If ` alpha , beta ,1` are roots of ` x^3 -2x^2 -5x +6=0 ( alpha gt 1)` then ` 3 alpha + beta=`

Options

A. 7
B. 5
C. 14
D. 10

Answer

To solve the problem, we need to find the value of \(3\alpha + \beta\) given that \(\alpha\), \(\beta\), and \(1\) are roots of the polynomial equation \(x^3 - 2x^2 - 5x + 6 = 0\) with the condition that \(\alpha > 1\).

### Step-by-Step Solution:

1. **Identify the coefficients of the polynomial:**
The polynomial is given as:
\[ x^3 - 2x^2 - 5x + 6 = 0 \]
Here, we can identify the coefficients:
- \(a = 1\) (coefficient of \(x^3\))
- \(b = -2\) (coefficient of \(x^2\))
- \(c = -5\) (coefficient of \(x\))
- \(d = 6\) (constant term)

2. **Use Vieta's formulas:**
According to Vieta's formulas, for a cubic equation \(x^3 + ax^2 + bx + c = 0\):
- The sum of the roots \(\alpha + \beta + 1 = -\frac{b}{a}\)
- The sum of the products of the roots taken two at a time \(\alpha\beta + \beta \cdot 1 + \alpha \cdot 1 = \frac{c}{a}\)
- The product of the roots \(\alpha \beta \cdot 1 = -\frac{d}{a}\)

Applying these formulas:
- From the sum of the roots:
\[ \alpha + \beta + 1 = -\frac{-2}{1} = 2 \implies \alpha + \beta = 2 - 1 = 1 \quad \text{(Equation 1)} \]
- From the sum of the products of the roots:
\[ \alpha\beta + \beta \cdot 1 + \alpha \cdot 1 = -\frac{-5}{1} = 5 \implies \alpha\beta + \alpha + \beta = 5 \]
Substituting \(\alpha + \beta = 1\):
\[ \alpha\beta + 1 = 5 \implies \alpha\beta = 5 - 1 = 4 \quad \text{(Equation 2)} \]
- From the product of the roots:
\[ \alpha \beta \cdot 1 = -\frac{6}{1} = -6 \implies \alpha \beta = -6 \quad \text{(Equation 3)} \]

3. **Solve the equations:**
From Equation 2 and Equation 3, we have:
\[ \alpha \beta = 4 \quad \text{and} \quad \alpha \beta = -6 \]
This indicates a contradiction, so let's re-evaluate the product of the roots:
The correct product of the roots should be:
\[ \alpha \beta \cdot 1 = -\frac{6}{1} = -6 \implies \alpha \beta = -6 \]

4. **Use the equations to find \(\alpha\) and \(\beta\):**
We have:
- \(\alpha + \beta = 1\)
- \(\alpha \beta = -6\)

Let \(\beta = 1 - \alpha\). Substituting into \(\alpha \beta = -6\):
\[ \alpha(1 - \alpha) = -6 \implies \alpha - \alpha^2 = -6 \implies \alpha^2 - \alpha - 6 = 0 \]
Factoring the quadratic:
\[ (\alpha - 3)(\alpha + 2) = 0 \]
Thus, \(\alpha = 3\) or \(\alpha = -2\). Since \(\alpha > 1\), we take \(\alpha = 3\).

5. **Find \(\beta\):**
Substitute \(\alpha = 3\) back into \(\alpha + \beta = 1\):
\[ 3 + \beta = 1 \implies \beta = 1 - 3 = -2 \]

6. **Calculate \(3\alpha + \beta\):**
Now we can calculate:
\[ 3\alpha + \beta = 3(3) + (-2) = 9 - 2 = 7 \]

### Final Answer:
\[ 3\alpha + \beta = 7 \]