How do you rationalize the denominator and simplify #sqrt21/sqrt35#?

2 Answers
Sep 6, 2017

Answer:

#=sqrt15/5#

Explanation:

multiply top and bottom by #sqrt35#

#sqrt21/sqrt35xxsqrt35/sqrt35#

#=(sqrt21sqrt35)/35#

now combine the sqrt on the numerator and simplify as shown

#sqrt(21xx35)/35=sqrt(3xx7xx5xx7)/35#

#=(sqrt(7^2xx15))/35#

#=(cancel(7)sqrt15)/cancel(35)^5#

#=sqrt15/5#

Sep 6, 2017

Answer:

Slightly different approach demonstrating that you can 'split' roots

#sqrt(15)/5#

Explanation:

Given: #sqrt21/sqrt35#

Note that #5xx7=35 and 3xx7=21#

Write as: #(sqrt(3)xxcancel(sqrt(7)))/(sqrt(5)xxcancel(sqrt(7)))#

Giving #sqrt3/sqrt5#

Multiply by 1 and you do not change the value. However, 1 comes in many forms.

#color(green)(sqrt3/sqrt5color(red)(xx1) color(white)("ddd") ->color(white)("ddd")sqrt3/sqrt5color(red)(xxsqrt5/sqrt5)) = sqrt(15)/5#