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Yay,I got it.One of the soloutions of the equation 64x^6-80x^4+8x^3+24x^2-4x-1=0 is the exact value of 

cos(2pi/11),which is a cosine value of central angle of hendecagon.(a regular 11-gon) 

 Aug 15, 2015

Best Answer 

 #2
avatar+33654 
+5

This is what I get, numerically, using MathCad:

 numerical solution of algebraic eqn.

.

 Aug 16, 2015
 #1
avatar+118696 
+5

Well that is great :)

 

64x^6-80x^4+8x^3+24x^2-4x-1=0

64*(cos(2pi/11))^6-80*(cos(2pi/11))^4+8*(cos(2pi/11))^3+24*(cos(2pi/11))^2-4*(cos(2pi/11))-1

 

64×cos360(36011)680×cos360(36011)4+8×cos360(36011)3+24×cos360(36011)24×cos360(36011)1=0.0000000000044649

 

Yes it looks like you might be right - it is certainly is very close to zero :)

 

I am not so sure that Wolfram|Alpha agrees with you though.  

http://www.wolframalpha.com/input/?i=%2864x%5E6-80x%5E4%2B8x%5E3%2B24x%5E2-4x-1%29%2F%28x-cos%282*pi%2F11%29

 

 

Why do you think it is exact?

 Aug 16, 2015
 #2
avatar+33654 
+5
Best Answer

This is what I get, numerically, using MathCad:

 numerical solution of algebraic eqn.

.

Alan Aug 16, 2015
 #3
avatar+118696 
0

Thanks Alan,

I'm still scratching my head on this one. (contemplation and confusion - not nits   LOL )

 Aug 16, 2015

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