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Question
If α,β are zeros of the polynomial f(x)=x2−c(x+1)−c, then (α+1)(β+1)=
If α,β are zeros of the polynomial f(x)=x2−c(x+1)−c, then (α+1)(β+1)=
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Solution
The correct option is B 1−c
If α, β are the zeros of the quadratic polynomial f(x)=x2−c(x+1)−c=x2−cx−c−c=x2−cx−2c, then,
(α+β)=−ba=−−c1=c,αβ=ca=−2c1=−2c
Now,
(α+1)(β+1)
=αβ+α+β+1
=−2c+c+1
=1−c
If α, β are the zeros of the quadratic polynomial f(x)=x2−c(x+1)−c=x2−cx−c−c=x2−cx−2c, then,
(α+β)=−ba=−−c1=c,αβ=ca=−2c1=−2c
Now,
(α+1)(β+1)
=αβ+α+β+1
=−2c+c+1
=1−c
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