How do you solve #(2sqrt3)/sqrt6#?

4 Answers
Jun 19, 2018

Answer:

#(2sqrt(3))/(sqrt(6))=color(blue)(sqrt(2))#

Explanation:

#(color(lime)2sqrt(3))/(color(magenta)sqrt(6)#

#color(white)("XXX")=(color(lime)(sqrt(2) * sqrt(2)) * sqrt(3))/(color(magenta)(sqrt(2) * sqrt(3)))#

#color(white)("XXX")=sqrt(2)#

Jun 19, 2018

#(2sqrt(3))/(sqrt(6))=(2sqrt(3))/(sqrt(3)sqrt(2))=(2)/(sqrt(2))=sqrt2#

Jun 19, 2018

Answer:

#sqrt(2)#

Explanation:

#(2sqrt3)/sqrt6#

#rArr (2sqrt3)/sqrt6 xxsqrt6/sqrt6#

#rArr =(2sqrt(18))/sqrt36#

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

#=> ( 6sqrt(2)) / 6 #

#rArr sqrt(2) #.

Jun 24, 2018

Answer:

#=>sqrt2#

Explanation:

Because of the radical law

#sqrtab=sqrta*sqrtb#, we can rewrite this expression as

#(2sqrt3)/(sqrt2*sqrt3)#

Cancelling out common terms

#(2cancel(sqrt3))/(sqrt2*cancel(sqrt3)#

#=>2/sqrt2#

The convention is to not have an irrational number in the denominator, so let's multiply the top and bottom by #sqrt2#.

#(2*sqrt2)/(sqrt2)^2#

#(2sqrt2)/2#

#(cancel2sqrt2)/cancel2#

#=>sqrt2#

Hope this helps!