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Math Help - quadratic equation reduction

  1. #1
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    quadratic equation reduction

    an example in my text says that if the values for a, b and c are 2π, 2πh and -S respectively

    with those values plugged into the quadratic equation you get


    -2πh πi^2 h^2 - 4(2π)(-S)
    --------------------------------------------------------------
    4
    π


    up to that point i follow what is going on but then the text reduces the formula to

    -2
    πh 2 π^2 h^2 - 2πS
    ------------------------------------------------------
    4
    π


    what I don't understand is how -4(2
    π)(-S) reduces to 2πS in the above equation.




    from there we divide everything by 2 and get

    -
    πh πi^2 h^2 + 2πiS
    ---------------------------------------------
    2
    π




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  2. #2
    MHF Contributor

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    Re: quadratic equation reduction

    Quote Originally Posted by kingsolomonsgrave View Post
    an example in my text says that if the values for a, b and c are 2π, 2πh and -S respectively

    with those values plugged into the quadratic equation you get


    -2πh πi^2 h^2 - 4(2π)(-S)
    --------------------------------------------------------------
    4
    π


    up to that point i follow what is going on but then the text reduces the formula to

    Unfortunately, it is this part, that you say you follow, that is wrong.
    a= 2\pi, b= 2\pi h, and c= -S so the "discriminant" inside the square root should be
    b^2- 4ac= (2\pi h)^2- 4(2\pi)(-S)= 4\pi^2h^2+ 8\pi S
    You are missing the "4" in first term.

    Once you have that, you can factor out a "4" so that \sqrt{4\pi^2h^2+ 8\pi S}= 2\sqrt{\pi^2h^2+ 2\pi h}.

    -2
    πh 2 π^2 h^2 - 2πS
    ------------------------------------------------------
    4
    π


    what I don't understand is how -4(2
    π)(-S) reduces to 2πS in the above equation.





    from there we divide everything by 2 and get

    -
    πh πi^2 h^2 + 2πiS
    ---------------------------------------------
    2
    π




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