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Root of f(x) = x^3-x-1 ?

  1. Apr 16, 2012 #1
    1. The problem statement, all variables and given/known data

    How does one find the root of f(x) = x^3 - x - 1 ? Quadratic Equation only works on power of 2. I can't factor out an x to get a first term of x^2 because then Quadratic equation still won't work because the middle and last term would be messed up, I think.

    What are the rules or the process?
     
  2. jcsd
  3. Apr 16, 2012 #2

    NascentOxygen

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    There is a formula, like the one you've memorized for quadratics, though the one for cubics is more frightening. Try a google search for it.

    Or you can sometimes see a solution by inspection, esp. integer solutions. Or you can use a numerical method to find an approximate solution, that approximation can be to whatever accuracy you desire. Or you could plot a graph and read off from there.

    Are you content to settle for a solution correct to 4 sig figs? :wink:
     
  4. Apr 16, 2012 #3

    NascentOxygen

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    It is always recommended that you sketch the curve so that you know what you are dealing with. There are plenty of online plotting resources, quickmath is a handy one. Here's a sample, but it's not of the exact function you have (I don't plan to do your work for you!) http://www.quickmath.com/webMathema...uation&v1=x^3+-+x+-+2=y&v2=-2&v3=2&v4=-4&v5=4

    Bookmark that site, it offers a lot more than just curve plotting. :smile:
     
  5. Apr 16, 2012 #4
    Use the rational root test, then test for each one using synthetic division.
     
  6. Apr 17, 2012 #5

    HallsofIvy

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    The rational root test won't give a root here because this polynomial has NO rational roots. The rational root test, for this polynomial, says that 1 and -1 are the only possible rational roots and they clearly are not roots.

    If a and b are any two numbers then [itex](a- b)^3= a^3- 3a^2b+ 3ab^2- b^3[/itex] and [itex]3ab(a- b)= 3a^2b- 3ab^2[/itex] so that [itex](a- b)^3+ 3ab(a- b)= a^3- b^3[/itex]. That means that if we let x= a- b, m= 3ab, and [itex]n= a^3- b^3[/itex], we have [itex]x^3+ mx= n[/itex].

    Now, suppose we know m and n. Can we solve for a and b and so find x? Yes, we can!

    From m= 3ab, we have b= m/3a. Putting that into [itex]n= a^3- b^3[/itex], we get [itex]n= a^3- m^3/3^3a^3[/itex]. Multiplying through by [itex]a^3[/itex] we have [itex]na^3= (a^3)^2- m^3/3^3[/itex] which is a quadratic [itex](a^3)^2- na^3- m^3/3^3= 0[/itex] for [itex]a^3[/itex].

    Solving that with the the quadratic formula,
    [tex]a^3= \frac{n\pm\sqrt{n^2+4\frac{m^3}{3^3}}}{2}= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}[/tex]

    Since [itex]a^3- b^3= n[/itex],
    [tex]b^3= a^3- n= -\frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2+ \left(\frac{m}{3}\right)^3}[/tex]
     
    Last edited: Apr 19, 2012
  7. Apr 18, 2012 #6
    You are right that in this case it wouldn't help, but it is a good rule of thumb in general.
     
  8. Apr 19, 2012 #7

    NascentOxygen

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    Very neat!

    Is it too late to fix the typo?
     
  9. Apr 19, 2012 #8

    HallsofIvy

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    Thanks. I have edited it.
     
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