Oh!!!Come On!!!!

Algebra Level 3

Find A


The answer is 3.14.

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2 solutions

Parth Lohomi
Nov 15, 2014

The roots of the equation will be -w(omega) & w 2 -w^{2} ..

Let alpha = -w & beta = w 2 -w^{2} ...

so ( a l p h a ) 2009 (alpha)^{2009} will be w 2009 -w^{2009} = - w 2007 w^{2007} * w 2 w^{2}

since 2007 is a multiple of 3 and w ( m u l t i p l e o f 3 ) w^{(multiple of 3)} is 1 so w 2007 w^{2007} =1.

..so ( a l p h a ) 2009 (alpha)^{2009} =- w 2 w^{2 } .

..Now ( b e t a ) 2009 (beta)^{2009} = - w 4018 w^{4018} = - w 4017 w^{4017} w

again 4017 is a multiple of 3 so w 4017 w^{4017} =1..

..so ( b e t a ) 2009 (beta)^{2009} = -w.

..therefore a l p h a 2009 alpha^{2009} + b e t a 2009 beta^{2009} = - w 2 w^{2} - w =(- w 2 w^{2} +w) = -(-1) ( since w 2 w^{2} +w+1=0) =1.

..so p i e a \frac{pie}{a} =1 so a= pie =3.14...

Ankush Gogoi
Sep 13, 2014

The roots of the equation will be : -w(omega) & -w^2(omega square).. Let alpha = -w & beta = -w^2....so (alpha)^2009 will be -w^2009 = - (w^(2007) *w^2 ) since 2007 is a multiple of 3 and w^(multiple of 3) is 1 so w^2007 =1...so (alpha)^2009 =-w^2 ...Now (beta)^2009 = -w^4018 = - w^4017 * w again 4017 is a multiple of 3 so w^4017 =1....so (beta)^2009 = -w...therefore alpha^(2009) + beta^(2009) = -w^2 - w = -(w^2 +w) = -(-1) ( since w^2+w+1=0) =1...so pie/a =1 so a= pie =3.14...

You should have used LATEX but nice solution

Parth Lohomi - 6 years, 9 months ago

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Thnkz...but I don't know how to use LATEX.

Ankush Gogoi - 6 years, 9 months ago

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No probs pal!!

Parth Lohomi - 6 years, 8 months ago

I have posted a laTexified solution

Parth Lohomi - 6 years, 7 months ago

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