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Question
If α,β are zeros of polynomial x2−p(x+1)−k such that (α+1)(β+1)=6, then value of k is?
If α,β are zeros of polynomial x2−p(x+1)−k such that (α+1)(β+1)=6, then value of k is?
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Solution
The correct option is C −5
p(x)=x2−p(x+1)−k
⇒p(x)=x2−px−p−k
Since, α and β are zeroes of p(x)
so α+β=−(−p)1=p
and αβ=−p−k1=−p−k
Given : (α+1)(β+1)=6
⇒αβ+α+β+1=6
⇒p−p−k+1=6
⇒k=−5 which is option (4).
p(x)=x2−p(x+1)−k
⇒p(x)=x2−px−p−k
Since, α and β are zeroes of p(x)
so α+β=−(−p)1=p
and αβ=−p−k1=−p−k
Given : (α+1)(β+1)=6
⇒αβ+α+β+1=6
⇒p−p−k+1=6
⇒k=−5 which is option (4).
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