How do you simplify #sqrt12+sqrt27-sqrt3#?

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Answer 1 · 2017-04-14T12:04:08

#4sqrt(3)#

#sqrt(12)+sqrt(27)-sqrt(3)#

By #12=2cdot2cdot3# and #27=3cdot3cdot3#,

#=sqrt(2cdot2cdot3)+sqrt(3cdot3cdot3)-sqrt(3)#

By grouping pairs of the same factors,

#=sqrt(2cdot2)sqrt(3)+sqrt(3cdot3)sqrt(3)-sqrt(3)#

By simplifying the square-roots of squares,

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

By factoring out #sqrt(3)#,

#=(2+3-1)sqrt(3)=4sqrt(3)#

I hope that this was clear.