How do you rationalize #2/(sqrt(72y))#?

1 Answer
Guilherme N.
May 16, 2015

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)#

That is because #sqrtf(x)=f(x)^(1/2)#, then #f(x)^(1/2)f(x)^(1/2)=f(x)^(1/2+1/2)=f(x)^1=f(x)#

So, for your function:

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

Now, let's just simplify!

#(6sqrt(2y))/(36y)=(sqrt(2y))/(6y)#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
1865 views around the world
You can reuse this answer
Creative Commons License