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How do you solve an equation from the third power?

  1. Jun 22, 2013 #1
    1. The problem statement, all variables and given/known data
    This is the equation:
    x^3 - 2*x^2 + 1 = 0


    2. Relevant equations



    3. The attempt at a solution
    If I take x out of brackets I get x*(x^2 - 2*x + 1/x) = 0 so either x = 0 or the thing inside it equals 0, however I'm not sure what to do inside the brackets now, I can't solve it as a quadric due to 1/x
     
  2. jcsd
  3. Jun 22, 2013 #2

    tiny-tim

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    Hi Hivoyer! :smile:

    (try using the X2 button just above the Reply box :wink:)
    Can't you instantly see one solution? :smile:
     
  4. Jun 22, 2013 #3
    Oh yeah thanks I didn't see that, but anyway, I'm not sure about the One.Are you referring to the '1'?
     
  5. Jun 22, 2013 #4

    SteamKing

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    Ve haff veys of solving zis eqvations.

    1. You can use trial and error to find a solution, and then use this solution to factor the original cubic equation and find a quadratic, which can then be solved by a formula.
    2. You can plot the cubic and determine at least one real solution.
    3. If you are really hard core, you can apply the formula for solving cubic equations. (Actually, several formulas) See: http://en.wikipedia.org/wiki/Cubic_function

    BTW, the factoring method used in your solution attempt is pretty useless for solving a cubic equation.
     
  6. Jun 22, 2013 #5

    Redbelly98

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    Here are a couple of tips for you:

    Tip #1. Always check x=1 first. It's super-easy to do, and is a solution to a disproportionately large number of homework problems.

    Tip #2. Always check your final answers against the original equation. What do you get when you substitute x=0 into [itex]x^3 - 2x^2 + 1 = 0 \ ?[/itex]

    As for your method, let's see how it works out in the following problem:
    Solve for x: x - 5 = 0​
    Obviously x=5 is the solution, but let's apply your method and see what happens.
    Factor out x: x(1 - 5/x) = 0
    x = 0
    Check by substituting x=0 into original equation:
    0 - 5 = 0
    -5 = 0 ← A false statement, so the solution x=0 is wrong.​
     
  7. Jun 22, 2013 #6

    tiny-tim

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    what are you referring to? :wink:

    divide by the factor you've seen, and that gives you a quadratic which you can solve in the usual way :smile:
     
  8. Jun 22, 2013 #7
    Find a factor first, do this by trying different values a for x. When you find a value for which f(a) = 0 then you know that x - a is a factor.

    Divide this by your cubic polynomial to obtain a quadratic equation and solve it in the usual way.
     
  9. Jun 22, 2013 #8

    Ray Vickson

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    In homework or exam questions you should always try using the Rational Root Test; see
    http://en.wikipedia.org/wiki/Rational_root_theorem .
    In your case the leading coefficient (the coefficient of x^4) is 1, so the test is a lot easier: if the equation has a rational root, that root must be an integer that divides +1 (the constant term), so must be either +1 or -1.

    Note: if the rational root test fails, the equation has only nasty, irrational roots; for example, the slightly different equation x^3 - 2*x^2 - 1 = 0 has no rational roots. You can see that by showing that both x = +1 and x = -1 fail to solve the equation.
     
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