How do you simplify #sqrt3(sqrt12-sqrt6)#?

1 Answer
Nov 9, 2015

First, expand the product: multiply #sqrt(3)#with #sqrt(12)# and with #sqrt(6)#, respectively:

#sqrt(3)(sqrt(12) - sqrt(6)) = sqrt(3) * sqrt(12) - sqrt(3) * sqrt(6)#

Now, use the rule #sqrt(a*b) = sqrt(a) * sqrt(b)#:

#sqrt(3) * sqrt(12) - sqrt(3) * sqrt(6)#
#= sqrt(3 * 12) - sqrt(3 * 6)#
#= sqrt(36) - sqrt(18)#
#= 6 - sqrt(18)#

The last one, #sqrt(18)#, can be simplifyed further since #18 = 2 * 9# and #9 = 3^2#:

#6 - sqrt(18)#
#= 6 - sqrt(2*9)#
#= 6 - sqrt(2)*sqrt(9)#
#= 6 - sqrt(2)*3#
#= 6 - 3sqrt(2)#

I hope that this helped!