \beta = \frac{17 + \sqrt{253}}{2} \)
- \( \gamma = 11 - 2\left(\frac{15 + \sqrt{253}}{2}\right) = -\frac{1 + \sqrt{253}}{2} \)

### Step 6: Calculate the absolute values and their sum
Now we need to find \( |\alpha| + |\beta| + |\gamma| \):
\[ |\alpha| = \frac{15 + \sqrt{253}}{2}, \quad |\beta| = \frac{17 + \sqrt{253}}{2}, \quad |\gamma| = \frac{1 + \sqrt{253}}{2} \]
Thus:
\[ |\alpha| + |\beta| + |\gamma| = \frac{15 + \sqrt{253}}{2} + \frac{17 + \sqrt{253}}{2} + \frac{1 + \sqrt{253}}{2} \]
Combining these:
\[ = \frac{33 + 3\sqrt{253}}{2} \]

### Final Result
The value of \( |\alpha| + |\beta| + |\gamma| \) is:
\[ \frac{33 + 3\sqrt{253}}{2} \]