How do you solve #(8sqrt(6mn) + 6sqrt(8mn))/ (2sqrt(2mn))#?

2 Answers
Aug 23, 2016

#4sqrt3 +6#

Explanation:

This is not an equation, but an expression or fraction, so it can be simplified rather than solved,

Write each term in the numerator with its own denominator.

#(8sqrt(6mn))/(2sqrt(2mn) ) + (6sqrt(8mn))/(2sqrt(2mn))#

Note that there is a factor of #sqrt(2mn)# in each part.

Try to isolate it in every part by splitting up the square roots into separate roots.

=# (cancel8^4sqrt(3xx2mn))/(cancel2sqrt(2mn)) +(cancel6^3sqrt(4xx2mn))/(cancel2sqrt(2mn))#

=# (4sqrt3 sqrt(2mn))/(sqrt(2mn)) +(3xx2 sqrt(2mn))/(sqrt(2mn))#

Now we can cancel the factors that are the same:

=# (4sqrt3 cancelsqrt(2mn))/cancel(sqrt(2mn)) +(3xx2 cancelsqrt(2mn))/(cancelsqrt(2mn))#

=#4sqrt3 +6#

Aug 23, 2016

#(8sqrt(6mn)+6sqrt(8mn))/(2sqrt(2mn))=4sqrt3+6#

Explanation:

#(8sqrt(6mn)+6sqrt(8mn))/(2sqrt(2mn))#

= #(8sqrt(6mn))/(2sqrt(2mn))+(6sqrt(8mn))/(2sqrt(2mn))#

= #8/2×sqrt((6mn)/(2mn))+6/2×sqrt((8mn)/(2mn))#

= #4sqrt3+3sqrt4=4sqrt3+3×2#

= #4sqrt3+6#