Question

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

• A. 1-2c
• B. 1+c
• C. c-2
• D. c

Answer 1

Answer: (B)

1+c

Answer 2

The given equation is $x^2 - p(x+1) + c = 0$. Expanding this, we get: $x^2 - px - p + c = 0$

This is a quadratic equation of the form $ax^2 + bx + d = 0$, where $a=1$, $b=-p$, and the constant term $d = c-p$. Let $\alpha$ and $\beta$ be the roots of this equation. According to Vieta's formulas: The sum of the roots is $\alpha + \beta = -\frac{b}{a} = -\frac{-p}{1} = p$. The product of the roots is $\alpha \beta = \frac{d}{a} = \frac{c-p}{1} = c-p$.

We need to find the value of $(\alpha + 1)(\beta + 1)$. Expanding this expression, we get: $(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1$

Now, substitute the values of $\alpha\beta$ and $\alpha+\beta$ obtained from Vieta's formulas: $(\alpha + 1)(\beta + 1) = (c-p) + (p) + 1$

Simplifying the expression: $(\alpha + 1)(\beta + 1) = c - p + p + 1 = c + 1$

Comparing this result with the given options: A) $1-2c$ B) $1+c$ C) $c-2$ D) $c$

The calculated value $c+1$ matches option B.