Proof Using Zeros of a Quadratic Polynomial

Video Explanation

Question

If \( \alpha \) and \( \beta \) are the zeros of the quadratic polynomial

\[ f(x) = x^2 – p(x + 1) – c, \]

show that

\[ (\alpha + 1)(\beta + 1) = 1 – c. \]

Solution

Step 1: Write the Polynomial in Standard Form

\[ f(x) = x^2 – px – p – c \]

Step 2: Use Relations Between Zeros and Coefficients

For a quadratic polynomial \( ax^2 + bx + d \),

\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{d}{a} \]

Here,

\[ a = 1,\quad b = -p,\quad d = -(p + c) \]

\[ \alpha + \beta = p \]

\[ \alpha\beta = -(p + c) \]

Step 3: Evaluate the Required Expression

\[ (\alpha + 1)(\beta + 1) \]

\[ = \alpha\beta + \alpha + \beta + 1 \]

Substitute the values:

\[ = (-(p + c)) + p + 1 \]

\[ = 1 – c \]

Conclusion

Hence,

\[ (\alpha + 1)(\beta + 1) = 1 – c \]

\[ \therefore \quad \text{The given result is proved.} \]