Question

How do you find the square root of `6724` ?

Answer 1

`sqrt(6724) = 82`

Explanation:

We can try factoring `6724` and identifying square factors:

`color(white)(000000)6724`
`color(white)(000000)"/"color(white)(00)"\"`
`color(white)(00000)2color(white)(000)3362`
`color(white)(000000000)"/"color(white)(00)"\"`
`color(white)(00000000)2color(white)(000)1681`
`color(white)(000000000000)"/"color(white)(00)"\"`
`color(white)(00000000000)41color(white)(00)41`

So:

`sqrt(6724) = sqrt(2^2*41^2) = 2*41=82`

That "worked", but I cheated slightly in knowing that `1681 = 41^2`

Let's try another approach...

Given `6724`, split off pairs of digits starting from the right to get:

`67|24`

Looking at the most significant pair of digits, note that:

`67 > 64 = 8^2`

Hence the square root of `6724` is a little larger than `80`

We can use the Babylonian method to find a better approximation:

Given a number `n` of which you want the square root and an approximation `a_i` to that root, a better approximation `a_(i+1)` is given by the formula:

`a_(i+1) = (a_i^2+n)/(2a_i)`

So in our case, with `n=6724` and `a_0 = 80` we find:

`a_1 = (80^2+6724)/(2*80) = (6400+6724)/160 = 82.025`

Hmmm - that's suspiciously close to `82`. Does `82` work?

`82^2 = (80+2)^2 = 80^2+2*80*2+2^2 = 6400+320+4 = 6724`