Question

a,ß are the roots of the quadratic equation x² -p(x + 1) -c=0, then (a +1)(ß+1) =
(a) c-1 (b) 1-c (c) c (d) 1+c​

Answer 1

Answer:

$option \: (b) \: is \: correct \: answer$

Step-by-step explanation:

$given \: it \: \\ \alpha \: \: \beta \: are \: the \: roots \: of \: the \: equation \: \\ {x}^{2} - p(x + 1) - c = 0 \\ {x}^{2} - px - p - c = 0 \\ {x}^{2} - px - (p + c) \\ \\ \alpha + \beta = p \: \: \: \alpha \beta = - p - c \\ \\ then( \alpha + 1)( \beta + 1) \\ \: \: \: \: \: = \alpha \beta + \alpha + \beta + 1 \\ \: \: \: \: \: = - p - c + p + 1 \\ \: \: \: \: \: = 1 - c \\ \\ hope \: it \: will \: help \: you.$