How do you simplify #sqrt(250/252)#?

1 Answer
Mar 14, 2017

Answer:

#(5*sqrt(5))/(3*sqrt(14))#

Explanation:

First, simplify the fraction in the square root:
#sqrt(125/126)#
Then, use the rule#sqrt(a/b)=sqrt(a)/sqrt(b)#:
#sqrt(125)/sqrt(126)#
Now, onto simplifying the radicals on the top and bottom: split the square roots using the rule #sqrt(ab)=sqrt(a)*sqrt(b)#:
#(sqrt(25)*sqrt(5))/(sqrt(9)*sqrt(14))# (see notes below if this doesn't make sense)
Simplify #sqrt(25)# and #sqrt(9)#:
#(5*sqrt(5))/(3*sqrt(14))#
And that's as far as you can go!

Note on simplification of #sqrt(125)# & #sqrt(126)#:
The way I used to simplify these is to find all the prime factors:
#sqrt(125)= sqrt(5) * sqrt(5) * sqrt(5)#
#sqrt(126) = sqrt(2) * sqrt(3) * sqrt(3) * sqrt(7)#
Since #sqrt(5) * sqrt(5) = sqrt(5)^2 = 5# and #sqrt(3) * sqrt(3) = sqrt(3)^2 = 3#, you can simplify these to:
#sqrt(125)= 5 * sqrt(5)#
#sqrt(126) = sqrt(2) * 3 * sqrt(7)#
Using the rule #sqrt(ab)=sqrt(a) * sqrt(b)# on the second expression:
#sqrt(125)= 5 * sqrt(5)#
#sqrt(126) = 3 * sqrt(14)#
And you end up in the same place as the first method