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 \(2x^4 - 3x^3 + 2x^2 - 7x - 1 = 0\) and find the value of \(\sum \frac{\alpha + 1}{\alpha}\) where the roots are \(\alpha, \beta, \gamma\), and \(5\), we will follow these steps:

### Step 1: Understand the expression
We need to evaluate:
\[ \sum \frac{\alpha + 1}{\alpha} = \sum \left(1 + \frac{1}{\alpha}\right) = 4 + \sum \frac{1}{\alpha} \]
Here, we have four roots: \(\alpha, \beta, \gamma, \delta\) (where \(\delta = 5\)).

### Step 2: Calculate the sum of the roots
Using Vieta's formulas, for the polynomial \(2x^4 - 3x^3 + 2x^2 - 7x - 1 = 0\):
- The sum of the roots \(\alpha + \beta + \gamma + \delta = -\frac{b}{a} = -\frac{-3}{2} = \frac{3}{2}\).

### Step 3: Substitute \(\delta = 5\)
We know one of the roots is \(5\):
\[ \alpha + \beta + \gamma + 5 = \frac{3}{2} \]
Thus,
\[ \alpha + \beta + \gamma = \frac{3}{2} - 5 = \frac{3}{2} - \frac{10}{2} = -\frac{7}{2} \]

### Step 4: Calculate the sum of the products of the roots taken three at a time
Using Vieta's again, the sum of the products of the roots taken three at a time is given by:
\[ \alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta = \frac{d}{a} = -\frac{-7}{2} = \frac{7}{2} \]
Substituting \(\delta = 5\):
\[ \alpha \beta \gamma + 5(\alpha + \beta + \gamma) = \frac{7}{2} \]
Substituting \(\alpha + \beta + \gamma = -\frac{7}{2}\):
\[ \alpha \beta \gamma + 5\left(-\frac{7}{2}\right) = \frac{7}{2} \]
\[ \alpha \beta \gamma - \frac{35}{2} = \frac{7}{2} \]
\[ \alpha \beta \gamma = \frac{7}{2} + \frac{35}{2} = \frac{42}{2} = 21 \]

### Step 5: Calculate \(\sum \frac{1}{\alpha}\)
Using the relationship:
\[ \sum \frac{1}{\alpha} = \frac{\alpha \beta \gamma + \alpha \beta \delta + \alpha \gamma \delta + \beta \gamma \delta}{\alpha \beta \gamma \delta} \]
We already have:
- \(\alpha \beta \gamma = 21\)
- \(\alpha \beta \gamma + 5(\alpha + \beta + \gamma) = \frac{7}{2}\)

The product of the roots:
\[ \alpha \beta \gamma \delta = \frac{e}{a} = -\frac{-1}{2} = \frac{1}{2} \]

### Step 6: Substitute values into the equation
Now substituting:
\[ \sum \frac{1}{\alpha} = \frac{21 + \frac{7}{2}}{\frac{1}{2}} = \frac{21 \cdot 2 + 7}{1} = 42 + 7 = 49 \]

### Step 7: Final calculation
Now we can compute:
\[ \sum \frac{\alpha + 1}{\alpha} = 4 + \sum \frac{1}{\alpha} = 4 + 49 = 53 \]

### Final Answer
Thus, the value of \(\sum \frac{\alpha + 1}{\alpha}\) is:
\[ \boxed{53} \]