# How do you find a one-decimal place approximation for #sqrt 50#?

##### 1 Answer

Oct 21, 2015

#### Explanation:

Since

#sqrt(50) = [7;bar(14)] = 7+1/(14+1/(14+1/(14+...)))#

A rough approximation can be made by truncating early:

#sqrt(50) ~~ [7;14] = 7+1/14 = 99/14 = 7.0dot(7)1428dot(5)#

If we want a better one, just include more terms:

#sqrt(50) ~~ [7;14;14] = 7+1/(14+1/14) = 7+14/197 = 1393/197 ~~ 7.071066#

Actually