How do you simplify the radical expression #sqrt72#?

2 Answers
Mar 20, 2017

#sqrt72 = sqrt(36xx26)=sqrt36xxsqrt2 = 6xxsqrt2#

Explanation:

This takes advantage of the relation

#sqrt(axxb)=sqrtaxxsqrtb#

Look for a factor in the original value that is a perfect square (4, 9, 16, 25, 36, etc.)

Mar 20, 2017

#6sqrt(2)#

Explanation:

(I think you meant as a simplified radical, so that's what I'm going to answer.)
#sqrt(72)# can also be thought of as #sqrt(36*2)#, or #sqrt(6*6*2)#.
Using the rule that #sqrt(ab)=sqrt(a)*sqrt(b)#, you end up with #sqrt(6)*sqrt(6)*sqrt(2)#.
This can also be expressed as #sqrt(6)^2*sqrt(2)#.
The square root and square cancel, leaving us with the simplified radical of #6sqrt(2)#.