Question:
If $\alpha, \beta$ are roots of the equation $x^{2}-a(x+1)-c=0$, then write the value of $(1+\alpha)(1+\beta)$
Solution:
Given: $x^{2}-a(x+1)-c=0$ or $x^{2}-a x-a-c=0$
Also, $\alpha$ and $\beta$ are the roots of the equation.
Sum of the roots $=\alpha+\beta=-\left(\frac{-a}{1}\right)=a$
Product of the roots $=\alpha \beta=\frac{-(a+c)}{1}=-(a+c)$
$\therefore(1+\alpha)(1+\beta)=1+\beta+\alpha+\alpha \beta$
$=1+(\alpha+\beta)+(\alpha \beta)$
$=1+a-a-c$
$=1-c$
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