How do you rationalize $\frac{2}{\sqrt{72 y}}$ $2/(sqrt(72y))$ ?

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How do you rationalize $\frac{2}{\sqrt{72 y}}$ $2/(sqrt(72y))$ ?

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Answer 1 · 2015-05-16T14:09:01

Let's remember that we use rationalization in order to remove roots from our denominator.

We proceed to do that by multiplying both numerator and denominator of your function by the same value as the root contained in the denominator. That way the proportion will be maintained and the root will be eliminated because

$\sqrt{f (x)} \sqrt{f (x)} = f (x)$ $sqrt(f(x))sqrt(f(x))=f(x)$

That is because $\sqrt{f (x)} = {f (x)}^{\frac{1}{2}}$ $sqrtf(x)=f(x)^(1/2)$ , then ${f (x)}^{\frac{1}{2}} {f (x)}^{\frac{1}{2}} = {f (x)}^{\frac{1}{2} + \frac{1}{2}} = {f (x)}^{1} = f (x)$ $f(x)^(1/2)f(x)^(1/2)=f(x)^(1/2+1/2)=f(x)^1=f(x)$

So, for your function:

$\frac{2}{\sqrt{72 y}} (\frac{\sqrt{72 y}}{\sqrt{72 y}}) = 2 \frac{\sqrt{72 y}}{72 y} = \frac{\sqrt{72 y}}{36 y}$ $2/(sqrt(72y))(sqrt(72y)/sqrt(72y)) = 2sqrt(72y)/(72y) = sqrt(72y)/(36y)$

Your function has been rationalized, but we can further simplify it.

$\sqrt{72 y}$ $sqrt(72y)$ is the same as $\sqrt{2 \cdot 36 y}$ $sqrt(2*36y)$ . We can take the root of $36$ $36$ out, like this: $6 \sqrt{2 y}$ $6sqrt(2y)$

Now, let's just simplify!

$\frac{6 \sqrt{2 y}}{36 y} = \frac{\sqrt{2 y}}{6 y}$ $(6sqrt(2y))/(36y)=(sqrt(2y))/(6y)$

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How do you rationalize $\frac{2}{\sqrt{72 y}}$ $2/(sqrt(72y))$ ?

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