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Thread: Solving Equations in Quadratic Form

  1. November 4th 2008, 02:55 AM #1
    Hi888
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    Solving Equations in Quadratic Form

    Hi I'm kinda confused by these equations.

    1.) 3[ (x+3) / (2x-1) ]^2 - 4(x+3) / (2x-1) + 1 = 0

    2.) 36x^4/3 - 25x^2/3 + 4 = 0

    Thanks!
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  2. November 4th 2008, 04:02 AM #2
    muks
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    3 (x+3 / 2x-1)^2 - 4(x+3 / 2x-1) + 1
    = 3a^2 - 4a + 1 { replacing x +3/2x-1) by a }
    = 3a^2 - 3a - a + 1
    =
    3a(a - 1) - 1(a-1)= (3a-1)(a-1)
    = {3(x+3 / 2x-1) - 1}{3(x+3 / 2x-1) - 1}
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  3. November 4th 2008, 04:26 AM #3
    muks
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    <br /> 36x^4/3 - 25x^2/3 + 4 = 0<br />
    Or 36a^2 - 25a + 4 = 0 {subtituting x^2/3 by a}
    Or 36a^2 - 16a -9a + 4=0
    Or 4a(9a-4) - 1(9a-4)=0
    Or (4a-1)(9a-4)=0
    Or a=1/4, 4/9
    Or x^2/3 = 1/4 , 4/9
    Or x^1/3=1/2 , 2/3
    Or x = 1/8 ,8/27.
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  4. November 5th 2008, 03:56 AM #4
    muks
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    You're welcome!
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