If alpha ,beta are the zeros of the polynomial P(x)=x^2-p(x+1)-c such that (alpha+1)(beta+1)=0 . What is the value of c

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Solution

Given that alpha and beta are the roots of the quadratic equation f(x) = x^2-p(x+1)-c
= x^2-px-p-c
= x^2 -px-(p+c),
comparing with ax^2 + bx + c, we have, a =1 , b= -p
& c= -(p+c)
alpha+beta = -b/a = -(-p)/1 = p
**& alphabeta = c/a = -(p+c)/1 = -(p+c)
Therefore,
(Alpha + 1)(beta+1)
= Alpha*beta + alpha + beta + 1
= -(p+c) + p + 1
= -p-c+p+1
= 1-c

or c=1**

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