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Simple polynomial question

  1. Dec 8, 2006 #1
    Hi, I was just reading up on some material for one of my exams and came across something. I realize this may seem like a trivial, obvious question but:
    What is the easiest way to find the solutions of the polynomial (for example):
    (x^2) - (1/6)x - 1/6 = 0
    Other than guessing and testing for the correct values of x, what is the quickest and most efficient method of solving for x when the coefficients of each term is a fraction rather than a whole number? Thanks in advance.
  2. jcsd
  3. Dec 8, 2006 #2

    Gib Z

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    Homework Helper

    well you just let your fraction, 1/6, equate to b, as the equation was a quadratic of the form [itex]ax^2 + bx + c = 0[/itex]. Then substitute your values for a, b and c into the quadratic formula, [tex]x=\frac {-b(+-)(b^2 - 4ac)^0.5}{2a}[/tex] If you mean in general, you it is quite hard to solve equations of degree 3 and 4 analytically, and after 5 it can only be approximated using methods such as Newtons Method.
  4. Dec 8, 2006 #3
    Given the equation

    [tex]x^2 + px + q = 0[/tex]

    the solutions are:

    [tex]x_{1} = - \frac{p}{2} - \sqrt{ \left( \frac{p}{2} \right)^2-q }[/tex]


    [tex]x_{2} = - \frac{p}{2} + \sqrt{ \left( \frac{p}{2} \right)^2-q }[/tex]

    So in your example:

    p=-1/6 and q=-1/6
    Last edited: Dec 8, 2006
  5. Dec 8, 2006 #4
    Thanks alot!
  6. Dec 8, 2006 #5
    one more thing: if im trying to find the solutions if the polynomial is of order 3, for example:
    x^3 + 0.5x^2 - 0.25x - 0.125 = 0

    Do I add 0.125 to both sides then factor an x or do must I do something else? Thanks
  7. Dec 8, 2006 #6
  8. Dec 9, 2006 #7
    Wow what a long and involved process :cry:
    There is no other process of finding them? oh well
  9. Dec 10, 2006 #8
    You can try to find (adjust) one solution ([tex]x=0.5[/tex]). After that you can divide [tex]x^3 + 0.5x^2 - 0.25x - 0.125[/tex] by [tex]x-0.5[/tex].
    The result is [tex]x^2+x+0.25[/tex].
    So [tex]x^3 + 0.5x^2 - 0.25x - 0.125=(x-0.5)(x^2+x+0.25)=0[/tex]
    You already can solve it.
    Last edited: Dec 10, 2006
  10. Dec 10, 2006 #9


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    Gold Member
    Dearly Missed

    Not all problems easily stated are easily solved.

    Most happen to be insoluble, I'm afraid.
  11. Dec 11, 2006 #10
    I don't think that ACLerock solve very difficult problems.
    It is possible to apply more easy methods when solving home and exam tasks.
    (for example the method from my post).
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