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Question

lf α, β are the roots of x2pxcp=0, then α2+2α+1α2+2α+c+β2+2β+1β2+2β+c=

A
3
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B
2
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C
1
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D
0
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Solution

The correct option is A 1
The given equation can be written as,
x2px(p+c)=0
The basic properties of quadratic equations says,
α+β=p and αβ=(p+c)
From above two equation we get,
c=αβ+αβ
Now according to the question,
(α2+2α+1)(α22α+c)+(β2+2β+1)(β2+2β+c)
Then expression gets reduced to:
(α2+2α+1)(α22ααβαβ)+(β2+2β+1)(β2+2ββααβ)

=(α+1)2(α+1)(αβ)(β+1)2(β+1)(αβ)

=(α+1)(αβ)(β+1)(αβ)

=(α+1β1)(αβ)

=(αβ)(αβ)

=1

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