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Find roots of the EQN: r^3-r^2+1=0

  1. Mar 28, 2017 #1
    1. The problem statement, all variables and given/known data
    Find roots of the EQN: r^3-r^2+1=0

    2. Relevant equations
    none

    3. The attempt at a solution
    r^2(r-1)+1=0
    from there i solved, r^2=-1 and r-1=-1 to find the following roots:
    r=+i,r=-i, r=0

    Is my method correct? Also, I don't think that synthetic division would work here since my possible rational zeroes are +1/-1, and those don't satisfy the equation when they are plugged in for 'r'.
     
  2. jcsd
  3. Mar 28, 2017 #2

    PeroK

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    If you plug your solutions into the polynomial do you get 0?
     
  4. Mar 28, 2017 #3

    Mark44

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    No, this is not correct. If a * b = 0, then you know that either a or b must be zero, but if a * b = 1, then there are an infinite number of possible solutions for a and b.
    Even though you haven't found the roots of the equation, you learned from doing synthetic division that if r = 1, r^3 - r^2 + 1 > 0 and if r = -1, then r^3 - r^2 + 1 < 0. That implies that there must be a root of the equation somewhere between -1 and +1.
     
  5. Mar 28, 2017 #4

    epenguin

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    We can always solve this equation algebraically by a general method for solving cubics. However the question seems to be, can we without that find an easier method to see solutions depending on the apparent special simplifying features of this equation?

    How does the problem arise? If it is an excercise from a book, what is the chapter and maybe section about? Are there any similar problems solved? Same if a course of lessons.

    I tried a few plausible approaches but got nothing. At this point I recall we here have often been able to solve seemingly intractable problems after asking the poster to check whether he had copied it out right. :oldbiggrin:
     
    Last edited: Mar 28, 2017
  6. Mar 28, 2017 #5

    symbolipoint

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    If synthetic division does not give you remainder of 0, then the three possible roots would be irrational(?), not sure if all Real. I would turn to graphing software to check.

    (Google shows a graph of the function having just one real root. Near -0.76)
     
  7. Mar 29, 2017 #6

    epenguin

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    By hand it's fairly easy to see that this function has a maximum at x = 0, and a minimum at positive x and f so can have only one real root which must be negative.

    We now need the OP's confirmation and explanation of the origin of this problem.
     
    Last edited: Mar 29, 2017
  8. Mar 29, 2017 #7

    Mark44

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    These extrema would be local maximum and local minimum.

    I agree we should wait for the OP to weigh in again.
     
  9. Mar 31, 2017 #8

    epenguin

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    :doh: Somebody who apparently doesn't want to be identified reminded of what I had forgotten trying to do algebra on this, looks like this was an exercise in the trigonometrical solution.
     
    Last edited: Mar 31, 2017
  10. Mar 31, 2017 #9
    How ?
     
  11. Apr 2, 2017 #10

    epenguin

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    Well as the OP has not come back I feel a sufficient answer is in the unanswered #4
    There is something about cos3 x and cubics in all the textbooks.
     
    Last edited: Apr 3, 2017
  12. Apr 4, 2017 #11
    r=-i or r=+i,,,,,r=0
     
  13. Apr 4, 2017 #12

    stevendaryl

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    There is a way to solve cubic equations (which I wrote an Insights article about). Skipping to the punch line...

    If you have the cubic equation [itex]r^3 + A r^2 + B r + C = 0[/itex], then find three numbers [itex]a, b, c[/itex] such that:
    1. [itex]a = -A/3[/itex]
    2. [itex]3a^2 - 3bc = B[/itex]
    3. [itex]a^3 + b^3 + c^3 - 3abc = -C[/itex]
    Then one solution is [itex]r_1 = a + b + c[/itex]. How do you find those three numbers? Well, the first equation gives you [itex]a[/itex]. Then the second equation gives you [itex]c = \frac{a^2 - B/3}{b}[/itex]. Plugging that into the third equation gives:

    [itex]a^3 + b^3 + \frac{(a^2 - B/3)^3}{b^3} - 3a (a^2 - B/3) + C = 0[/itex]

    That seems like a mess, but it becomes simpler if you multiply through by [itex]b^3[/itex] it becomes:

    [itex]b^6 + (a^3 - 3a (a^2 - B/3) + C) b^3 + (a^2 - B/3)^3 = 0[/itex]

    Then you subtitute [itex]z = b^3[/itex], and it becomes a quadratic equation for [itex]z[/itex]. Solve for [itex]z[/itex], then get [itex]b[/itex] and then get [itex]c[/itex].

    My calculations gave a solution of [itex]a = 0.333, b = -0.114, c = -0.974[/itex], which gives a solution [itex]r_1 = -.755[/itex] (to 3 digits).
     
  14. Apr 4, 2017 #13
    What is the point in copying what already written.
     
  15. Apr 4, 2017 #14
    But those are just approximations.
     
  16. Apr 4, 2017 #15

    stevendaryl

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    Well, I sketched how to get an exact answer. What more do you want?
     
  17. Apr 4, 2017 #16

    stevendaryl

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    Letting [itex]K = (\frac{25 - \sqrt{621}}{54})^{\frac{1}{3}}[/itex], one solution is:

    [itex]r_1 = \frac{1}{3} - \frac{1}{9K} - K[/itex]
     
  18. Apr 5, 2017 #17
    Referring to stevendaryl's solution:0.755

    I believe the 3, I might be convinced the solution has a 9, but are you sure the cubic equation solution really has a square root of what, (a 621). A cube root of
    (25 - square root divided by 54) ???

    The cubic equation is so simple r cube - r squared + 1 = 0.

    PS. I used MATLAB and your solution works! I can't believe it. There must be another universe somewhere where the solution to this equation is simpler.

    0.755 is about pi /4. Are you sure the answer isn't really just pi / 4, and we are in the wrong universe.
     
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