How do you simplify #\frac { \sqrt { 63} + 24} { 3}#?

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Answer 1 · 2018-04-05T03:53:22

#8+sqrt7#

#(sqrt63+24)/3#
=#((sqrt9xxsqrt7)+24)/3#
=#((3xxsqrt7)+24)/3#
=#(3(sqrt7+8))/3#
=#8+sqrt7#

Answer 2 · 2018-04-05T04:00:20

#sqrt(7) + 8#

First, you would simplify the square root of #63# by splitting it into #sqrt(7)# multiplied by #sqrt(9)# (because #9xx7=63#).
Then #sqrt(9)# reduces to #3#, so you have #3sqrt(7) + 24# (ignoring the dividing by #3# for now).
Then take #3sqrt(7) + 24# and divide each term by #3#.
The #3#'s in #3sqrt(7)# would cancel and give you #sqrt(7)#.
The #24# divided by #3# would give you #8#.
So your final answer is #sqrt(7) + 8#