How do you simplify #sqrt27+sqrt48#?

2 Answers
Jun 9, 2018

Answer:

#7sqrt(3)#

Explanation:

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)#

Jun 9, 2018

Answer:

#7sqrt3#

Explanation:

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