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# ​ Dividing Radicals

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4 I really don't understand these, please help >.<

Oct 24, 2018

### 4+0 Answers

#1
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Second one only

Oct 24, 2018
#2
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$$\dfrac{\sqrt{4r^3}}{\sqrt{9r}} = \dfrac{r\sqrt{4}}{\sqrt{9r}} \\ \\ \text{This is really as far as it can be simplified}\\ \text{but it's common to want the denominator to be free of radicals}\\ \dfrac{r\sqrt{4}}{\sqrt{9r}} =\dfrac{r\sqrt{4}}{\sqrt{9r}} \dfrac{\sqrt{9^2r^2}}{\sqrt{9^2r^2}} = \dfrac{r\sqrt{324r^2}}{\sqrt{9^3r^3}}=\dfrac{r\sqrt{324r^2}}{9r}= \dfrac{\sqrt{324r^2}}{9}$$

Rom  Oct 24, 2018
#3
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What about the 2? Where does that go?

RainbowPanda  Oct 24, 2018
#4
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eh b****y h**l I missed it.

It goes in the denominator so the denominator becomes 18 rather than 9.

Rom  Oct 25, 2018