Algebra

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How do I find x for this equation? I think I should get e^(something).
Thanks in advance!

3x^2+4lnx+4=0

  • Algebra -

    There's no algebraic way to do it. A numeric method or a graphical analysis will be required. See

    http://www.wolframalpha.com/input/?i=solve+3x^2%2B4lnx%2B4%3D0

  • Algebra -

    For a complicated equation such as yours, the most powerful method is probably Newton's Method.

    let y = 3x^2+4lnx+4
    dy/dx = 6x + 4/x

    newx = x - y/(dy/dx)

    = x - (3x^2+4lnx+4)/(6x + 4/x)
    which simplifies to

    newx = (3x^2 - 4lnx)/(6x + 4/x)

    now use an iteration process, I started with x = 1

    x | newx
    1 -- .3
    .3 -- .336072..
    .336072 -- .337716
    .337716 -- .337719143
    .337719143 -- .337719143
    since the last x used returned the same value of x, we cannot improve on that answer.

    Amazing that after only 4 interations, I got
    x =.337719143 correct to 9 decimal places

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