How do you find the square root of 43?

If you are looking at approximation methods that you can employ using pen, paper and some mental arithmetic, you can try a Binomial Expansion.

If you start at 36, a square number, so that you are looking for `sqrt(36 +7)`, you can now play with that a little:

`sqrt(36 +7) = sqrt 36 sqrt(1 +7/36) = 6 sqrt(1 +7/36)`

You can then use the Binomial Expansion , ie:

`(1+x)^alpha = 1 + alpha x + (alpha (alpha - 1))/(2!)x^2 + ...`:

In this case:

`6 (1 +7/36)^(1/2)`

`=6 (color(green)(1 +1/2 * 7/36) + 1/2 (-1/2)(1/(2!)) (7/36)^2 + ...)`

Even just the first two terms give `79/12 approx 6.583` and `6.583^2 approx 43.34`

We could get a little closer by using a different square number. If you start at 49, another square number, you are now looking at:

`sqrt(49 -6) = sqrt 49 sqrt(1 - 6/49) = 7 sqrt(1 - 6/49)`

Using just the first 2 terms of the Binomial Expansion:

`= 7 (1 - 1/2 * 6/49 ) = 46/7 approx 6.571`

And `6.571^2 = 43.18`

`43` is a prime number, so its square root is irrational.

We can find approximations to it as follows...

Note that `43` is roughly half way between `36=6^2` and `49=7^2`

So a good first approximation for `sqrt(43)` would be `13/2`.

We find:

`(13/2)^2 = 169/4 = 42.25`

Given `n > 0` and `0 < a < sqrt(n)` we can write `sqrt(n)` as a generalised continued fraction:

`sqrt(n) = a+b/(2a+b/(2a+b/(2a+...)))`

where `b = n-a^2`

So in our example:

`n = 43`, `a = 13/2` and `b=43-169/4 = 3/4`

So:

`sqrt(43) = 13/2+(3/4)/(13+(3/4)/(13+(3/4)/(13+...)))`

We can truncate this to get rational approximations.

For example:

`sqrt(43) ~~ 13/2+(3/4)/13 = 341/52 ~~ 6.5577`

`sqrt(43) ~~ 13/2+(3/4)/(13+(3/4)/13) = 8905/1358 ~~ 6.557437`

A calculator tells me:

`sqrt(43) ~~ 6.5574385243`

See https://socratic.org/s/aCh3Xasm for another example and explanation of this method.