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Math Help - [SOLVED] Can't Figure out this simplification

  1. #1
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    [SOLVED] Can't Figure out this simplification

    I came across a math related problem in my science text book I can't figure out on my own.

    The first part of the problem has led me to this equation:
     \frac{(2-Y)Y}{(\frac{3}{2}-\frac{3}{2}Y)^\frac{3}{2}(\frac{1}{2}-\frac{1}{2}Y)^\frac{1}{2}}=\kappa

    I am 100% positive that the first part is right but the second part of the problem requires the equation to be in terms of Y.

    The textbook has an answer in the back of the book, but so far I haven't been able to figure it out. There are a couple of problems like this so if someone could help me figure out this problem it would be a huge help.

    The details: the answer in the back of the book reads Y=1-(1+1.3\kappa^\frac{-1}{2})
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    Quote Originally Posted by 360modina View Post
     \frac{(2-Y)Y}{(\frac{3}{2}-\frac{3}{2}Y)^\frac{3}{2}(\frac{1}{2}-\frac{1}{2}Y)^\frac{1}{2}}=\kappa
    ...
    Y=1-(1+1.3\kappa^\frac{-1}{2})
    Your solution does not work. I think that you meant Y=1-(1+1.3\kappa)^{-1/2}, which does seem to satisfy the equation.

    The solution is long and tedious, and I really do not have time right now, but I wanted to point out this mistake as it could confuse other posters.
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  3. #3
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    Yes your correction is right, I apologize for that although I am still stuck on the problem...
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  4. #4
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    Quote Originally Posted by 360modina View Post
    I came across a math related problem in my science text book I can't figure out on my own.

    The first part of the problem has led me to this equation:
     \frac{(2-Y)Y}{(\frac{3}{2}-\frac{3}{2}Y)^\frac{3}{2}(\frac{1}{2}-\frac{1}{2}Y)^\frac{1}{2}}=\kappa

    I am 100% positive that the first part is right but the second part of the problem requires the equation to be in terms of Y.

    The textbook has an answer in the back of the book, but so far I haven't been able to figure it out. There are a couple of problems like this so if someone could help me figure out this problem it would be a huge help.

    The details: the answer in the back of the book reads Y=1-(1+1.3\kappa^\frac{-1}{2})

     Denominator~= (\frac{3}{2}-\frac{3}{2}Y)^\frac{3}{2}(\frac{1}{2}-\frac{1}{2}Y)^\frac{1}{2})

    =(\frac{3}{2})^{\frac{3}{2}}(1-Y)^{\frac{3}{2}}(\frac{1}{2})^{\frac{1}{2}} (1-Y)^\frac{1}{2}

     <br />
=(3)^{\frac{3}{2}}(1-Y)^{\frac{3}{2}}(\frac{1}{4}) (1-Y)^\frac{1}{2}<br />

     <br />
=\frac{(3)^{\frac{3}{2}}(1-Y)^2}{4} <br />

    Thus fraction is

     \frac{(2-Y)Y}{\frac{(3)^{\frac{3}{2}}(1-Y)^2}{4}}=\kappa


     \frac{(2-Y)Y }{(1-Y)^2}=\frac{3^{3/2} \kappa}{4}

    Divide the polynomials on LHS

    Now go ahead
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  5. #5
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    Solved: Can't Figure out this simplification

    OK, from the top then:

    \frac{(2-Y)Y}{(\frac{3}{2}-\frac{3}{2}Y)^\frac{3}{2}(\frac{1}{2}-\frac{1}{2}Y)^\frac{1}{2}}=\kappa


    (\frac{3}{2}-\frac{3}{2}Y)^\frac{3}{2}(\frac{1}{2}-\frac{1}{2}Y)^\frac{1}{2}

    (\frac{3}{2})^\frac{3}{2}(1-Y)(\frac{1}{2})^\frac{1}{2}(1-Y)^\frac{1}{2}

    \frac{(3)^\frac{3}{2}(1-Y)^2}{4}

    \frac{(2-Y)Y}{\frac{(3)^{\frac{3}{2}}(1-Y)^2}{4}}=\kappa


    \frac{(2-Y)Y }{(1-Y)^2}=\frac{3^{3/2} \kappa}{4}

    \frac{(2-Y)Y}{(1-Y)^2}=1.3\kappa

    \frac{(1+2Y-Y^2-1)}{(1-Y)^2}=1.3\kappa

    \frac{(1-(Y^2-2Y+1))}{(Y-1)^2}=1.3\kappa

    \frac{(1-(Y-1)^2)}{(Y-1)^2}=1.3\kappa

    \frac{1}{(Y-1)^2}-1=1.3\kappa

    \frac{1}{(Y-1)^2}=1.3\kappa +1

    Let \Omega = 1 + 1.3\kappa

    \frac{1}{(Y-1)^2} = \Omega

    1=\Omega(Y^2-2Y+1)

    1=\Omega Y^2 - 2\Omega Y + \Omega

    -\Omega Y^2+2\Omega Y -\Omega + 1 = 0

    Y=\frac{-(-2\Omega)\pm \sqrt{(2\Omega )^2-4\Omega (-(-\Omega +1))}}{2\Omega}

    Y=\frac{2\Omega \pm \sqrt{4\Omega}}{2\Omega}

    Y=\frac{2(\Omega \pm \sqrt{\Omega})}{2\Omega}

    Y=\frac{\Omega \pm \sqrt{\Omega}}{\Omega}

    Y=\frac{\Omega}{\Omega} \pm \frac{\sqrt{\Omega}}{\Omega}

    Y=1 \pm \frac{\Omega ^ \frac{1}{2}}{\Omega}

    Y=1 \pm \frac{1}{\sqrt{\Omega}}

    Y=1 \pm \Omega ^ \frac{-1}{2}

    Y=1 \pm (1+1.3\kappa)^\frac{-1}{2}
    Last edited by 360modina; March 22nd 2009 at 01:33 PM.
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