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Math Help - solve for x

  1. #1
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    solve for x

    i need a help solving this algebraic equation
    4(x-3)^2 +1=√((x-1)/4) +3

    plz i need some quick answers
    i would really appreciate whoever solves this question
    thanks
    Last edited by hawtmonkey; September 17th 2008 at 04:20 PM. Reason: plz i need some quick answers
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  2. #2
    Junior Member
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    Yeesh

    4(x-3)^2 +1=√((x-1)/4) +3
    4(x-3)^2 - 2 = √((x-1)/4) = 1/2 * √(x-1)
    8(x-3)^2 - 4 = √(x-1)
    8 x^2 - 48 x + 68 = √(x-1) = t
    8 (t^2 + 1)^2 - 48 (t^2 + 1) + 68 = t (because x = t^2 + 1)
    8 t^4 - 32 t^2 - t + 28 = 0

    I am stuck here.
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  3. #3
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by hawtmonkey View Post
    i need a help solving this algebraic equation
    4(x-3)^2 +1=√((x-1)/4) +3

    4(x-3)^2+1= \sqrt{\frac{x-1}{4}} + 3

    4(x^2-6x+9) +1 = \sqrt{\frac{x-1}{4}} +3

    4x^2- 24x +36 + 1 = \sqrt{\frac{x-1}{4}} +3

    4x^2 -24x + 34 = \sqrt{\frac{x-1}{4}}

    (4x^2 -24x + 34)^2 = \frac{x-1}{4}

    16x^4 -192x^3 +848x^2-1632x + 1156 = \frac{x-1}{4}

    64x^4-768x^3 + 3392x^2 -6528x +4624 = x -1

    64x^4-768x^3 + 3392x^2 -6529x +4625 = 0

    which factors out to

    (4x^2-25x +37)(16x^2-92x+125) = 0

    according to TI89

    so

    (4x^2-25x +37)=0

    (16x^2-92x+125)=0

    You should be able to finish up from here.
    Last edited by 11rdc11; September 17th 2008 at 06:10 PM.
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  4. #4
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    thanks for the solution
    thanks a lot
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  5. #5
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    can you show me how you factored 64x^4-768x^3+3392x^2-6529x+4625=0

    to:

    (4x^2-25x=37)(16x^2-92+125)=0
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