# Optimal way to pick the starting $x$ when computing the square root with the Babylonian method

I was reading the example given in the Wikipedia article for the Babylonian method of computing the square root, and I wondered why did they set the starting $$x_0$$ to 600:

To calculate $$\sqrt S$$, where S = 125348, to six significant figures, use the rough estimation method above to get:

$$x_0 = 6 * 10^2$$ = 600.000

Did they pick 6 because the number $$S$$ has 6 digits, and multiplied by $$10^2$$ to get a number that has half as many digits as $$S$$?

Update: another reason I'm asking is that I've seen this algorithm that calculates the square root efficiently, but which lacks explanatory comments. I am fairly confident that what it does is compute the square root as if the input was a perfect square of a power of two, but I'm not sure if that's the optimal way in a wider mathematical sense.

In that section, it says that the number $$6$$ is for numbers in the form $$\sqrt{S} = \sqrt{a} \cdot 10^n$$, where $$10 < a < 100$$.
To minimise the absolute error (say $$5$$ is $$3$$ more than $$2$$), we take the arithmetic mean $$\frac{\sqrt{10} + \sqrt{100}}{2} \approx 6.58$$, and to minimise the relative error (say $$5$$ is $$2.5$$ times that of $$2$$), we take the geometric mean $$\sqrt{\sqrt{10} \cdot \sqrt{100}} \approx 5.62$$. Therefore, as a rough estimate to $$1$$ significant figure, $$6$$ is the value that minimises both the arithmetic and the geometric mean.