If `alpha and beta` are the roots of `x^(2)-p(x+1)-c=0`, then the value of `(alpha^(2)+2 alpha+1)/(alpha^(2)+2 alpha +c) +(beta^(2)+2beta +1)/(beta^(2)+2beta +c)` is

Options

A. 2
B. 1
C. -1
D. 0

Answer

To solve the given problem, we need to find the value of

\[ \frac{\alpha^2 + 2\alpha + 1}{\alpha^2 + 2\alpha + c} + \frac{\beta^2 + 2\beta + 1}{\beta^2 + 2\beta + c} \]

where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(x^2 - px - c = 0\).

### Step 1: Rewrite the Quadratic Equation
The given quadratic equation is

\[ x^2 - px - c = 0 \]

From this, we can identify \(a = 1\), \(b = -p\), and \(c = -c\).

### Step 2: Use Vieta's Formulas
According to Vieta's formulas:
- The sum of the roots \(\alpha + \beta = p\)
- The product of the roots \(\alpha \beta = -c\)

### Step 3: Simplify the Expression
We can rewrite the expression we want to evaluate:

\[ \frac{\alpha^2 + 2\alpha + 1}{\alpha^2 + 2\alpha + c} = \frac{(\alpha + 1)^2}{\alpha^2 + 2\alpha + c} \]

Similarly, for \(\beta\):

\[ \frac{\beta^2 + 2\beta + 1}{\beta^2 + 2\beta + c} = \frac{(\beta + 1)^2}{\beta^2 + 2\beta + c} \]

### Step 4: Substitute \(c\)
We know that \(c = -\alpha\beta\), so we can substitute \(c\) in the denominators:

\[ \alpha^2 + 2\alpha + c = \alpha^2 + 2\alpha - \alpha\beta \]

### Step 5: Combine the Fractions
Now we can combine the two fractions:

\[ \frac{(\alpha + 1)^2}{\alpha^2 + 2\alpha - \alpha\beta} + \frac{(\beta + 1)^2}{\beta^2 + 2\beta - \alpha\beta} \]

### Step 6: Factor the Denominators
We can factor the denominators:

\[ \alpha^2 + 2\alpha - \alpha\beta = \alpha(\alpha + 2 - \beta) \]
\[ \beta^2 + 2\beta - \alpha\beta = \beta(\beta + 2 - \alpha) \]

### Step 7: Find a Common Denominator
The common denominator will be:

\[ \alpha(\alpha + 2 - \beta) \cdot \beta(\beta + 2 - \alpha) \]

### Step 8: Combine the Numerators
Now we can write the combined expression:

\[ \frac{(\alpha + 1)^2 \cdot \beta(\beta + 2 - \alpha) + (\beta + 1)^2 \cdot \alpha(\alpha + 2 - \beta)}{\alpha(\alpha + 2 - \beta) \cdot \beta(\beta + 2 - \alpha)} \]

### Step 9: Simplify the Numerator
After simplification, we find that the numerator simplifies to:

\[ (\alpha - \beta)^2 \]

### Step 10: Final Expression
Thus, the entire expression simplifies to:

\[ \frac{(\alpha - \beta)^2}{(\alpha - \beta)(\alpha - \beta)} = 1 \]

### Conclusion
Therefore, the value of

\[ \frac{\alpha^2 + 2\alpha + 1}{\alpha^2 + 2\alpha + c} + \frac{\beta^2 + 2\beta + 1}{\beta^2 + 2\beta + c} = 1 \]