Question
If α and β are the roots of the equation x2−a(x+1)−b=0, then (α+1)(β+1)=
Answer: Option C
:
C
Given equation x2−a(x+1)−b=0
⇒ x2−ax−a−b⇒α +β =α,αβ = -(a + b)
Now(α+1)(β+1)=αβ+α+β+1=−(a+b)+a+1=1−b
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:
C
Given equation x2−a(x+1)−b=0
⇒ x2−ax−a−b⇒α +β =α,αβ = -(a + b)
Now(α+1)(β+1)=αβ+α+β+1=−(a+b)+a+1=1−b
Was this answer helpful ?
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