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

**jamesrb** · December 11th 2012, 08:27 PM

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

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.

**Soroban** · December 11th 2012, 08:58 PM

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

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