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

Oct 18, 2015

Use one step of Newton Raphson method to find:

$\sqrt{37} \approx \frac{73}{12} = 6.08 \dot{3} \approx 6.1$

#### Explanation:

To find the square root of a number $n$, choose a reasonable first approximation ${a}_{0}$ and use the formula:

${a}_{i + 1} = \frac{{a}_{i}^{2} + n}{2 {a}_{i}}$

Repeat to get more accuracy.

For our purposes, $n = 37$ and let ${a}_{0} = 6$ since ${6}^{2} = 36$.

Then:

${a}_{1} = \frac{{a}_{0}^{2} + n}{2 {a}_{0}} = \frac{{6}^{2} + 37}{2 \cdot 6} = \frac{36 + 37}{12} = \frac{73}{12} = 6.08 \dot{3}$

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