Question

If alpha and bita are the zeroes of the polynomial f(x) =x2 - p(x+1) - c show that (alpha +1 ) (Bita + 1 ) =1-c​

Answer 1

$\sf\huge\underline{Answer:}$

Given zeroes of the polynomial f(x) are alpha and beta

$f(x) = {x}^{2} - px -( p + c)$

Finding sum and product of the zeroes,

$→sum \: ( \alpha + \beta ) = \frac{ - b}{a}$

$\alpha + \beta = p$

$→product( \alpha \beta ) = \frac{c}{a}$

$\alpha \beta = - (p + c)$

Required solution is

$(\alpha + 1)( \beta + 1) = \alpha \beta + \alpha + \beta + 1$

$\alpha \beta + \alpha + \beta + 1 = - (p + c) + p + 1$

$= - p - c + p + 1$

$= 1 - c$

Hence proved,

$(\alpha + 1)( \beta + 1) = 1 - c$

Have a good day!