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Constructible roots

  1. Apr 8, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    see above

    3. 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 cant find a suitable substitution in terms of x which would change this into a proper cubic equation and see if it has rational roots.
     
  2. jcsd
  3. Apr 8, 2009 #2
    Typo? It has no real roots at all.
     
  4. Apr 9, 2009 #3
    how do i show it then? by assuming x=p/q as a rational root and then going on from there?
     
  5. Apr 9, 2009 #4
    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 dont know how.
     
  6. Apr 9, 2009 #5

    Hurkyl

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    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)
     
  7. Apr 9, 2009 #6
    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.
     
  8. Apr 9, 2009 #7
    No, you have to use a cubic for the theorem to apply.

    How about w=x^2?
     
  9. Apr 9, 2009 #8

    HallsofIvy

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    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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