Question

How do you find the square root of 74889?

Answer 1

The simplest form of the square root is `3sqrt(8321)`

We can find approximations such as:

`sqrt(74889) ~~ 273.659`

Explanation:

Given:

`74889`

Notice that the sum of the digits is divisible by `9`. That is:

`7+4+8+8+9 = 36 = 4*9`

So `74889` is divisible by `9` (`=3^2`) too:

`74889 = 9 * 8321`

`8321` is less easy to factorise, but eventually you might find:

`8321=53*157`

In fact, to check for square factors we only needed to look for prime factors up to `19`, since `20` and `21` are composite and:

`21^3 = 9261 > 8321`

Since there are no more square factors, the simplest form of the square root is given by:

`sqrt(74889) = sqrt(3^2*8321) = sqrt(3^2)*sqrt(8321) = 3sqrt(8321)`

This is an irrational number, not expressible as a fraction, but we can find rational approximations:

Given:

`74889`

First split into pairs of digits from the right:

`7"|"48"|"89"`

Note that:

`2^2 = 4 < 7 < 9 = 3^2`

Hence:

`2 < sqrt(7) < 3`

and:

`200 < sqrt(74889) < 300`

For a better estimate, if we know a few more square roots we can include the next two digits and note that:

`27^2 = 729 < 748 < 784 = 28^2`

Hence:

`270 < sqrt(74889) < 280`

We can linearly interpolate between these limits to find:

`sqrt(74889) ~~ 270+10*(74889 - 72900)/(78400-72900) = 270+10*1989/5500 ~~ 273.6`

Let us choose `274` as an approximation.

Given an approximation `a` for the square root of a number `n`, a better approximation is given by:

`(a^2+n)/(2a)`

So in our case, putting `n=74889` and `a=274` we find:

`sqrt(74889) ~~ (274^2+74889)/(2*274) = (75076+74889)/548 = 149965/548 ~~ 273.659`

If we want more accuracy, then repeat with this new approximation. Each iteration will roughly double the number of significant digits which are correct.

Answer 2

`sqrt74889~~273.658`

Explanation:

As `74889` is not a perfect square, to find the square root of `74889`, we should do a special long division, where we pair, the numbers in two, starting from decimal point in either direction. When we group them we do so starting from `89`, then `48` and then `7`.

Here for `7`, the number whose square is just less than it, is `2`, whose square is `4` and so we write `4` below `7`. The difference is `3` and now we bring down next two digits `48`. As a divisor we first write double of `2` i.e. `4` and then find a number `x` so that `4x` (here `x` stands for single digit in units place) multiplied by `x` is just less than the number, here `348`. We find for `x=7`, we have `47xx7=329` and get the difference as `19`.

Next we bring down next two digits `89` and we have `1989`. Recall we had brought as divisor `2xx2=4`, but this time we have `27` double of which is `54`, so we make the divisor as `54x`, selecting an `x` so that `54x` multiplied with `x`, just goes in `1989` and this is `3`, for which we get `543xx3=1629` and have a remainder `360`.

Now as we still have a remainder of `360`, for more accuracy, we bring `00` after decimal point. Also put decimal after `273`.

We continue in similar way by bringing down `00`, which makes it `36000` and continue as shown till we have desired accuracy. You can see the details below, where we have gone up to three decimal places.

`color(white)(xX)2color(white)(xx)7color(white)(xx)3color(white)(xx).6color(white)(xx)5color(white)(xx)8`
`ul2|bar(7)color(white)(.)bar(48)color(white)(.)bar(89).color(white)(.)bar(00)color(white)(.)bar(00)color(white)(.)bar(00)`
`color(white)(XX)ul(4)color(white)(.)darr`
`color(red)(4)7|3color(white)(X)48`
`color(white)(xxx)ul(3color(white)(..)29)`
`color(white)(x)color(red)(54)3|color(white)(.)19color(white)(x)89`
`color(white)(xxx.xx)ul(16color(white)(.)29)`
`color(white)(x)color(red)(546)6|color(white)(.)3color(white)(.)60color(white)(.)00`
`color(white)(xxxxxx)ul(3color(white)(.)27color(white)(.)96)`
`color(white)(xx)color(red)(5472)5|color(white)()32color(white)(.)04color(white)(.)00`
`color(white)(xxXxxxx)ul(27color(white)(.)36color(white)(.)25`
`color(white)(xx)color(red)(54730)8|4color(white)(.)67color(white)(.)75color(white)(.)00`
`color(white)(xxxxxxxx)ul(4color(white)(.)37color(white)(.)84color(white)(.)64`
`color(white)(xxxxxxxxx.)29color(white)(.)90color(white)(.)36`

Hence `sqrt74889~~273.658`

Answer 3

Here's another method for finding rational approximations...

Explanation:

For interest, here's another idea for finding rational approximations to `sqrt(74889)`.

Start by noting that:

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

Given that `274` is the closest integer approximation, with `274^2 = 75076 = 74889 + 187`, we can write a generalised continued fraction for the square root, putting `a=274` and `b=-187` to get:

`sqrt(74889) = 274-187/(548-187/(548-187/(548-187/(548-...))))`

This is related to `274+sqrt(74889)` being one of the zeros of:

`(x-274-sqrt(74889))(x-274+sqrt(74889)) = x^2-548x+187`

Now consider a sequence defined recursively as follows:

`{ (a_0 = 0), (a_1 = 1), (a_(n+2) = 548a_(n+1)-187a_n" for "n >= 0):}`

The first few terms are:

`0, 1, 548, 300117, 164361640, 90014056841, 49296967522188`

Because of the way it is constructed, the ratio between pairs of successive terms tends to `274+sqrt(74889)`. In fact each successive ratio corresponds to truncating the continued fraction after one more term.

So we can use this sequence to get successively better approximations to `sqrt(74889)`, such as:

`sqrt(74889) ~~ 164361640/300117 - 274 ~~ 273.6585465`