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Problem of solving the cubic function

  1. Apr 14, 2014 #1
    Guys, I may need your help. There is a question saying that how to solve the cubic function in general form, which means that y=ax^3+bx^2+cx+d. How do you guys solve for x? To be honest, I have no idea of this question. Probably, it uses the same way as the quartic function. Thanks!
     
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  3. Apr 14, 2014 #2

    micromass

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  4. Apr 14, 2014 #3
    There are numerous ways of solving cubic functions, but the most efficient way would be to plot it using some type of software--most easily maple, MATLAB, or Mathematica. Once you do that, locate one of the roots. Use the root as a factor, and divide the cubic by that factor to obtain a quadratic. Quadratics are easy to solve, thus you can easily find the remaining two roots.
     
  5. Apr 15, 2014 #4

    micromass

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    Well, if you're going to be content with numerical answers, then there are many good techniques to approximate the roots to a very high degree: http://en.wikipedia.org/wiki/Newton's_method
     
  6. Apr 16, 2014 #5

    disregardthat

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    For the cubic equation [itex]ax^3+bx^2+cx+d=0[/itex] (in your case the constant term is [itex]d-y[/itex], not [itex]d[/itex]), try substituting [itex]x = z +\frac{\gamma}{z}[/itex], and solve for [itex]z[/itex] by choosing the constant [itex]\gamma[/itex] correctly. If fairly certain that for a good choice of [itex]\gamma[/itex] (it will become apparant what [itex]\gamma[/itex] must be) you will end up with a quadratic function in [itex]z^2[/itex].

    This way you may arrive at the formula yourself, it's a neat exercise. You probably need to be careful verifying your solution afterwards, as [itex]z +\frac{\gamma}{z}[/itex] is not defined everywhere, and does not attain all values. To make calculations easier, you can assume [itex]a = 1[/itex] first, and make the necessary modification afterwards.
     
    Last edited: Apr 16, 2014
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