How do you find a one-decimal place approximation for $sqrt 37$ ?

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Answer 1 · 2015-10-18T15:08:37

Use one step of Newton Raphson method to find:

$sqrt(37) ~~ 73/12 = 6.08dot(3) ~~ 6.1$

To find the square root of a number $n$ , choose a reasonable first approximation $a_0$ and use the formula:

$a_(i+1) = (a_i^2+n)/(2a_i)$

Repeat to get more accuracy.

For our purposes, $n = 37$ and let $a_0 = 6$ since $6^2=36$ .

Then:

$a_1 = (a_0^2+n)/(2a_0) = (6^2+37)/(2*6) = (36+37)/12 = 73/12 = 6.08dot(3)$

We don't need any more steps to get the first decimal place, since we were pretty close to start.

Actually $sqrt(37)$ is expressible as something called a continued fraction:

$sqrt(37) = [6;bar(12)] = 6+1/(12+1/(12+1/(12+...)))$

So you can also get approximations for $sqrt(37)$ by just truncating this continued fraction and working out the value.

For example:

$sqrt(37) ~~ [6;12,12] = 6+1/(12+1/12) = 6 + 12/145 = 882/145 ~~ 6.08276$

How do you find a one-decimal place approximation for sqrt 37sqrt 37?