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

Options

A. 1+ c
B. 1-c
C. p + c
D. p-c

Answer

To solve the problem, we need to find the value of \( (1 + \alpha)(1 + \beta) \) given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - p(x + 1) + c = 0 \).

### Step-by-step Solution:

1. **Identify the Coefficients**:
The given quadratic equation can be rewritten as:
\[ x^2 - p \cdot x - p + c = 0 \]
From this, we can identify the coefficients:
- Coefficient of \( x^2 \) is \( 1 \)
- Coefficient of \( x \) is \( -p \)
- Constant term is \( c - p \)

2. **Use Vieta's Formulas**:
According to Vieta's formulas, for a quadratic equation \( ax^2 + bx + c = 0 \):
- The sum of the roots \( \alpha + \beta = -\frac{b}{a} \)
- The product of the roots \( \alpha \beta = \frac{c}{a} \)

Applying this to our equation:
- \( \alpha + \beta = p \)
- \( \alpha \beta = c - p \)

3. **Expand \( (1 + \alpha)(1 + \beta) \)**:
We can expand the expression:
\[ (1 + \alpha)(1 + \beta) = 1 + \alpha + \beta + \alpha \beta \]

4. **Substitute the Values**:
Now, substitute the values we found using Vieta's formulas:
- \( \alpha + \beta = p \)
- \( \alpha \beta = c - p \)

Thus, we have:
\[ (1 + \alpha)(1 + \beta) = 1 + p + (c - p) \]

5. **Simplify the Expression**:
Now, simplify the expression:
\[ 1 + p + c - p = 1 + c \]

6. **Final Result**:
Therefore, we conclude that:
\[ (1 + \alpha)(1 + \beta) = c + 1 \]

### Answer:
The value of \( (1 + \alpha)(1 + \beta) \) is \( c + 1 \).

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