If alpha, beta are the zeroes of a quadratic polynomial x^2-p(x+2)-c, then prove that (alpha+2)(beta+2)-4+c=0
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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
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
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