How do you rationalize the denominator and simplify #sqrt24/sqrt3#?

1 Answer
May 13, 2016

#sqrt(24)/sqrt(3) = 2sqrt(2)#

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(3)# first:

#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)#