How do you find the square root of 1242?

As we can see, 1242 is not a perfect square, and so therefore, will not be able to be simplified into a whole number. What we can do, though, is simplify `sqrt1242` until it's in the form `a*sqrtb`, where `sqrtb` cannot be simplified further.

Let's start by dividing `sqrt1242` into its factors:

`sqrt1242=sqrt2*sqrt621=sqrt2*sqrt3*sqrt207=sqrt2*sqrt3*sqrt3*sqrt69=sqrt2*sqrt3*sqrt3*sqrt3*sqrt23`

That last expression can be rewritten as:

`sqrt2*sqrt3*sqrt9*sqrt23`

And that can be rewritten as:

`3*sqrt2*sqrt3*sqrt23`

Further simplification:

`3*sqrt138`

To find rational approximations to `sqrt(1242)` we can proceed as follows:

First split `1242` into pairs of digits starting from the right:

`12"|"42`

Examining the leftmost group of digits, note that:

`3^2 = 9 < 12 < 16 = 4^2`

So:

`3 < sqrt(12) < 4`

and:

`30 < sqrt(1242) < 40`

In fact note that `12` is close to the average of `9` and `16`, so a reasonable approximation to `sqrt(12)` is `3.5` and to `sqrt(1242)` is `35`.

Let's use a variant on the Babylonian method:

Given a rational approximation `p_i/q_i` to `sqrt(n)`, a better approximation is given by `p_(i+1)/q_(i+1)` where:

`{ (p_(i+1) = p_i^2+n q_i^2), (q_(i+1) = 2 p_i q_i) :}`

Starting with `p_0/q_0` repeatedly apply these formulae to get better approximations.

Let `n=1242` and `p_0/q_0 = 35/1`

Then:

`{ (p_1 = p_0^2+n q_0^2 = 35^2+1242 * 1^2 = 1225+1242 = 2467), (q_1 = 2 p_1 q_1 = 2 * 35 * 1 = 70) :}`

`{ (p_2 = p_1^2+n q_1^2 = 2467^2 + 1242 * 70^2 = 6086089 + 6085800 = 12171889), (q_2 = 2 p_1 q_1 = 2 * 2467 * 70 = 345380) :}`

So:

`sqrt(1242) ~~ 12171889/345380 ~~ 35.2420204`

`sqrt(12'42.00'00'00'00')=35.2420`
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