Question

if alpha and beeta are the zeroes of the polynomialx^2-p(x+1)+c such that (alpha+1)(beeta+1)=0 then find value of c

Answer 1

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 
& alpha*beta = 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