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Question

If α and β are the roots of the equation x2+3x+1=0, then the value of (α1+β)2+(βα+1)2 is equal to

A
18
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B
19
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C
20
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D
21
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Solution

The correct option is A 18
x2+3x+1=0
α+β=3; αβ=1
Also, α2+3α+1=0
and β2+3β+1=0
α2=(3α+1)
and β2=(3β+1)

Now, (α1+β)2+(βα+1)2
=α2(1+β)2+β2(α+1)2
=α21+2β+β2+β21+2α+α2
=((1+3α)β)+((1+3β)α)
=1+3αβ+1+3βα
=α(1+3α)+β(1+3β)αβ
=3(α2+β2)+(α+β) [αβ=1]
=3[(α+β)22αβ]+(α+β)
=3[92]+(3)=213=18

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