How do you rationalize the denominator and simplify #sqrt(9xy)/(sqrt(3x^2y)#?

1 Answer
Mar 31, 2016

Answer:

#+-sqrt(3x)/x#

Explanation:

Disregarding the possibilities of #+-#

Consider the example
#sqrt(16)/(sqrt(4))=4/2=2#

Now view it as

#sqrt(16/4)= sqrt(4)=2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given: #sqrt(9xy)/sqrt(3x^2y)#

Write as

#sqrt( (9xy)/(3x^2y))" "=" "sqrt( 3/x)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is format is frowned upon so we need to 'get rid' of the root in the denominator.

Multiply by 1 but in the form of #1=sqrt(x)/sqrt(x)#

#sqrt(3)/sqrt(x)xxsqrt(x)/(sqrt(x)) = sqrt(3x)/x#

Now think about #(-2)xx(-2)=+4=(+2)xx(+2)#

So our answer needs to be #+-sqrt(3x)/x#