1.

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

Answer»

ANSWER:

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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