Question

If $\alpha$ and $\beta$ are the zeroes of quadratic polynomial $f\left(x\right)=x^2-p(x+1)-c$, show that $\left(\alpha +1\right)\left(\beta +1\right)=1-c$

Answer 1

Given $\propto and\beta$ are the roots of

the quad ration equation

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

$=x^2-px-(p+c)$

Comparing with $a{x}^2+bx+c,$ we have

$a=1b=-pc=-(p+c)$

$\propto +\beta =\frac{-b}{a}=-\frac{(-p)}{1}$

$\propto \beta =\frac{c}{a}=-(p+c)$

$\because (\propto +1)(\beta +1)$

$=\propto \beta +\propto +\beta +1$

$=-p-c+p+1$

$=1-c$

$\because (\propto +1)(\beta +1)=1-c$