How do you simplify #sqrt(72)#?

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Answer 1 · 2015-07-01T12:39:25

#sqrt72 = 6sqrt2#

Check to see if #72# is divisible by a perfect square:
#2^2 = 4# and #72 = 4*18#
So #sqrt72 = sqrt(4*18) = sqrt4*sqrt18 =2sqrt18#

Continue be checking to see is #18# is divisible by a perfect square:
#2^2 = 4#, and #18# is not divisible by #4#
#3^2=9# and #18=9*2#

so we get:
#sqrt72 = 2sqrt18 = 2sqrt(9*2)=2sqrt9sqrt2=2*3sqrt3=6sqrt2#

There are many ways of writing/working this simplification. Here are a few more:

#sqrt72 = sqrt(4*9*2) = 2*3*sqrt2 = 6sqrt2#

#sqrt72 = sqrt (36*2) = sqrt36sqrt2=6sqrt2#

#sqrt72 = sqrt(2^3*3^2) = sqrt(2^2*2*3^2) = 2*sqrt2*3 = 6sqrt2#