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Math Help - [SOLVED] Hard Quadratic Equation

  1. #1
    audiogeek
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    [SOLVED] Hard Quadratic Equation

    This is a straightforward math question to solve for R.

    R^2(1/((Y^2)(X^2)))+R((2/(YX))-1)=1

    I've spent a while trying to figure this out. Any help would be greatly appreciated! Thanks!

    -Mike
    Last edited by audiogeek; August 30th 2007 at 06:36 AM. Reason: Posted wrong
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  2. #2
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    Hello, Mike!

    It seems to be a "simple" quadratic equation . . .


    Solve for R\!:\;\;R^2\left(\frac{1}{x^2y^2}\right) <br />
+R\left(\frac{2}{xy}-1\right)\;=\;1

    Multiply through by x^2y^2\!:\;\;R^2 + xy(2-xy)R - x^2y^2 \;=\;0

    We have a quadratic equation with: . a \,= \,1,\;\;b \,= \,xy(2-xy),\;\;c \,= \,\text{-}x^2y^2


    Then: . x \;=\;\frac{-xy(2-xy) \pm \sqrt{[xy(2-xy)]^2 - 4(1)(\text{-}x^2y^2)}}{2(1)}

    . . . . . x \;=\;\frac{xy(xy-2) \pm\sqrt{x^2y^2(2-xy)^2 + 4x^2y^2}}{2}

    . . . . . x \;=\;\frac{xy(xy-2) \pm xy\sqrt{x^2y^2 - 4xy + 8}}{2}

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  3. #3
    Senior Member DivideBy0's Avatar
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    Alternatively just move the 1 over to the left and proceed with the quadratic formula
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