Question

If $\alpha ,\beta$ are the zeros of the polynomial $f\left(x\right)=x^2-p(x+1)-c=0$ such $(\alpha +1)(\beta +1)=0$, then $c=$

• A. $1$
• B. $0$
• C. $-1$
• D. $2$

Answer 1

The correct option is A $1$

We have,

$f\left(x\right)=x^2-p(x+1)-c=0$

$f\left(x\right)=x^2-px-(p+c)=0$ .............$\left(1\right)$

Since,

$\alpha ,\beta$ are the zeroes of the above polynomial.

So,

$\alpha +\beta =p$

$\alpha \beta =-(p+c)$

Since,

$(\alpha +1)(\beta +1)=0$

$\alpha \beta +\alpha +\beta +1=0$

$-p-c+p+1=0$

$-c+1=0$

$c=1$

Hence, this is the answer.