If 'a'&'b'are the zeroes of the quadratic polynomial f(x)=x²-p(x+1)-c,show that (a+1)(b+1)=1-c
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x^2 - p(x + 1) - c
= x^2 - px - p - c
Given that x^2 - px - p - c has zeroes a and b
Sum of roots = a + b = -(-p)/1 = p
Product of roots = ab = (-p - c)/1 = -p -c
We have to prove that (a + 1)(b + 1) = 1 - c
LHS
= (a + 1)(b + 1)
= ab+ a + b + 1
Substitute the values of αβ and a + b
= - p - c + p + 1
= 1 - c
= RHS
Hence Proved
= x^2 - px - p - c
Given that x^2 - px - p - c has zeroes a and b
Sum of roots = a + b = -(-p)/1 = p
Product of roots = ab = (-p - c)/1 = -p -c
We have to prove that (a + 1)(b + 1) = 1 - c
LHS
= (a + 1)(b + 1)
= ab+ a + b + 1
Substitute the values of αβ and a + b
= - p - c + p + 1
= 1 - c
= RHS
Hence Proved
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