Let-be-a-root-of-x-5-x-3-x-2-0-Then-prove-that-6-3-where-denotes-greatest-integer-less-than-or-equal-

Question Number 121099 by Anuragkar last updated on 05/Nov/20

L e t α b e a r o o t o f x^{5} - x^{3} + x - 2 = 0

T h e n p r o v e t h a t [α^{6}] = 3 w h e r e [λ] d e n o t e s g r e a t e s t i n t e g e r

l e s s t h a n o r e q u a l λ

Answered by TANMAY PANACEA last updated on 05/Nov/20

f (x) = x^{5} - x^{3} + x - 2

f (0) < 0

f (1) < 0

f (2) > 0

s o r o o t 2 > α > 1

u s i n g g r a p h a p p . . α = 1.206 \to α^{6} \approx 3.08

[α^{6}]

= [3.08] = 3

Commented by TANMAY PANACEA last updated on 05/Nov/20

Commented by Anuragkar last updated on 13/Nov/20

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