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Solving 4th degree equation

  1. Jul 26, 2012 #1
    1. The problem statement, all variables and given/known data
    From the higher order linear homogeneous differential equation, I get the characteristic equation:
    r^4 + r^3 -7r^2 -r + 6 = 0
    solve for r.


    2. Relevant equations
    how do you do this?


    3. The attempt at a solution
    Even if I factor out the r from the first 4 terms on the LH, i get no where. I wish in highschool they went over this stuff, but they seriously didn't
     
  2. jcsd
  3. Jul 26, 2012 #2

    SammyS

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    It's fairly easy to see that r = 1 is a solution.

    Almost as easy to see that r = -1 is also a solution.
     
  4. Jul 26, 2012 #3

    Simon Bridge

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    nmbr42: the usual approach is to guess, using your experience of how the equations work, then use long division to remove the guessed factors. It is usually taught in junior High School. Possibly you were sick that day or more interested in other things.

    How easy it is to realize that 1 and -1 are roots does kinda depend on your experience with these things... but you do learn how to make good guesses as you become familiar with the way polynomials behave.

    In a pinch you can use Newton/Raphson to get the first root, or just use a computer to plot the graph :) and there are general solutions online.
     
  5. Jul 26, 2012 #4

    HallsofIvy

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    One important guide to making "guesses" is the "rational root theorem".

    "Any rational number, satisfying [itex]a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0[/itex], is of the form m/n where m evenly divides the "constant term", [itex]a_0[/itex] and n evenly divides the "leading coefficient", [itex]a_n[/itex]."

    Of course, it is always possible that a polynomial equation does not have any rational roots but we can tell that if this equation has rational roots, they must be integers (because the leading coefficent is 1 and only 1 divides 1) and must be 1, -1, 2, -2, 3, -3, 6, or -6 (because the constant term is 6 and only those numbers divide 6).
     
  6. Jul 26, 2012 #5
    I actually used this link to help me since Simon Bridge mentioned about long division:
    http://www.purplemath.com/modules/polydiv3.htm

    quite simple straight forward method.
    the roots are -3,2,1, and -1.
    Since -1 and 1 are guesses that actually worked, I first divided the equation by (r-1) then (r+1). Then at last you get the equation
    r^2 + r -6 =0
     
  7. Jul 26, 2012 #6

    Simon Bridge

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    no worries... it's all good and glad we could help :)
     
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