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Archimedes and Numerical Roots

The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?

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Archimedes and Numerical Roots

Age 14 to 16
Challenge Level

This problem builds on Approximating Pi. This brilliant man Archimedes managed to establish that $3 1/10 < \pi < 3 1/7$.

The problem is how did he calculate the lengths of the sides of the polygons, which he needed to be able to calculate square roots? He didn't have a calculator but needed to work to an appropriate degree of accuracy. To do this he used what we now call numerical roots.

How might he have calculated $\sqrt{3}$?

This must be somewhere between 1 and 2. How do I know this?

Now calculate the average of $3/2$ and $2$ (which is 1.75) - this is a second approximation to $\sqrt{3}$. i.e. we are saying that a better approximation to $\sqrt{3}$ is $(3/n + n)/2$ where $n$ is an approximation to $\sqrt{3}$.

We then repeat the process to find the new (third) approximation to $\sqrt{3}$. $$\sqrt{3} \approx {(3 / 1.75 + 1.75) \over {2}} = 1.73214\dots$$ to find a fourth approximation repeat this process using 1.73214 and so on...

How many approximations do I have to make before I can find $\sqrt{3}$ correct to five decimal places?

Why do you think it works?

Will it always work no matter what I take as my first approximation and does the same apply to finding other roots?