Question

if alpha and beta are the zeros of the quadratic polynomial f(x) = x²-p(x+1)-c, show that ( alpha + 1)(beta + 1) = 1-c

Answer 1

If alpha and beeta are zeroes of polynomial x2-p(x+1)+c such that (alpha+1)

If alpha and beeta are zeroes of polynomial x2-p(x+1)+c such that (alpha+1)(beeta+1)=0. Then find the value of c.

A)

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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

Answer 2

Hey mate,

From the above equation we conclude,

Alpha + beta = p....(1)

Alpha*beta = - p - c..(2)

Then ATQ,

L.H.S = (alpha+1)(beta+1)

= alpha*beta + alpha + beta + 1

From (1) and (2),

= - c - p + p + 1

= 1 - c

= R. H. S

Hope this helps you out!