How do you simplify #(5sqrt15) /( 3sqrt27) #?

1 Answer
Mar 11, 2016

Answer:

Same thing. Just a very slightly different presentation!

#(5sqrt(5))/9#

Explanation:

The answer given was stopped at #(5sqrt(5))/(3sqrt(9))#

#color(red)("'~~~~~~~~~~~~~~~~ Something to think about! ~~~~~~~~~~~")#

It is considered better mathematical practice if you 'get rid' of any roots that are in the denominator (bottom number)

The method is based on: Multiply any number by 1 and you do not change its inherent value

So #(5sqrt(5))/(3sqrt(9))xx1# does not change its value

Suppose we wrote 1 as #(sqrt(9))/(sqrt(9))#

This is still 1

#color(red)("'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#
#color(blue)("Back to solving the question")#

Multiply by 1 but in the form of #(sqrt(9))/(sqrt(9))# giving

#(5sqrt(5))/(3sqrt(9))xx(sqrt(9))/(sqrt(9))" "=" "(5sqrt(5)sqrt(9))/(3sqrt(9)sqrt(9))#

#(5sqrt(5)sqrt(9))/(3xx9)#

But #sqrt(9)=3# giving:

#(5xx3xxsqrt(5))/(3xx9)" "=" " 3/3xx(5sqrt(5))/9#

But #3/3=1# giving:

#(5sqrt(5))/9#