How do you simplify #\sqrt { \frac { 80} { 3} }#?

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Answer 1 · 2017-12-19T17:49:04

#=(4sqrt(5))/3#

It is unclear if the radicand includes the #3# or not. Assuming you mean

#(sqrt(80))/3#

Then you have

#(sqrt(16*5))/3#

#=(4sqrt(5))/3#

This does not simplify any further and thus is the final answer.

Answer 2 · 2017-12-19T17:51:26

See a solution process below:

First, rewrite the expression as:

#sqrt(80)/sqrt(3)#

Next, simplify the numerator:

#sqrt(80)/sqrt(3) => sqrt(16 * 5)/sqrt(3) => (sqrt(16)sqrt(5))/sqrt(3) => (4sqrt(5))/sqrt(3)#

Now, we can rationalize the denominator:

#(4sqrt(5))/sqrt(3) => (4sqrt(5))/sqrt(3) xx sqrt(3)/sqrt(3) => (4sqrt(5)sqrt(3))/(sqrt(3)sqrt(3)) =>#

#(4sqrt(5 xx 3))/3 => (4sqrt(15))/3#