How do you simplify #3sqrt5 (times) sqrt5 + 3sqrt5 (times) 2sqrt75#?

1 Answer
Sep 30, 2017

Answer:

Depending on how the question is interpreted you have:

#45+sqrt(15)" or "120sqrt(5)#

Explanation:

#color(blue)("Assumption: you really meant the question to be as written")#

#color(brown)("3sqrt(5)xxsqrt(5)+3sqrt(5)xx2sqrt(75)#

As you there is no grouping by brackets we have to look at priority of action (add, divide, multiply etc). Multiplication has a higher priority than add so we have to apply that first. Thus we have:

#(3sqrt(5)xxsqrt(5))+(3sqrt5xx2sqrt(75))#

#(3xx(sqrt(5))^2)+(3xx2xxsqrt(5)xxsqrt(75))#

note that #75->5xx15# so #sqrt(5)xxsqrt(75)->(sqrt(5))^2xxsqrt(15)#

#(3xx(sqrt(5))^2)+(3xx2xx(sqrt(5))^2xxsqrt(15))#

#color(white)("dddd")(15)color(white)("ddd")+color(white)("ddd")(30+sqrt(15))#

#45+sqrt(15)#

Note that #15=3xx5# and both 3 and 5 prime numbers so it is simpler to leave the #sqrt(15)# as it is.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Assumption: the brackets are the wrong way round")#
Having the brackets round the way you have is very unexpected.

#color(brown)( (3sqrt(5))xx(sqrt(5)+3sqrt(5))xx(2sqrt(5)))#

#(3sqrt(5))xxcolor(white)("dd")(4sqrt(5))color(white)("ddd")xx(2sqrt(5))#

Dealing with the whole numbers part #->3xx4xx2=color(red)(24)#

Dealing with the square roots part #color(white)("d")->(sqrt(5)xxsqrt(5))xxsqrt5#

#color(white)("dddddddddddddddddddddddddd")->color(white)("dddd") 5color(white)("dddd")xxsqrt(5)=color(purple)(5sqrt(5))#

Putting it all together we have: #color(red)(24)color(purple)(xx5sqrt(5)) = 120sqrt(5)#