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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 \( C \). |
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Answer» 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 |
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