How do you evaluate #\sqrt { 48} + \sqrt { 3} - \sqrt { 27}#?

1 Answer
Mic
May 14, 2017

You want to have the number inside the square root all be the same for all the numbers. Since since #sqrt(3)# is the smallest irrational number in a square root, we want to have the other numbers have the #sqrt(3)# in order to solve.

So first expand to get:

#sqrt(16*3)# + #sqrt(3 * 1)# - #sqrt(9*3)#

Then:

4#sqrt(3)# + 1#sqrt(3)# - 3 #sqrt(3)#

That will give you the answer:
2#sqrt(3)#

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