Is a Heptagon Constructible with Straightedge and Compass?

  • Thread starter Thread starter saadsarfraz
  • Start date Start date
saadsarfraz
Messages
86
Reaction score
1
constructible angles-->help please

Homework Statement



x=2(pi)/7 we will show that this is not constructible and therefore 7-gon is not constructible.

a) show cos4x = cos3x
b) Use the above equation to find a rational quartic polynomial f(y)
where f(cos x) = 0.
c)From f, find a cubic rational polynomial g(y) where g(cos x) = 0

Homework Equations



see above

The Attempt at a Solution



im having trouble in part b). i expanded cos4x - cos3x = 0 in terms of cos(x) and I made the substitution y= cos(x) i got the quartic equation 8y^4 + 4y^3 - 8y^(2) -3y + 1 =0.
but when i put y= cos(2(pi)/7)) it dosent come out to 0.
 
Physics news on Phys.org


You found cos 4x + cos 3x instead of minus. LOL, I made the same error.
 


oh right it should be -4 and +3 in the above equation, thanks billy do u know how to change this into a cubic equation.
 


Try long division? Divide by y-c, where c is a root of the quartic. Hopefully c is easy to find, by graphing, or guessing.
 


i tried doing that the root is 1 so i divided it by y-1 I am getting 8y^3 + 4y^2 - 4y +7 + some remainder?
 


oh i got it, thank you
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top