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Roots of a polynomial

  1. Oct 16, 2004 #1
    Hey guys can you help me solve this problem:

    Find all real numbers [tex]a[/tex] with the property that the polynomial equation [tex]x^{10} + ax + 1 = 0[/tex] has a real solution [tex]r[/tex] such that [tex]1 / r[/tex] is also a solution.

    Thank you for helping me :smile:
    Sry I have posted this problem once before, but nobody helped me :frown: and I need to know how to solve the problem badly. Please help me :smile:
     
    Last edited: Oct 16, 2004
  2. jcsd
  3. Oct 16, 2004 #2
    R is the solutions to this equation:

    [tex]x^{10} - \frac{x}{x^{-1}+x^9} +1 = 0[/tex]
     
  4. Oct 16, 2004 #3
    Thanks for helping me, you r awesome :smile: But I dont understand how you get there, can u explain?
     
    Last edited: Oct 16, 2004
  5. Oct 16, 2004 #4

    krab

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    plug 1/r into the equation, multiply the result by r^10. Now you have two 10th-order equations for r. Subtract them. The rest is easy.
     
  6. Oct 16, 2004 #5
    You have two equations in two unknowns, so use systems of equations or some other such thing to solve.

    For your conditions to be met both of the following must be true.

    x^10 + rx + 1 = 0

    x^10 +1/rx + 1 = 0

    solve for r in the second equation.

    r = -1/(x^-1 + x^9)

    plug in r into the first equation, and you get what I got.
     
  7. Oct 16, 2004 #6
    Thanks for helping me Krab, your reply is refering to the orginial thread right?
     
    Last edited: Oct 16, 2004
  8. Oct 16, 2004 #7
    First, Thanks for helping me JonF. But you r saying that [tex]r = a[/tex] and [tex]1 / r = a[/tex]. However, the problem said that the polynomial [tex]x^{10} + ax + 1 = 0[/tex] has roots [tex]r[/tex] and [tex]1 / r[/tex]. Which implies that [tex]x = r[/tex] and [tex]x = 1 / r[/tex].
    I think Krab is right
     
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