Question

If $\alpha \text{and}\beta$ are zeroes of $x^2-p\left(x+1\right)+a$ such that $\left(\alpha +1\right)\left(\beta +1\right)=0$, find the value of 'a'.

Answer 1

$x^2-p(x+1)+a=0$

$x^2-px+a-p=0$

Roots are $\alpha$ and $\beta$

Sum of Roots $=\alpha +\beta =p$

Product of roots$=\alpha \beta =a-p$

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

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

$a-p+p+1=0$

$\therefore a=-1$