How do you simplify #sqrt27*sqrt33#?

2 Answers
Feb 6, 2016

Answer:

9#sqrt#11

Explanation:

#sqrt#27 * #sqrt#33

= [#sqrt#39] * [#sqrt#311]
= [3#sqrt#3] * [#sqrt#311] ; #sqrt#27 becomes 3#sqrt#3 because the sqrt of 9 is 3
= [3#sqrt#3] * [#sqrt#3] * [#sqrt#11] ; we separate #sqrt#3 and #sqrt#11
= 3#sqrt#3
#sqrt#11 ; we multiply those with #sqrt#3
= 3#sqrt#9 * #sqrt#11
= 3 * 3 * #sqrt#11 ; there are two 3s now because again the sqrt of 9 is 3
= 9#sqrt#11 ; FINAL ANSWER

Feb 6, 2016

Answer:

First, before multiplying, you can simplify the √27.

Explanation:

#sqrt(27) = sqrt(9 xx 3)#

= #3sqrt(3)#

Now we can multiply, multiplying radicals with radicals and whole numbers with whole numbers.

#3sqrt(3) xx sqrt(33)#

= #3sqrt(99)#

= #3sqrt(9 xx 11)#

= #3(3)sqrt(11)#

= #9sqrt(11)#

So, #9sqrt(11)# is your answer, in simplest form.

Practice exercises:

  1. Simplify:

a) #3sqrt(5) xx 4sqrt(7)#

b) #sqrt(24) xx 2sqrt(48)#

2 . Solve for x in #sqrt(6x) xx sqrt(2) = 2x#

Good Luck!