How do you simplify #sqrt(80)#?

1 Answer
George C.
Jul 23, 2015

If #a, b >= 0# then #sqrt(ab) = sqrt(a)sqrt(b)#
If #a >= 0# then #sqrt(a^2) = a#

Hence #sqrt(80) = sqrt(16*5) = 4sqrt(5)#

Explanation:

#sqrt(80) = sqrt(16*5) = sqrt(4^2*5) = sqrt(4^2)sqrt(5) = 4sqrt(5)#

Note that since #sqrt(81) = sqrt(9^2) = 9# then

#9/4 = sqrt(81)/(sqrt(80)/sqrt(5)) = sqrt(81/80)sqrt(5)# is a useful approximation for #sqrt(5)#.

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