How do you rationalize the denominator and simplify #(sqrt (77))/ (sqrt (35))#?

Question

2350 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-19T18:49:17

In this way:

#sqrt77/sqrt35=sqrt77/sqrt35*sqrt35/sqrt35=sqrt(77*35)/35=#

#=sqrt(7*11*5*7)/35=(7*sqrt55)/35=sqrt55/5#.

OR:

#sqrt77/sqrt35=sqrt(77/35)=sqrt(11/5)=sqrt11/sqrt5*sqrt5/sqrt5=sqrt55/5#.

OR:

#sqrt77/sqrt35=(sqrt7*sqrt11)/(sqrt7*sqrt5)=sqrt11/sqrt5*sqrt5/sqrt5=sqrt55/5#.

Answer 2 · 2015-04-19T21:03:32

#sqrt(77)/sqrt(35)#

#=(sqrt(11)sqrt(7))/(sqrt(5)sqrt(7))#

#=sqrt(11)/sqrt(5)#

#=sqrt(11)/sqrt(5)*sqrt(5)/sqrt(5)#

#=(sqrt(11)sqrt(5))/5#

#=sqrt(55)/5#