How do you simplify #sqrt(16/4)#?

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Answer 1 · 2016-02-16T21:30:19

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Note that #4^2=16" and " 2^2=4#

Write #sqrt(16/4)" "as" " sqrt(4^2/2^2)#

This is the same as:

#(sqrt(4^2))/(sqrt(2^2)#

#sqrt(4^2)=4" and "sqrt(2^2)=2#

So we end up with:

#4/2 =2#