How do you simplify #sqrt 20div sqrt 15#?

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Answer 1 · 2018-07-23T20:22:05

#(2sqrt3)/3#

We have the following:

#sqrt20/sqrt15#

#20=4*5# and #15=3*5#. With this in mind, we can rewrite the radicals in this way. We now have

#(sqrt4sqrt5)/(sqrt3sqrt5)#

Common factors in the numerator and denominator cancel, and we're left with

#(sqrt4cancelsqrt5)/(sqrt3cancelsqrt5)=>2/sqrt3#

The convention is to not have an irrational number in the denominator, so we can multiply this by #sqrt3/sqrt3#.

We are essentially multiplying by #1#, and we get

#(2sqrt3)/3#

Hope this helps!