Question

If \( \alpha \) and \( \beta \) are zeroes of the polynomial \( x^{2}-p(x+1)+c \) Such that \( (\alpha+1)(\beta+1)=0 \), then find the value of \( C \).

Answer 1

We have,

f(x) = x² −p(x+1)−c = 0

f(x) = x²−px−(p+c) = 0

Since,

α,β are the zeroes of the above polynomial.

So,

α + β = p

αβ = −(p + c) 

Since,

(α + 1)(β + 1) = 0

αβ + α + β + 1 = 0

−p − c + p + 1 = 0

−c + 1 = 0

c = 1