simplifying square roots

**donnagirl** · March 15th 2011, 03:05 AM

If $r>0$ and $s>0$ and $\sqrt{\frac{r}{s}} + \sqrt{\frac{s}{r}}$ show that $\frac{r+s}{\sqrt{rs}}$ is equivalent.

**Soroban** · March 15th 2011, 03:33 AM

Hello, donnagirl!

$\text{If }r>0\text{ and }s>0\text{, show that: }\,\sqrt{\dfrac{r}{s}} + \sqrt{\dfrac{s}{r}} \:=\:\dfrac{r+s}{\sqrt{rs}}$

Just add the radicals . . .

. . $\displaystyle \sqrt{\frac{r}{s}} + \sqrt{\frac{s}{r}} \;=\;\frac{\sqrt{r}}{\sqrt{s}} + \frac{\sqrt{s}}{\sqrt{r}} \;=\;\frac{\sqrt{r}\!\cdot\!\sqrt{r} + \sqrt{s}\!\cdot\!\sqrt{s}}{\sqrt{s}\,\sqrt{r}} \;=\;\frac{r + s}{\sqrt{rs}}$

**donnagirl** · March 15th 2011, 03:35 AM

OMG soroban, thank you that was just too easy! I was going about it all wrong, I was messing around with the conjugate and such LOL!

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