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Math Help - Need help understanding how this was simplified. Fraction with multiple square roots

  1. #1
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    Need help understanding how this was simplified. Fraction with multiple square roots

    \frac{\sqrt{x^2-9}-[x(\frac{1}{2}(x^2-9)(2x)]}\sqrt({x^2-9})^2

    I was able to clean this up:
    \frac{\sqrt{x^2-9}-\frac{x^2}{\sqrt{x^2-9}}}\sqrt({x^2-9})^2

    I don't know how to proceed any further. I know the simplified answer is:
     \frac{-9}{(x^2-9)^\frac{3}{2}}

    But I really don't know how to get there.
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  2. #2
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    Re: Need help understanding how this was simplified. Fraction with multiple square ro

    Hello, jamesrb!

    \frac{\sqrt{x^2-9}-[x(\frac{1}{2}(x^2-9)^{-\frac{1}{2}}(2x)]}{(\sqrt{x^2-9})^2}

    I was able to clean this up: . \frac{\sqrt{x^2-9}-\frac{x^2}{\sqrt{x^2-9}}}{{\color{blue}x^2-9}}

    I don't know how to proceed any further.
    I know the simplified answer is: .  \frac{\text{-}9}{(x^2-9)^\frac{3}{2}}

    You have: . \frac{\sqrt{x^2-9}-\dfrac{x^2}{\sqrt{x^2-9}}}{x^2-9}


    Multiply top and bottom by \sqrt{x^2-9}

    . . \frac{\sqrt{x^2-9}\cdot\left(\sqrt{x^2-9}-\dfrac{x^2}{\sqrt{x^2-9}}\right)}{\sqrt{x^2-9}\cdot(x^2-9)} \;=\;\frac{x^2-9-x^2}{(x^2-9)^{\frac{3}{2}}} \;=\;\frac{\text{-}9}{(x^2-9)^{\frac{3}{2}}}
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