Theory of Equations 1 Question 20

21. Let α and β be the roots of x26x2=0, with α>β. If an=αnβn for n1, then the value of a102a82a9 is

(a) 1

(b) 2

(c) 3

(d) 4 (2011)

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Solution:

  1. a102a82a9=(α10β10)2(α8β8)2(α9β9)

=α8(α22)β8(β22)2(α9β9)

α is root of x26x2=0α22=6α

[and β is root of x26x2=0β22=6β ]

=α8(6α)β8(6β)2(α9β9)=6(α9β9)2(α9β9)=3



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