# Rationalizing denominators

• Jan 24th 2009, 04:50 PM
hershey93
Rationalizing denominators
$\displaystyle \frac{5 \sqrt{2}}{3 \sqrt{7x}}$
• Jan 24th 2009, 05:00 PM
Chris L T521
Quote:

Originally Posted by hershey93
$\displaystyle \frac{5 \sqrt{2}}{3 \sqrt{7x}}$

You need to multiply by "1", which in our case would be $\displaystyle \frac{\sqrt{7x}}{\sqrt{7x}}$

So, you would have $\displaystyle \frac{5\sqrt{2}}{3\sqrt{7x}}\cdot\frac{\sqrt{7x}}{ \sqrt{7x}}=\frac{5\sqrt{2}\cdot\sqrt{7x}}{3\left(\ sqrt{7x}\right)^2}=\frac{5\sqrt{14x}}{3\cdot 7x}=\color{red}\boxed{\frac{5\sqrt{14x}}{21x}}$

Does this make sense?
• Jan 24th 2009, 05:04 PM
hershey93
Quote:

Originally Posted by Chris L T521
You need to multiply by "1", which in our case would be $\displaystyle \frac{\sqrt{7x}}{\sqrt{7x}}$

So, you would have $\displaystyle \frac{5\sqrt{2}}{3\sqrt{7x}}\cdot\frac{\sqrt{7x}}{ \sqrt{7x}}=\frac{5\sqrt{2}\cdot\sqrt{7x}}{3\left(\ sqrt{7x}\right)^2}=\frac{5\sqrt{14x}}{3\cdot 7x}=\color{red}\boxed{\frac{5\sqrt{14x}}{21x}}$

Does this make sense?

Ya that seems right...that was exactly what i thought it would look like.
Thanx for your help Chris..Have a good night :)