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### Babylonian method

Using the **Babylonian method** you can approximate the value of a positive square root *x* = √*a* by iterations of

##### Explanation

We want to calculate *x* = √*a* and search for an equivalent equation to do so

We can make an iteration of this, where √*a* always lies between *x _{n}* and

*a*/

*x*. Then you can assume that the average of these two values gives a better value, so that

_{n}

##### Example 1

If we start for *x* = √2 with *a* = 2 and *x*_{0} = 2, we get the following iterations

That's even more accurate than the value 1,414213562… your calculator shows.

##### Example 2

If we start for *x* = √9 with *a* = 9 and *x*_{0} = 2, we get the following iterations

If we start with *x*_{0} = 3, then the result is immediately

because √9 = 3, but we already knew that.