Question:
If $\alpha, \beta$ are the zeros of the polynomial $f(x)=x^{2}-p(x+1)-c$ such that $(\alpha+1)(\beta+1)=0$, then $c=$
(a) 1
(b) 0
(c) $-1$
(d) 2
Solution:
Since $\alpha$ and $\beta$ are the zeros of quadratic polynomial
$f(x)=x^{2}-p(x+1)-c$
$f(x)=x^{2}-p x-p-c$
$\alpha+\beta=\frac{-\text { Coefficient of } x}{\text { Coefficient of } x^{2}}$
$=-\left(\frac{-p}{1}\right)$
$=p$
$\alpha \beta=\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}$
$=\frac{-p-c}{1}$
$=-p-c$
We have
$0=(\alpha+1)(\beta+1)$
$0=\alpha \beta+(\alpha+\beta)+1$
$0=-c+1$
$c=1$
The value of $c$ is 1 .
Hence, the correct alternative is $(a)$
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