What is 7 over the square root of 27?

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Answer 1 · 2015-07-24T16:18:21

#(7 sqrt(3))/9#

Start by writing your expression, which features #7# in the numerator and #sqrt(27)# in the denominator.

#7/(sqrt(27)#

Now, the important thing to realize here is that you can write #27# as

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

This means that the denominator becomes

#sqrt(27) = sqrt(3""^2 * 3) = sqrt(3""^2) * sqrt(3) = 3sqrt(3)#

The expression is now

#7/(3 * sqrt(3))#

Next, you have to rationalize the denominator, which you can do by multiplying the numerator and the denominator by #sqrt(3)#, to get

#(7 * sqrt(3))/(3 * underbrace(sqrt(3) * sqrt(3))_(color(blue)("=3))) = (7 * sqrt(3))/(3 * color(blue)(3)) = color(green)((7 sqrt(3))/9)#