# Rationalizing denominator

• April 23rd 2012, 09:21 AM
jimdec23
Rationalizing denominator
Hi,
I need some help in determining how the following equation develops from this: T= 2 Pi r / (Sq. root of r X g) to: T= 2 Pi X (sq root r/g). How do we get the quantity (Sq. root r X g) in the denominator to (Sq. root r/g) in the numerator? I multiplied the numerator and denominator of the right side of the original equation by Sq. root r X g to get T= 2 Pi r X Sq root r X g in the numerator, and r X g in the denominator (eliminating radical in denominator). Now, the r in the denominator cancels the r in the numerator, giving T= 2 Pi Sq root r X g / g. I don't understand how the equation becomes T= 2 Pi X Sq. root r/g (now no denominator). Any help will be appreciated.
• April 23rd 2012, 12:13 PM
MathoMan
Re: Rationalizing denominator
$\frac{2\pi r}{\sqrt{r g}}=\frac{2 \pi}{\frac{\sqrt{rg}}{r}}=\frac{2 \pi}{\sqrt{\frac{rg}{r^2}}}=\frac{2 \pi}{\sqrt{\frac{g}{r}}}=\frac{2\pi}{\frac{\sqrt{g }}{\sqrt{r}}}=\frac{2 \pi \sqrt{r}}{\sqrt{g}}=2\pi \frac{\sqrt{r}}{\sqrt{g}}=2 \pi \sqrt{\frac{r}{g}}$
• April 23rd 2012, 12:30 PM
ignite
Re: Rationalizing denominator
${r=\sqrt{r}\sqrt{r};}\\\frac{2\pi r}{\sqrt{r g}}=\frac{2 \pi\sqrt{r}\sqrt{r}}{\sqrt{r}\sqrt{g}}=2\pi \frac{\sqrt{r}}{\sqrt{g}}=2 \pi \sqrt{\frac{r}{g}}$