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

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Answer 1 · 2015-10-21T23:51:19

$sqrt(50) ~~ 7 + 1/14 ~~ 7.1$

Since $50 = 7^2 + 1$ is of the form $n^2 + 1$ , its square root is a very simple continued fraction:

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

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 $sqrt(50) ~~ 7.07106781186547524400$

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