What is #sqrt(27/16)#?

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Answer 1 · 2015-07-26T22:55:15

I found: #3/4sqrt(3)#

You can write your square root as:
#sqrt(27)/sqrt(16)=sqrt(3*9)/4=(sqrt(9)sqrt(3))/4=3/4sqrt(3)#

Answer 2 · 2015-07-26T22:58:03

#(3sqrt(3))/4#

Use the quotient property of radicals to rewrite your expression as

#sqrt(27/16) = sqrt(27)/sqrt(16) = sqrt(27)/4#

Since #27# is not a perfect square, you're going to have to see if you can write it as a product of a perfect square and another number

#27 = 9 * 3 = 3^2 * 3#

This means that your expression will be

#sqrt(3^2 * 3)/4 = (sqrt(3^2) * sqrt(3))/4 = color(green)((3 sqrt(3))/4)#