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Thread: Solving cubic equation

  1. March 24th 2009, 07:58 PM #1
    U-God
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    Solving cubic equation

    Hi, I am really out of touch with solving these.

    -t^3+10t^2-30t+32

    solving for t.
    If I can break it down to a quadratic, I should be able to do it with no problems. But I don't know how to find which factor to take out.
    Any help is great!
    Thanks in advance,
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  2. March 24th 2009, 10:02 PM #2
    ADARSH
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    Quote Originally Posted by U-God View Post
    Hi, I am really out of touch with solving these.

    -t^3+10t^2-30t+32

    solving for t.
    If I can break it down to a quadratic, I should be able to do it with no problems. But I don't know how to find which factor to take out.
    Any help is great!
    Thanks in advance,
    Roots are not real, use a calculator, the other methods will not be easy at all
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  3. March 25th 2009, 12:32 AM #3
    mr fantastic
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    Quote Originally Posted by U-God View Post
    Hi, I am really out of touch with solving these.

    -t^3+10t^2-30t+32

    solving for t.
    If I can break it down to a quadratic, I should be able to do it with no problems. But I don't know how to find which factor to take out.
    Any help is great!
    Thanks in advance,
    There is only 1 real root and its exact value is found with difficulty using Cardano's method. Why are you trying to solve this cubic, what is the whole question that it has come from?
    Last edited by mr fantastic; March 25th 2009 at 04:50 AM. Reason: Added exact value qualifier
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  4. March 25th 2009, 02:21 AM #4
    chisigma
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    The only real solution can be found using the Newton-Raphson method. If tou have an equation in the form...

    f(x)=0

    ... the method consists in searching better approximation of a solution in iterative way as follows...

    x_{n+1} = x_{n} - \frac{f(x_{n})}{f ' (x_{n})}

    In our case is...

    f(x)= x^{3} -10\cdot x^{2} + 30\cdot x -32

    In few itrations you can find the only real solution x \simeq 5.75110070238\dots and after that you can reduce the equation in quadratic form...

    Kind regards

    \chi \sigma
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