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

2 Answers
Jun 9, 2018

Answer:

#(6sqrt5)/15#

Explanation:

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#

Jun 9, 2018

Answer:

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

Explanation:

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#