1. ## Simplification

Can anybody show me how this

$\displaystyle \frac{-4x}{9\sqrt{4-\frac{4x^2}{9}}}$

winds up simplifying to this:

$\displaystyle \frac{-2x}{3\sqrt{9-x^2}}$

Here's what I have done, but I get a different answer:

$\displaystyle \frac{-4x}{9\sqrt{4-\frac{4x^2}{9}}}$
$\displaystyle \frac{-4x}{9\sqrt{36-4x^2}}$
$\displaystyle \frac{-4x}{9\sqrt{4(9-x^2)}}$
$\displaystyle \frac{-4x}{9(2)\sqrt{9-x^2}}$
$\displaystyle \frac{-2x}{9\sqrt{9-x^2}}$

I just don't see how that 3 winds up in the denominator. I get a 9.

Thanks.

2. Originally Posted by joatmon
Can anybody show me how this

$\displaystyle \frac{-4x}{9\sqrt{4-\frac{4x^2}{9}}}$

winds up simplifying to this:

$\displaystyle \frac{-2x}{3\sqrt{9-x^2}}$

Here's what I have done, but I get a different answer:

$\displaystyle \frac{-4x}{9\sqrt{4-\frac{4x^2}{9}}}$
$\displaystyle \frac{-4x}{9\sqrt{36-4x^2}}$ Wrong
$\displaystyle \frac{-4x}{9\sqrt{4(9-x^2)}}$
$\displaystyle \frac{-4x}{9(2)\sqrt{9-x^2}}$
$\displaystyle \frac{-2x}{9\sqrt{9-x^2}}$

I just don't see how that 3 winds up in the denominator. I get a 9.

Thanks.
$\displaystyle \displaystyle \frac{-4x}{9\sqrt{4 - \frac{4x^2}{9}}} = \frac{-4x}{9\sqrt{\frac{36 - 4x^2}{9}}}$

$\displaystyle \displaystyle = \frac{-4x}{\frac{9\sqrt{36-4x^2}}{3}}$...

Go from here...

3. Got it. Thanks.