How do you simplify #sqrt(2/3)*sqrt(6/5)#?

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Answer 1 · 2018-06-09T18:21:49

#(6sqrt5)/15#

Use the rule for radicals of the same degree:

#sqrta * sqrtb=sqrt(a*b)#

#sqrt(2/3)*sqrt(6/5)=sqrt(2/3*6/5)=sqrt(12/15)#

now use the rule for radicals of the same degree:

#sqrt(a/b) = sqrta/sqrtb#

#sqrt(12/15)=sqrt12/sqrt15#

now rationalize the denominator:

#sqrt12/sqrt15*sqrt15/sqrt15 = sqrt180/15 = (6sqrt5)/15#

Answer 2 · 2018-06-09T18:22:15

#(2*sqrt(5))/5#

We get by
#sqrt(a)*sqrt(b)=sqrt(ab)# for #a,b>=0#
so we obtain

#sqrt(2/3)*sqrt(6/5)=sqrt(4/5)=2/sqrt(5)=(2*sqrt(5))/5#