How do you rationalize the denominator and simplify #sqrt15/sqrt35#?

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Answer 1 · 2016-04-07T09:32:24

Very slightly different approach

#(+-sqrt(21))/7#

Given:#" "sqrt(15)/sqrt(35)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Consider the example: #sqrt(16) =4#
(actually it is #+-4#)

Write as #sqrt(4)xxsqrt(4) = 2xx2 =4#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Using the principle demonstrated in the example:

Write as:#" "(sqrt(3)xxsqrt(5))/(sqrt(5)xxsqrt(7))#

#=> sqrt(3)/sqrt(7)xxsqrt(5)/sqrt(5)" " =" " sqrt(3)/sqrt(7)#

Multiply by 1 but in the form of #1=sqrt(7)/sqrt(7)#

#=> sqrt(3)/sqrt(7)xxsqrt(7)/sqrt(7)#

Both 3 and 7 are prime so we have

#(+-sqrt(3xx7))/7 =(+-sqrt(21))/7#