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Question

If lf α, β are the roots of the equation :
(π1)λ2+2λ+1=0
αβ+βα=?

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Solution

(π1)λ2+2λ+1=0
Here, a=π1,b=2,c=1
αβ=ca=1π1 ----- ( 1 )

α+β=ba=2π1 ------- ( 2 )

(α+β)2=α2+β2+2αβ
(2π1)2=α2+β2+2×1π1 [ Using ( 1 ) and ( 2 ) ]

4(π1)22π1=α2+β2

42π+2(π1)2=α2+β2

α2+β2=2π+6(π1)2 ----- ( 3 )
Now,
αβ+βα=α2+β2αβ

=2π+6(π1)21π1

=2π+6(π1)2×(π1)1

=2π+6π1

=2×227+62271

=44+42227

=215

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