Question:
If $\alpha, \beta$ are the roots of the equation $x^{2}-p(x+1)-c=0$, then $(\alpha+1)(\beta+1)=$
(a) c
(b) c − 1
(c) 1 − c
(d) none of these
Solution:
(c) 1 − c
Given equation: $x^{2}-p(x+1)-c=0$
or $\quad x^{2}-p x-p-c=0$
Also $\alpha$ and $\beta$ are the roots of the equation.
Sum of the roots $=\alpha+\beta=p$
Product of the roots $=\alpha \beta=-(c+p)$
Then, $(\alpha+1)(\beta+1)=\alpha \beta+\alpha+\beta+1$
$=-(c+p)+p+1$
$=-c-p+p+1$
$=1-c$
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