How do you simplify: #sqrt12+2sqrt27-5sqrt48# ?

Question

967 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-06T05:59:08

#-12sqrt3#

Let A#=sqrt12+2sqrt27-5sqrt48#

When simplifying roots of integers it if often helpful to express the integer in terms of its prime factors.

#sqrt12 = sqrt(2xx2xx3) = 2sqrt3#

#sqrt27= sqrt(3xx3xx3) = 3sqrt3#

#sqrt48 = sqrt(2xx2xx2xx2xx3) = 2xx2sqrt3#

Hence A#= 2sqrt3 + 2xx3sqrt3 - 5xx4sqrt3#

#= 2sqrt3+6sqrt3-20sqrt3#

#=-12sqrt3#