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Math Help - having trouble factoring

  1. #1
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    having trouble factoring

    although technically I am working on a calculus problem the problem I'm having is with algebra. a step in my problem requires me to factor this equation:

    4x-2(x^2-1)^-1/2=0


    I just end up going in circles even if I simplify it and use the quadratic formula. help would be much appreciated.
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  2. #2
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    Re: having trouble factoring

    Hello, deviantxane!

    Where did this equation come from?
    Are you sure it's correct?


    \text{Solve: }\:4x - 2(x^2-1)^{-\frac{1}{2}}\:=\:0

    Note that x \ne \pm1.

    We have: . 4x - \frac{2}{\sqrt{x^2-1}}\:=\:0 \quad\Rightarrow\quad 4x \:=\:\frac{2}{\sqrt{x^2-1}}

    Multiply by \sqrt{x^2-1}:\;\;4x\sqrt{x^2-1} \:=\: 2 \quad\Rightarrow\quad 2x\sqrt{x^2-1} \:=\: 1

    Square both sides: . 4x^2(x^2-1) \:=\:1 \quad\Rightarrow\quad 4x^4 - 4x^2 - 1 \:=\:0

    Quadratic Formula: . x^2 \;=\;\frac{4\pm\sqrt{(\text{-}4)^2 - 4(4)(\text{-}1)}}{2(4)} \;=\;\frac{4\pm4\sqrt{2}}{8} \;=\;\frac{1\pm\sqrt{2}}{8}

    For a real root: . x \;=\;\pm\sqrt{\frac{1 + \sqrt{2}}{8}}
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  3. #3
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    Re: having trouble factoring

    thanks, I looked back over my problem today and realized I had forgotten a calculus rule that lead me to the equation I was trying to factor thanks so much or your help.
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