If roots of equation `2x^(4) - 3x^(3) + 2x^2 - 7x -1 = 0` are `alpha, beta, gamma` and 5 then value of `sum(alpha+1)/alpha` is equal to:

Answer

To solve the equation \(2x^4 - 3x^3 + 2x^2 - 7x - 1 = 0\) with roots \(\alpha, \beta, \gamma,\) and \(5\), we need to find the value of the expression \(\sum \frac{\alpha + 1}{\alpha}\).

### Step-by-step Solution:

1. **Understanding the Expression**:
We need to evaluate:
\[ \sum \frac{\alpha + 1}{\alpha} = \frac{\alpha + 1}{\alpha} + \frac{\beta + 1}{\beta} + \frac{\gamma + 1}{\gamma} + \frac{5 + 1}{5} \]
This can be rewritten as:
\[ = \left(1 + \frac{1}{\alpha}\right) + \left(1 + \frac{1}{\beta}\right) + \left(1 + \frac{1}{\gamma}\right) + \left(1 + \frac{1}{5}\right) \]

2. **Simplifying the Expression**:
This simplifies to:
\[ = 3 + \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right) + \frac{6}{5} \]
Thus, we have:
\[ = 3 + \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{6}{5} \]

3. **Using Vieta's Formulas**:
According to Vieta's formulas for a polynomial \(ax^n + bx^{n-1} + cx^{n-2} + \ldots = 0\):
- The sum of the roots \(\alpha + \beta + \gamma + 5 = -\frac{b}{a} = \frac{3}{2}\)
- The sum of the products of the roots taken two at a time \(\alpha\beta + \alpha\gamma + \beta\gamma + 5(\alpha + \beta + \gamma) = \frac{c}{a} = 1\)
- The product of the roots \(\alpha\beta\gamma \cdot 5 = -\frac{d}{a} = \frac{7}{2}\)

4. **Finding \(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\)**:
We know:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma} \]
From Vieta's, we can find:
- \(\alpha + \beta + \gamma = \frac{3}{2} - 5 = -\frac{7}{2}\)
- \(\alpha\beta + \alpha\gamma + \beta\gamma = 1 - 5(-\frac{7}{2}) = 1 + \frac{35}{2} = \frac{37}{2}\)
- \(\alpha\beta\gamma = \frac{7}{10}\)

5. **Calculating the Final Value**:
Thus:
\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\frac{37}{2}}{\frac{7}{10}} = \frac{37 \cdot 10}{2 \cdot 7} = \frac{370}{14} = \frac{185}{7} \]

6. **Putting it All Together**:
Now substituting back:
\[ = 3 + \frac{185}{7} + \frac{6}{5} \]
To combine these, we need a common denominator. The least common multiple of \(7\) and \(5\) is \(35\):
\[ 3 = \frac{105}{35}, \quad \frac{185}{7} = \frac{925}{35}, \quad \frac{6}{5} = \frac{42}{35} \]
Thus:
\[ = \frac{105 + 925 + 42}{35} = \frac{1072}{35} \]

### Final Answer:
The value of \(\sum \frac{\alpha + 1}{\alpha}\) is \(\frac{1072}{35}\).