How do you simplify #(8sqrt3 )/ sqrt6#?

3 Answers
Mar 27, 2018

#8/sqrt(2)#

Explanation:

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

So, we're looking to simplify #8sqrt(3/6)#

#3/6=1/2,# so we have

#8sqrt(1/2)=(8sqrt(1))/sqrt(2)=8/sqrt(2)#

Mar 27, 2018

#4 sqrt2#

Explanation:

#(8 sqrt3)/sqrt6#

Rationalize Denominator

#(8 sqrt3sqrt6)/6#

Combine Roots

#(8 sqrt18)/6#

Factor

#(8 sqrt(9 * 2))/6#

Take Out The Nine

#(8 * 3 sqrt2)/6#

Simplify

#(24 sqrt2)/6#

Simplify

#4 sqrt2#

Mar 27, 2018

#(8 sqrt(3))/(sqrt(6)) = 4 sqrt2#

Explanation:

Given:

#(8 sqrt(3))/(sqrt(6))#

Rationalize the denominator:

#rArr (8 sqrt(3))/(sqrt(6))*(sqrt(6))/(sqrt(6))#

#rArr [8 sqrt(3) sqrt(6)]/(sqrt(6) sqrt(6)#

Observe that #color(blue)(sqrt(m) * sqrt(m)=m# and

#rArr color(blue)(sqrt(mn)=sqrt(m)*sqrt(n)#

#rArr [8 sqrt(3) sqrt(3*2)]/6#

#rArr (2*4 sqrt(3) sqrt(3) sqrt(2)]/(2*3)#

#rArr [2*4*3*sqrt(2)]/(2*3)#

#rArr [cancel (2)*4*cancel 3*sqrt(2)]/(cancel(2)*cancel 3)#

#rArr 4 sqrt(2)#

Hence,

#color(red)((8 sqrt(3))/(sqrt(6)) = 4 sqrt2#