# simplifying square roots

• Mar 15th 2011, 04:05 AM
donnagirl
simplifying square roots
If $r>0$ and $s>0$ and $\sqrt{\frac{r}{s}} + \sqrt{\frac{s}{r}}$ show that $\frac{r+s}{\sqrt{rs}}$ is equivalent.
• Mar 15th 2011, 04:33 AM
Soroban
Hello, donnagirl!

Quote:

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

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