Is a Heptagon Constructible with Straightedge and Compass?

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Homework Help Overview

The discussion revolves around the constructibility of a heptagon using straightedge and compass, specifically focusing on the angle \( x = \frac{2\pi}{7} \). Participants are tasked with demonstrating that this angle is not constructible by deriving a polynomial equation related to cosine values.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss deriving a quartic polynomial from the equation \( \cos 4x = \cos 3x \) and the subsequent steps to find a cubic polynomial. There is an exploration of errors in the initial setup and attempts to correct them.

Discussion Status

The discussion is active with participants correcting each other's mistakes and providing suggestions for polynomial manipulation. Some guidance has been offered regarding polynomial division and finding roots, indicating a collaborative effort to clarify the problem.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the type of assistance they can provide to one another. There is an ongoing examination of the assumptions related to the constructibility of angles and the properties of the derived polynomials.

saadsarfraz
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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.
 
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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
 

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