How do you simplify #sqrt27+sqrt48#?

Question

2368 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-09T05:58:42

#7sqrt(3)#

Writing
#sqrt(3*9)+sqrt(3*16)#
and usting that
#sqrt(ab)=sqrt(a)*sqrt(b)# for #a,b>=0#
we get
#3sqrt(3)+4*sqrt(3)=7sqrt(3)#

Answer 2 · 2018-06-09T06:57:22

#7sqrt3#

Neither #27# nor #48# are perfect squares, but they have factors which are squares.

Find the roots of the factors where possible and then give the answer in surd form.

#sqrt27 +sqrt48#

#= sqrt(color(blue)(9)xx3) + sqrt(color(blue)(16)xx3)#

#=color(blue)(3)sqrt3 + color(blue)(4)sqrt3#

#=7sqrt3" "larr# add like terms