Question

If α and β are the zeroes of the quadratic polynomial f(x) = x2 – p(x + 1) – c, show that (α + 1)(β + 1) = 1 – c.


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Answer 1

Solution:

Since, α and β are the zeroes of the quadratic polynomial

f(x) = x2 – p(x + 1)– c

Now,

Sum of the zeroes = α + β = p

Product of the zeroes = α × β = (- p – c)

So,

(α + 1)(β + 1)

= αβ + α + β + 1

= αβ + (α + β) + 1

= (− p – c) + p + 1

= 1 – c = RHS

So, LHS = RHS

Hence, proved.

Answer 2

Answer:-

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

Hope it helps