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Root of the symmetric equation

  1. Mar 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve this equation, and find x.

    [tex]6x^5-5x^4-29x^2-5x+6=0[/tex]

    2. Relevant equations

    if [tex]x= \alpha[/tex] is root of the symmetric equation, then [tex]x= \frac{1}{\alpha}[/tex], is also root of the symmetric equation

    3. The attempt at a solution

    I tried first to write like this

    [tex]6x^5-5x^4-30x^2+x^2-5x+6=0[/tex]

    and then multiplying by five

    [tex]30x^5-25x^4-150x^2+5x^2-25x+30=0[/tex]

    then trying to divide with [tex]x^3[/tex]

    [tex]30x^2-25x-150 \frac{1}{x} +5 \frac{1}{x} -25 \frac{1}{x^2}+30 \frac{1}{x^3}=0[/tex]

    then adding [tex]-30x^3+30x^3[/tex]

    [tex]-30x^3+30x^3 + 30x^2-25x-150 \frac{1}{x} +5 \frac{1}{x} -25 \frac{1}{x^2}+30 \frac{1}{x^3}=0[/tex]

    And I am stuck in here. Please help me. Thank you.
     
  2. jcsd
  3. Mar 9, 2008 #2

    tiny-tim

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    Hi Physicsissuef!

    Be logical. :cool:

    1. How many linear factors (ie, of the form x+a) are there?

    2. Using the hint they gaave you, what can you say about all those linear factors? :smile:
     
  4. Mar 9, 2008 #3
    Hmm... I don't know, what you want to say.
     
  5. Mar 9, 2008 #4

    tiny-tim

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    [tex]6x^5-5x^4-29x^2-5x+6[/tex] is fifth-order (in other words, it starts with x^5).

    So it has five linear factors, which means it equals:
    (x + a)(x + b)(x + c)(x + d)(x + e), for some values of a b c d and e.

    So, using the hint they gaave you, what can you say about a b c d and e? :smile:
     
  6. Mar 9, 2008 #5
    [tex](x^5 + \frac{1}{x^5})(x^4+ \frac{1}{x^4})+(x^3+ \frac{1}{x^3})+(x^2+ \frac{1}{x^2})+(x+ \frac{1}{x})[/tex]
    like this? What's next? :smile:
     
  7. Mar 9, 2008 #6

    tiny-tim

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    No - I meant with a b c d and e constants.

    So (x + a) etc are first-order in x (linear).

    like this: (x + a)(x + b)(x + c)(x + d)(x + e).

    (so a b c d and e are like the [tex]\alpha[/tex] in the hint.)

    Try again - what can you say about a b c d and e? :smile:
     
  8. Mar 9, 2008 #7
    but how I will find a,b,c,d,e? This equation obviously doesn't have real roots.
    a,b,c,d,e are complex numbers.
     
  9. Mar 9, 2008 #8

    tiny-tim

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    (i) That doesn't seem obvious to me. Why do you think that?

    (ii) The hint they gave you applies to complex numbers also.

    (iii) I'm not asking you to find a b c d and e yet - I'm only asking what can you say about them?

    Can they all be complex numbers?

    And if, say, a and b are complex numbers, what does that tell you about a + b or a - b? :smile:
     
  10. Mar 9, 2008 #9
    I don't know. Can we please start solving this equation?
     
  11. Mar 9, 2008 #10

    HallsofIvy

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    You can start any time you like.

    But why would you simple assert that "This equation obviously doesn't have real roots."
    Any odd degree polynomial equation "obviously" has a least one real root! Do you see why?

    It is also true that complex roots of polynomial equations with real coefficients come in pairs. That was tiny-tim's point.
     
  12. Mar 9, 2008 #11
    And how can I start? Should I combine the elements somehow?
     
  13. Mar 9, 2008 #12

    Hootenanny

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    I would start by expanding the brackets, then you can compare the coefficients.
     
  14. Mar 9, 2008 #13
    If you sow above, I tried with -30x^2+x^2, can you please give me, some hint how to expand the brackets, please?
     
  15. Mar 9, 2008 #14

    tiny-tim

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    (btw, thanks, HallsofIvy!)
    No, don't start that way … it's horrible! :frown:

    Start with the hint:
    If (x + a) is a factor, then so is (x + 1/a).​

    Now, what does that tell you about b c d and e? :smile:
     
  16. Mar 9, 2008 #15

    Hootenanny

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    Oops, the hint does make it a little easier... :blushing:

    [Hides in the corner]
     
  17. Mar 9, 2008 #16
    tiny-tim can you please start, with solving the equation, just start... I will continue...
     
  18. Mar 9, 2008 #17

    Tom Mattson

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    Um...guys? This equation doesn't look too "symmetric" to me. When you plug in [itex]x=\frac{1}{\alpha}[/itex], you get the following.

    [tex]\frac{6}{\alpha^5}-\frac{5}{\alpha^4}-\frac{29}{\alpha^2}-\frac{5}{\alpha}+6=0[/tex]

    Multiplying both sides by [itex]\alpha^5[/itex] gives the following.

    [tex]6-5\alpha-29\alpha^3-5\alpha^4+6\alpha^5=0[/tex].

    That's not the same equation.

    Physicsissuef, can you please look carefully at the exercise you were given to make sure that there is not supposed to be another term in there, namely [itex]-29x^3[/itex]?
     
  19. Mar 9, 2008 #18
    No, there isn't. I asked also my professor, but also he told me that there is some problem with -29. I don't know even if this task is possible to solve. I tried the others, and solved them. Also I sow the results in the book, and they wrote [tex]x_1[/tex]=1 which is not correct. So maybe they have typo error.
     
  20. Mar 9, 2008 #19

    tiny-tim

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    hmm … never spotted that … :redface:

    It should be [tex]6x^5-5x^4-29x^3-29x^2-5x+6=0[/tex].

    Take my word for it, this factors very nicely. :smile:

    (So nicely, you could actually guess it!)
    (But no telling, please, except of course for Physicsissuef!)
     
  21. Mar 9, 2008 #20

    Tom Mattson

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    There's no "maybe", if the hint that was given to you is supossed to mean anything, then that term should be there.
     
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