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Roots of a cubic polynomial

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Hello there!

    I'm trying to find the roots of the following cubic polynomial

    x^3 - 10x + 18 = 0

    3. The attempt at a solution
    I did the following: I rewrite 18 as

    18 = - (x^3 - 10x)

    then I did back substitution and factored out

    x^3 - 10x - x^3 + 10x = 0 or x(x^2-10) - x(x^2-10) = 0

    => (x - x)(x^2-10) = 0

    => x = sqrt(10)

    I'm not sure if what I did is legit and correct. I think is wrong.

    Is it valid?
     
    Last edited: Feb 21, 2012
  2. jcsd
  3. Feb 22, 2012 #2
    No it is not at all legit .
    You are equating zero to zero , fundamentally wrong :\
     
  4. Feb 22, 2012 #3
    Ok.

    Any suggestion?
     
  5. Feb 22, 2012 #4

    Deveno

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    we have x3 - 10x = -18

    let x = w + 10/(3w)

    after doing the arithmetic, you should get a quadratic in w3, which you can then solve.
     
  6. Feb 22, 2012 #5
    Hello there!

    To be honest, I don't understand from where does the "x = w + 10/(3w)" comes from... Could you be a little bit more explicit please?

    ...
     
    Last edited: Feb 22, 2012
  7. Feb 22, 2012 #6

    Deveno

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    it's magic!

    (really, it's just a substitution)

    here. i'll get you started:

    (w + 10/(3w))3 - 10(w + 10/(3w) = -18, so

    w3 + 3w2(10/(3w)) + 3w(100/(9w2) + 1000/(27w3) - 10w - 100/(3w) = -18

    continue?
     
  8. Feb 22, 2012 #7
    The function x^3 - 10x + 18 has 3 dangerous roots , which will require hours of calculation , and wont be given as homework , check if you copied it correctly
     
  9. Feb 22, 2012 #8

    vela

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    There's nothing wrong with setting 0 equal to 0. It's true, after all.

    The problem here is that the first equation is of the form ##0 \times a = 0##. Obviously, this statement is true for any a. In your case, this means x2-10 can be equal to anything, so you can't assume it's equal to 0 and solve for x.
     
  10. Feb 22, 2012 #9
    yea i kind of meant that :tongue2:
     
  11. Feb 22, 2012 #10
    There's a cubic formula, you could try memorising that :3

    And as kushan said, the roots of this are not nice at all, are you sure you copied it out correctly?
    Homework problems generally have nice, friendly solutions
     
  12. Feb 22, 2012 #11
    Hello there!

    The original problem statement is:

    (x^3)/90 - (x^2)/9 + 1/5

    I put the whole thing equal to zero ( for the roots), and then multiply by 90

    x^3 - 10x + 18 = 0

    This question was on a test. The problem also asks for local extrema, but to find them is simple.

    I mean the "x = w + 10/(3w)" from where does that comes from?
     
    Last edited: Feb 22, 2012
  13. Feb 22, 2012 #12
    is the power 3/90 ?
     
  14. Feb 22, 2012 #13
    no:

    (x^3)/90

    Same with the others...
     
  15. Feb 22, 2012 #14
    I am afraid there is no other way you need to find the roots by the cubic equation
    and for your information the cube roots are
    And yes they are all real
    beautiful i would say :cry: :yuck:
     

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  16. Feb 22, 2012 #15
    My worst fear is being realized... I'm still skeptical, though...
     
  17. Feb 22, 2012 #16
    hello there again!

    I tried the following but, again, it's a litlle bit fishy...

    I started from:

    x^3 - 10x = -18

    => sqrt(x^3 - 10x) = sqrt(-18)

    => sqrt[x^2 (x - 10/x)] + sqrt(-18)

    => x * sqrt{(x^2 - 10)/x)} + sqrt(-18)

    I divide by sqrt(-18)

    => x * sqrt{(x^2 - 10)/-18x} + 1

    Then I factor 1/x

    => 1/x * {sqrt[(x^2 - 10)/-18x] + x}

    From here, I say that P(x) = 0, when x = (x^2 - 10)/-18x

    Is this ok?

    No, this is absurd. I don't know: I refuse to use the cubic equation...

    All this is plain wrong... I guess "x" is not allowed to be as a "b" or "d" on a expression like (a + b)(c + d) = 0

    Right?
     
    Last edited: Feb 22, 2012
  18. Feb 22, 2012 #17

    Ray Vickson

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    I don't know whether one of your roots simplifies to a real (it should), but Maple gets different results from yours:

    root1 = -A/3 - 10/A and
    root2 = A/6 + 5/A + I*(sqrt(3)/2)*[10/A - A/3],
    where A = [243 + 3*sqrt(3561)]^(1/3), and I = sqrt(-1)

    root3 = conjugate of root 2.

    RGV
     
  19. Feb 22, 2012 #18

    SammyS

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    OK !

    Which are you trying to solve?

    x3 - 10x + 18 = 0

    or

    x3 - 10x2 + 18 = 0

     ?
     
  20. Feb 22, 2012 #19
    hello! Oh, my bad:

    This one:

    x^3 - 10x + 18 = 0
     
  21. Feb 23, 2012 #20
    yea all of them are real :D
     
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