How do you simplify #(\sqrt { 12} + \sqrt { 2} ) ( \sqrt { 2} \sqrt { 12} - \sqrt { 6} )#?

2 Answers
IDKwhatName · EZ as pi
Jun 26, 2017

#6sqrt(2)+6sqrt(3)#

Explanation:

#(sqrt(12) + sqrt(2))(sqrt(2)sqrt(12)-sqrt(6))#

The FOIL method will be used for this.
First
Outside
Inside
Last

F - #sqrt(12)*sqrt(2)*sqrt(12) = 12*sqrt(2) = 12sqrt(2)#
O - #sqrt(12)*-sqrt(6) = -sqrt(12*6) = -sqrt(72)#
I - #sqrt(2)*sqrt(2)*sqrt(12) = 2*sqrt(12) = 2sqrt(12)#
L - #sqrt(2)*sqrt(6) = sqrt(2*6) = sqrt(12)#

Now add each term, #12sqrt(2) - sqrt(72) + 2sqrt(12) + sqrt(12)#

#3sqrt(12)+12sqrt(2)+sqrt(72)#

#sqrt(12)=sqrt(4*3)=(sqrt(4)*sqrt(3))=2*sqrt(3)=2sqrt(3)#

Therefore, #3sqrt(12) = 3*(2sqrt(3)) = 6sqrt(3)#

#-sqrt(72)=-sqrt(2*36)=-(sqrt(2)*sqrt(36))=(-6*sqrt(2))=-6sqrt(2)#

#(sqrt(12) + sqrt(2))(sqrt(2)sqrt(12)-sqrt(6))# = #12sqrt(2)+6sqrt(3)-6sqrt(2) = 6sqrt(2)+6sqrt(3)#

EZ as pi
Jun 26, 2017

#=6sqrt2+2sqrt3#

Explanation:

Write all the radicands as the product of prime factors first, then we know what we are dealing with.

#(sqrt12+sqrt2)(sqrt2sqrt12-sqrt6)#

#=(sqrt(2^2*3) +sqrt2)(sqrt2sqrt(2^2*3)-sqrt(2*3))#

Now find the product of the brackets - 'F O I L '

#=sqrt(2^5*3^2)-sqrt(2^3*3^2)+sqrt(2^4*3) -sqrt(2^2*3)#

Find the roots where possible:

#=2^2*3sqrt2-2*3sqrt2+2^2sqrt3-2sqrt3#

#=12sqrt2-6sqrt2+4sqrt3-2sqrt3#

#=6sqrt2+2sqrt3#

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