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Math Help - Solve 2/[(x^2)-x] = (1/x) + 4/(x^2-1)

  1. #1
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    Solve 2/[(x^2)-x] = (1/x) + 4/(x^2-1)

    Can somebody help me solve this equation? I have gotten to a certain point then get stuck. Thank you!


    2/x^2-x = (1/x) + (4/x^2-1)
    Last edited by mr fantastic; May 11th 2010 at 06:53 PM. Reason: Re-titled.
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    Quote Originally Posted by connroy17 View Post
    Can somebody help me solve this equation? I have gotten to a certain point then get stuck. Thank you!


    2/x^2-x = (1/x) + (4/x^2-1)
    Are the terms in the red part of exponent or not?
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    2/((x^2)-x) = (1/x) + (4/((x^2)-1)

    Hopefully that makes it easier to understand. The exponents are in red.
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    \frac{2}{x^2}-x=\frac{1}{x}+\frac{4}{x^2}-1

    Move to one side and simplify:

    \frac{2}{x^2}+x+\frac{1}{x}-1=0

    Obtain same common denominator.

    You can do that one.
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  5. #5
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    Quote Originally Posted by connroy17 View Post
    2/((x^2)-x) = (1/x) + (4/((x^2)-1)

    Hopefully that makes it easier to understand. The exponents are in red.
    \frac{2}{x^2-x} = \frac{1}{x} + \frac{4}{x^2-1}<br />

    \frac{2}{x(x-1)} = \frac{1}{x} + \frac{4}{(x+1)(x-1)}

    common denominator is x(x+1)(x-1) ...

    \frac{2(x+1)}{x(x-1)(x+1)} = \frac{1(x+1)(x-1)}{x(x+1)(x-1)} + \frac{4x}{x(x+1)(x-1)}

    note that x cannot equal 0 , 1 , or -1 (why?)

    with a common denominator, the numerators form the equation ...

    2(x+1) = (x+1)(x-1) + 4x

    solve this quadratic for x
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