Proving No Constructible Roots for x^(6) - x^(2) + 2 = 0

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Homework Help Overview

The discussion revolves around proving that the polynomial equation x^(6) - x^(2) + 2 = 0 has no constructible roots. Participants explore the nature of the roots and the implications of constructibility in the context of algebraic equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various substitutions and transformations of the original equation, including dividing by x^(3) and substituting x with y + 1. Questions arise about the nature of constructible numbers and how to demonstrate the irrationality of roots.

Discussion Status

The discussion is ongoing, with participants offering different perspectives on how to approach the problem. Some suggest using properties of constructible numbers and algebraic orders, while others express uncertainty about the next steps in their reasoning.

Contextual Notes

There is a mention of the need to show that the polynomial has no rational roots, and participants are considering the implications of irrational roots in relation to constructibility. The definitions and assumptions regarding constructible numbers are also being questioned.

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Homework Statement



prove that x^(6) - x^(2) +2 =0 has no constructible roots

Homework Equations



see above

The Attempt at a Solution



I have to divide the equation by x^(3) which would give me x^(3) - x^(-1) + 2x^(-2)= 0
I can't find a suitable substitution in terms of x which would change this into a proper cubic equation and see if it has rational roots.
 
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saadsarfraz said:
prove that x^(6) - x^(2) +2 =0 has no constructible roots

Typo? It has no real roots at all.
 
how do i show it then? by assuming x=p/q as a rational root and then going on from there?
 
ok i made the substitution x=y+1 and now I have this polynomial y^6 + 6y^5 + 15y^4 + 20y^3 + 14y^2 + 4y + 2 = 0, all i have to do is show that this has irrational roots but i just don't know how.
 
What precisely do you mean by "constructible"? In the version I'm familiar with, many complex numbers are constructible. (In particular, anything with constructible real and imaginary parts is constructible)
 
anything that can be drawn using a straight edge and a compass, the only problem i have is to show that the equation i have written has no rational roots.
 
saadsarfraz said:
ok i made the substitution x=y+1 and now I have this polynomial y^6 + 6y^5 + 15y^4 + 20y^3 + 14y^2 + 4y + 2 = 0, all i have to do is show that this has irrational roots but i just don't know how.

No, you have to use a cubic for the theorem to apply.

How about w=x^2?
 
Why would irrational roots tell you anything about constructible roots? Many irrational numbers, for example [itex]\sqrt{2}[/itex], are constructible.

It's hard to tell you how to proceed without know what fact, theorems, etc. you have to work with. If it were me, I would just use the fact that all constructible numbers are algebraic of order a power of 2. Then show that none of the roots of [itex]x^6- x^2+ 2= 0[/itex] (which is equivalent to [itex]y^3- y+ 2= 0[/itex] with [itex]y= x^2[/itex]) are algebraic of order 2 or 0.
 

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