How do you find the square root of 16562?

The ancient greeks used to compute square roots by sucessive approximations.

Given a number `N` for which they need to compute the square root and given an initial approximation `q_0` they proceed as follows:

`(q_0+ delta q_0)^2=N` or
`q_0+2 q_0 delta q + (delta q_0)^2 = N`

So they were searching for an approximation variation `delta q_0` with the purpose of correct the initial guess `q_0`. They supposed also that `delta q_0` being small, much smaller would be `(delta q_0)^2` so they used the approximation

`q_0+2 q_0 delta q_0 approx N`

solving for `delta q_0` they got

`delta q_0 = ((N/q_0) - q_0)/2`

once corrected `q_0` they got `q_1 = q_0 + delta q_0`
and then follow with

`delta q_1 = ((N/q_1) - q_1)/2` etc.

Let us apply that process for calculation of square root of

`N =16562`

our initial guess will be `q_0 = 400`

`delta q_0 =((16562/400)-400)/2 =-179.298`

so `q_1 = 400-179.298` and calculating `delta q_1`

`delta q_1=-72.83015` so `q_2 = 400-179.298-72.83015`

calculating `delta q_2`

`delta q_2 = -17.93516` so `q_3 = 400-179.298-72.83015-17.93516`

In the third iteration we get

`delta_3 = -1.23779` obtaining a result of

`sqrt(16562) approx 128.699` which is a satisfactory result.

To find square root of `16562`, we should first factorize it.

From divisibility rules, it is apparent that it is divisible by `2` and dividing by `2`, we get `8281`.

`8281` is clearly not divisible by `3` and `5`, but is divisible by `7`. Dividing by `7`, we get `1183`, which is again divisible by `7`and dividing it by `7` we get `169`, which is `13xx13`.

Hence, `16562=2xx7xx7xx13xx13` and hence

`sqrt16562=sqrt(2xxbar(7xx7)xxbar(13xx13))`

= `7xx13xxsqrt2=91sqrt2=91xx1.4142=128.6922`

If you are not sure of the numbers build a factor tree to find the squared prime numbers.

Thus `sqrt(16562) =sqrt(2xx7^2xx13^2)`

`=7xx13xxsqrt(2) = 91sqrt(2)~~128.69` to 2 decimal places