How do you simplify #sqrt12 +sqrt 27#?
1 Answer
Apr 26, 2016
Explanation:
#sqrt(12)+sqrt(27)=sqrt(2^2*3)+sqrt(3^2*3)=2sqrt(3)+3sqrt(3)=5sqrt(3)#
Note that:
-
If
#a >= 0# then#sqrt(a^2) = a# -
If
#a, b >= 0# then#sqrt(ab) = sqrt(a) sqrt(b)#
So if
#sqrt(a^2 b) = sqrt(a^2)sqrt(b) = a sqrt(b)#