Question

if alpha and beta are zeroes of x2-p(x+1)-c prove (alpha+1)(beta+1)=1-c

Answer 1

Answer:

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Step-by-step explanation:

= x² - p(x+1)-c = x² -px-p -c

compare it with ax²+bx+c =0

a= 1, b= -p , c= -p-c  

α and β are two zeroes

i) sum of the zeros= -b/a

α+β = - (-p)/1= p -----(1)

ii) product of the zeroes = c/a

αβ = (-p-c) /1 = -p-c -----(2)

now take  

lhs = (α+1)(β+1)

 = α(β+1) +1(β+1)

=αβ +α +β +1

=-p-c+p+1  [from (2) and (1) ]

= -c+1

=1 -c

=RHS