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Question
If α and β are zeroes of the polynomial x2−a(x+1)−b such that (α+1)(β+1)=0, find the value of b.
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Solution
Given polynomial is x2−a(x+1)−b=0 x2−ax−a−b=0
x2−ax−(a+b)=0
α+β=a,αβ=−a−b.......... given(α+1)(β+1)=αβ+α+β+1=αβ+(α+β)+1=−a−b+a+1=0⇒b=1
Given polynomial is
x2−a(x+1)−b=0
x2−ax−a−b=0
x2−ax−(a+b)=0
α+β=a,αβ=−a−b.......... given
(α+1)(β+1)=αβ+α+β+1
=αβ+(α+β)+1
=−a−b+a+1=0⇒b=1
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