Rationalizing denominators

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January 24th 2009, 04:50 PM
hershey93

Rationalizing denominators

$\frac{5 \sqrt{2}}{3 \sqrt{7x}}$
January 24th 2009, 05:00 PM
Chris L T521

Quote:

Originally Posted by hershey93

$\frac{5 \sqrt{2}}{3 \sqrt{7x}}$

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

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

Does this make sense?
January 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 $\frac{\sqrt{7x}}{\sqrt{7x}}$

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

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 :)

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