Question

If alpha and beta are zeroes of x^2-a(x+1)-b such that (alpha+1)(beta+1)=0

Answer 1

Given equation:$x^{2} -a(x+1)-b \implies x^2 -ax-a-b \implies x^2 -ax + (-a-b)$
$\therefore p(x) = x^2 -ax +(-a-b)$
Let α,β be zeroes of p(x)
We know sum of zeroes = $\frac{-B}{A}$
$\implies \alpha + \beta = a$ ....(i)
& Product of zereos = $\frac{C}{A}$
$\implies \alpha \beta = -a-b$ ..... (ii)

Given condition $( \alpha +1 ) ( \beta +1) =0 \implies \alpha \beta + \alpha + \beta + 1 =0$
From equation (i) and (ii)
$-a-b + (a) = -1\implies b=1$
Substitute b in (i)