How do you rationalize the denominator and simplify #sqrt24/sqrt3#?
1 Answer
May 13, 2016
Explanation:
We will use:
-
If
#a, b >= 0# then#sqrt(ab) = sqrt(a)sqrt(b)# -
If
#b > 0# then#sqrt(a/b) = sqrt(a)/sqrt(b)#
The normal way to solve this example is to multiply both the numerator and denominator by
#sqrt(24)/sqrt(3) = (sqrt(24)*sqrt(3))/(sqrt(3)*sqrt(3)) = sqrt(72)/3 = sqrt(6^2*2)/3 = (sqrt(6^2)*sqrt(2))/3 = (6sqrt(2))/3 = 2sqrt(2)#
Alternatively, we can combine the numerator and denominator inside the square root:
#sqrt(24)/sqrt(3) = sqrt(24/3) = sqrt(8) = sqrt(2^2)*sqrt(2) = 2sqrt(2)#
