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Math Help - Finding x

  1. #1
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    Finding x

    ( 1 / (x^2 + 2) ) - x = 0

    I am searchin
    g for x, but having a hard time solving the equation.
    Have multiplied the left side of the equal sign with
    (x^2 + 2) and got:

    1 - (x^3 + 2x) = 0

    The same thing with the equation
    ln(x^2 + 2) - x = 0

    How do i solve x
    ?
    Last edited by Tikitac; October 12th 2007 at 06:26 AM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Tikitac View Post
    ( 1 / (x^2 + 2) ) - x = 0

    I am searchin
    g for x, but having a hard time solving the equation.
    Have multiplied the left side of the equal sign with
    (x^2 + 2) and got:

    1 - (x^3 + 2x) = 0

    The same thing with the equation
    ln(x^2 + 2) - x = 0

    How do i solve x
    ?
    So far so good with the first one. The problem with cubics is that it is generally hard to factor them.
    x^3 + 2x - 1 = 0

    The rational root theorem says that the possible rational roots of this equation are \pm 1. Subbing them in we find that neither of them are roots. So there are no rational roots to this equation. The best we can do from here is numeric approximation. (Well, there's Cardano's method, but it's butt-ugly!) I get 0.453398 is the only real root.

    For
    ln(x^2 + 2) - x = 0
    we are in a similar situation.

    ln(x^2 + 2) = x

    x^2 + 2 = e^x

    x^2 + 2 - e^x = 0

    I believe there is a way to use the Lambert W function to solve this, but again we are stuck with numerical approximation. I get x = 1.31907 as the only solution.

    -Dan
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  3. #3
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    Is it ok if I ask how you did the numerical calculation on the first formula?
    I am looking in my math book and see the formula ( f(a + h) - f(a) ) / h and have tried to use it in different ways, but always mess up somehow.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Tikitac View Post
    Is it ok if I ask how you did the numerical calculation on the first formula?
    I am looking in my math book and see the formula ( f(a + h) - f(a) ) / h and have tried to use it in different ways, but always mess up somehow.
    Unless you are trying to use something like Newton's method I don't see how the derivative would be pertinent. I'm not sure what your text is up to.

    I used my calculator to automatically solve for the roots, and I'm not sure what method it uses, but you can always do something like a binary search: take a guess at what the root might be and keep trying successive guesses until your expression becomes close to 0.

    -Dan
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