1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rational Root Thorem

  1. Jun 22, 2013 #1

    Nugso

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    x4 - 4(x3) + 3(x2) -2x +1 = 0


    2. Relevant equations

    Rational Root Theorem, q/p



    3. The attempt at a solution

    Hello everyone. Today, I've learnt the rational root theorem( it's a bit late, isn't it? :( ) and thus wanted to see how it works. According to the rational root theorem the roots must be ± 1, but wolframalpha says the roots are x ≈ 0.672378 and x ≈ 3.23402. ( It also says there're also complex roots, but it's not what I'm looking for).

    What am I doing wrong?
     
  2. jcsd
  3. Jun 22, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    The example equation is not the best candidate for using the RRT to suggest possible roots, because both a0 and an = 1. Once you test both x = 1 and x = -1 as possible solutions and find that they are not, then the RRT doesn't provide much further help in suggesting solutions.
     
  4. Jun 22, 2013 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hello Nugso! :smile:
    Ah, but ≈ 0.672378 and ≈ 3.2340 aren't rational! :wink:
     
  5. Jun 22, 2013 #4

    Nugso

    User Avatar
    Gold Member

    @SteamKing

    Well, let's assume for x=1 the equation is fine, then would it be OK to say the only root is 1? Or do I have to seek more?

    @tiny-tim

    It's because the "≈", right? Sorry for being sort of dummy. :shy:


    Thanks for the replies SteamKing and tiny-tim, by the way.
     
  6. Jun 22, 2013 #5

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Yup!

    Those roots are irrational, hence the "≈" :smile:
     
  7. Jun 22, 2013 #6

    Nugso

    User Avatar
    Gold Member

    Thank you very much!
     
  8. Jun 22, 2013 #7

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    When both a0 and an are equal to 1, the only thing about the roots which the RRT says is that if 1 or -1 are NOT the solutions, then 1 or -1 will be a factor of the true solution. This doesn't help, since 1 or -1 is a factor of any number. Once 1 or -1 are shown not to be solutions, then the RRT really provides no further help in suggesting possible solutions.
     
  9. Jun 22, 2013 #8

    Nugso

    User Avatar
    Gold Member

    Thank you very much SteamKing. I think I understand now. If I may, I'd like to ask one more question; how could I solve such problems without RRT? I know it heavily depends upon the problems, but are there any other theories out there to find roots?
     
  10. Jun 22, 2013 #9

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

  11. Jun 23, 2013 #10

    Nugso

    User Avatar
    Gold Member

    Thanks again SteamKing. The links, especially Decartes Rule of Sign, were of help.
     
  12. Jun 23, 2013 #11

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Although there are formulas for the roots of third and 4th degree polynomials, they are rarely used in practice. Usually we solve such problems numerically, using any one of a number of effective computational algorithms.

    Just so you know: for equations of 5th or higher order there are no nice formulas for finding roots, and in such cases (if the rational root theorem fails) we are forced to fall back on numerical methods. Interestingly, it is NOT just the case that nobody has been smart enough to find a formula for 5th degree polynomials; it has actually been *proven* that no such formula can possibly exist!
     
  13. Jun 23, 2013 #12

    Nugso

    User Avatar
    Gold Member

    Aha! I recall seeing that there's no formula for 5th degree polnomials. Though I did not know it actually had been proven that no such formula can possibly exist. Sounds interesting to me.

    I just looked up the formula for 4th degree polynomials and it looks rather difficult to remember. Thank god they're rarely used as you said. :biggrin:

    Thank you very much for the reply.
     
  14. Jun 23, 2013 #13

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    French Mathematician Evariste Galois is the man who proved that no solutions exist to polynomial equations of degree 5 and higher involving the extraction of roots.

    http://en.wikipedia.org/wiki/Évariste_Galois

    He was 21 when he did so and reportedly wrote his proof down the night before his execution during one of the French revolutions/purges in the 19th century. Galois' proof was one of the signal events which started the study of groups and group theory in mathematics.
     
  15. Jun 23, 2013 #14

    Nugso

    User Avatar
    Gold Member

    Wow! These forums never cease to amaze me. Last week I was reading a book by Jerry P. King and in the book he tells a story of Evariste Galois, how much he loved maths, how he died(duel) and now I encountered him on these forums!

    Sorry if I'm being off-topic. Thank you very much!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted