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

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- Jan 24th 2009, 04:50 PMhershey93Rationalizing denominators
$\displaystyle \frac{5 \sqrt{2}}{3 \sqrt{7x}}$

- Jan 24th 2009, 05:00 PMChris 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? - Jan 24th 2009, 05:04 PMhershey93