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Roots of equation

  1. May 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the number and position of real roots of x4+4x3-4x-13=0.

    2. Relevant equations

    3. The attempt at a solution
    I found the number of real roots using the Descarte's rule of signs. One is positive and other is negative. Now about the position of roots, i don't have any idea. Substituting the values of x and checking if f(x)=0 won't be a good idea. Also i can't factorize the given expression. I can't think of any other way for finding the roots.

    Is it possible to find the roots using calculus? If yes, please tell me, i would be happy to know about that.

    Thanks! :smile:
  2. jcsd
  3. May 16, 2012 #2
    Use the rational root test.
  4. May 16, 2012 #3
    I don't think it will work here.
  5. May 16, 2012 #4

    I like Serena

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    Hey Pranav!

    There are a couple of algorithms to find non-nice roots in a quartic polynomial.
    You can find a selection of them on wiki:

    However, to calculate them is quite involved.
    And in the end they will give you the solutions that wolframalpha also gives.
    If you take a look at wolframalpha, you'll see that the exact solutions are quite complex.

    Alternatively there are other ways, like Newton-Raphson to approximate the roots.
    Or else you can "bracket" the roots.
    That is, find a (small) interval in which each of the roots must be.
  6. May 16, 2012 #5
    Thank you ILS for the reply but is there any calculus based way to find the roots?
  7. May 16, 2012 #6

    Ray Vickson

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    No. However, you could regard Newton's method (which is an iterative algorithm to find successively better approximations to a root) as calculus-based.

  8. May 16, 2012 #7
    Thanks for the clarification, i will look into that. :smile:
  9. May 16, 2012 #8


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    The question asked for "number" and "position" of real roots. You might be able to simplify your work a teensy tiny bit by doing a rough sketch of the curve. Even this is tough, when you differentiate to find the turning points, you get a cubic with a discriminant greater than zero, meaning there are 3 real distinct turning points, none of which can be found "easily" (e.g. with RRT). You'll definitely need to use Newton's method here. Or you can just roll up your sleeves and go through the tedious algebraic process for an exact solution (which, after you're done, will likely look ugly enough that you'll be wondering why you even bothered!). :biggrin:
  10. May 16, 2012 #9
    Graphing? I am always poor at that. :biggrin:
    I still need to look into the Newton's method on Wikipedia, there is a huge information about that on Wikipedia. I guess i won't look into the algebraic process as you have already made me aware of the results. :P

    I am lucky enough as i have found a way of solving equations of degree 4, i found these in one of my elder brother's engineering books so i won't have to refer Wikipedia for those. :tongue2:
    The methods given in the book are Ferrari's method and Descarte's method.
  11. May 16, 2012 #10

    Ray Vickson

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    There might only be one turning point, since it is possible that two of the roots of the cubic are complex.

  12. May 16, 2012 #11

    Ray Vickson

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    Nowadays there are easier ways. For example, you can use Wolfram Alpha http://www.wolframalpha.com . Just type the formula x^4 +4*x^3 - 4*x - 13 and press 'enter'---see what you get! Note: there is no cost other than internet connection charges.

  13. May 16, 2012 #12
    But there are two real roots here in this case.
  14. May 16, 2012 #13


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    Not in this case. The cubic discriminant of the derivative is positive. http://en.wikipedia.org/wiki/Discriminant#Cubic
  15. May 16, 2012 #14
  16. May 16, 2012 #15


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    A general iteration that converges nicely comes from the re-arrangement:

    4x³ = 13 + 4x - x⁴

    i.e., x = ∛( (13 + 4x - x⁴)/4 )

    1.0 → 1.588 → 1.48 → .....

    I understand you wanting a method with a better guarantee than the hit and miss of a general iteration re-arrangement, but the lure of its "lucky dip" outcome always beckons me to give it a try. :smile:

    If you need certainty: http://m.wolframalpha.com/input/?i=x^4++4*x^3+-+4*x+-+13=0&x=6&y=4&js=off
  17. May 17, 2012 #16


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    Yes, but how much does that really help you? You can tell that there's exactly one positive root, but there may be either 3 or 1 negative root(s).
  18. May 17, 2012 #17
    We can find the number of negative roots by the same rule. Just plugin -x at the place of x and see how many times the sign changes from negative to positive. ;)
  19. May 17, 2012 #18


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    Yes, but the rule does not fix the number of negative roots down in this case. There are 3 sign changes - that means there can be *either* 3 *or* 1 negative root. In this case, it's the latter (if you graph the curve), but there's no way to exclude the former possibility (3 negative roots) without graphing the curve.
    Last edited: May 17, 2012
  20. May 17, 2012 #19
    How?? :confused:
    I only see one sign change.
    Substituting -x i get x4-4x3+4x-13=0.
    I see only one sign change, from negative sign to positive sign. There is only one negative root.
  21. May 17, 2012 #20


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    +x^4 -> -4x^3 : one sign change
    -4x^3 -> +4x: one sign change
    +4x -> -13: one sign change

    3 sign changes, implying 3 or 1 negative root(s).
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