How do you find the square root of 50?

#sqrt50#

= #sqrt(2xxul(5xx5))#

= #5sqrt2#

The square root of #50# is not a whole number, or even a rational number. It is an irrational number, but you can simplify it or find rational approximations for it.

First note that

#50 = 2 xx 5 xx 5#

contains a square factor #5^2#. We can use this to simplify the square root:

#sqrt(50) = sqrt(5^2*2) = sqrt(5^2)*sqrt(2) = 5 sqrt(2)#

Apart from simplifying it algebraically, what is its numerical value?

Note that #7^2 = 49#, so #sqrt(49) = 7# and #sqrt(50)# will be slightly larger than #7#.

In fact, since #50=7^2+1#, the square root of #50# is expressible as a very regular continued fraction:

#sqrt(50) = 7+1/(14+1/(14+1/(14+1/(14+1/(14+1/(14+...))))))#

This can be written as #sqrt(50) = [7;bar(14)]# where the bar over the #14# indicates the repeating part of the continued fraction.

We can terminate the continued fraction early to give us rational approximations for #sqrt(50)#.

For example:

#sqrt(50) ~~ [7;14] = 7+1/14 = 7.0bar(714285)#

#sqrt(50) ~~ [7;14,14] = 7+1/(14+1/14) = 7+14/197 ~~ 7.071066#

In fact:

#sqrt(50) ~~ 7.071067811865475244#