What is #sqrt(48)# ?

2 Answers
Apr 28, 2017

#4sqrt3#

Explanation:

Like Asma said you can calculate the square root on a calculator for the exact amount. But sometimes a professor will ask you to simplify. This will be very useful in the future.

#sqrt48#

We want to check if there is a number with a perfect square that is a multiple of the number in our square root. In this case, there is one.
#16*3=48#

Since #16# has a perfect square root of #4# we can solve this like this...

#sqrt48#

#sqrt(16*3)# #<---# [Move square root of #16# outside of the root]

#4sqrt3#

May 13, 2017

#sqrt(48) = 4sqrt(3) ~~ 18817/2716 ~~ 6.92820324#

Explanation:

Note that:

#48 = 4^2*3#

So we find:

#sqrt(48) = sqrt(4^2*3) = sqrt(4^2)sqrt(3) = 4sqrt(3)#

That is the "simplest" form.

#sqrt(48)# is an irrational number a little less than #7#, since #7^2 = 49#.

In fact, since #48=7^2-1# is in the form #n^2-1# its square root can be expressed as a continued fraction with a simple pattern:

#sqrt(48) = [6;bar(1,12)] = 6+1/(1+1/(12+1/(1+1/(12+1/(1+1/(12+..))))))#

To get rational approximations, we can truncate this continued fraction early (preferably just before a "#12#")...

For example:

#sqrt(48) ~~ 6+1/(1+1/(12+1/1)) = 97/14 = 6.9bar(285714)#

#sqrt(48) ~~ 6+1/(1+1/(12+1/(1+1/(12+1/1)))) = 1351/195 = 6.9bar(282051)#

#sqrt(48) ~~ 6+1/(1+1/(12+1/(1+1/(12+1/(1+1/(12+1/1)))))) = 18817/2716 ~~ 6.92820324#

A calculator tells me that it is closer to #6.92820323#, but that's not bad.