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Q.
If $\alpha, \beta$ are the roots of $x^{2}-p(x+1)-c=0, c \neq 1$, then the value of $(\alpha+1)(\beta+1)$ is equal to:
Complex Numbers and Quadratic Equations
Solution:
The given equation may be written as
$x ^{2}- px -( p + c )=0$
$\therefore \alpha+\beta= p$ and $ \alpha \beta=-( p + c )$
Now, $(\alpha+1)(\beta+1)=\alpha \beta+\alpha+\beta+1$
$=- p - c + p +1=1- c$