Question

$If \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: the \: polynomial \: f(x) = x2 - P(x +1) - c then\ \ \ \ ( \alpha + 1)( \beta + 1) =$
Answer with explanation ​

Answer 1

Step-by-step explanation:

We have,

$f(x) = {x}^{2} - p(x + 1) - c$

$\implies \: f(x) = {x}^{2} - px - p - c$

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

Now,

$\sf { \rarr \: sum \: of \: roots,} \: ( \alpha + \beta ) = \frac{ - (- p)}{1} = p$

$\sf { \rarr \: product \: of \: roots,} \: ( \alpha \beta ) = \frac{ - (p + c)}{1} = - (p + c)$

So,

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

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

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

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