Question

How can we find square root of `34.667895`?

Answer 1

`sqrt34.667895~~5.8879`

Explanation:

As `34.667895` is not a perfect square, to find the square root of `34.667895`, we should do a special long division, where we pair, the numbers in two's, starting from decimal point in either direction.

Here first pair is `34` and the number whose square is just less than it is `5`, so we write `25` below `34` and the difference is `9`. After this we move to the right of decimal point and we bring down next two digits (after decimal point) `66`. As a divisor we first write double of `5` i.e. `10` and then find a number `x` so that `10x` (here `x` stands for single digit in units place) multiplied by `x` is just less than the number, here `966`. We find for `x=8`, we have `108xx8=864` and get the difference as `102`.

Now with a remainder of `102`, we bring down next two digits i.e. `78`, which makes it `10278`. Also put decimal after `5` and this makes it `5.8`.

Now recall we had brought as divisor `5xx2=10`, but this time we have `58` (excluding decimal sign) so we make the divisor as `116x` and identify an `x` so that `116x` multiplied by `x` is just less than `10278`. Now if we choose `9` we have `1169xx9=10521`, which is more than `10278`, so we choose `8`, which makes the product as `1168xx8=9344<10278`.

We continue to do this till we have desired accuracy. If numbers end, we can go on adding `00`s.

`color(white)(xxxx)5 .color(white)(x)8color(white)(xx)8color(white)(xx)7color(white)(xx)9`
`ul5|bar(34).color(white)(.)bar(66)color(white)(.)bar(78)color(white)(.)bar(95)color(white)(.)bar(00)`
`color(white)(xxx)ul(25)color(white)(X)darr`
`color(red)(10)8|color(white)(.)9color(white)(.)66`
`color(white)(xxxx.)ul(8color(white)(.)64)`
`color(white)(.)color(red)(116)8|1color(white)(x)02color(white)(.)78`
`color(white)(xxxxxxx)ul(93color(white)(.)44)`
`color(white)(xx)color(red)(1176)7|color(white)(.)9color(white)(.)34color(white)(.)95`
`color(white)(xxxxxxxx)ul(8color(white)(.)23color(white)(.)69)`
`color(white)(xx)color(red)(11774)9|color(white)()1color(white)(.)11color(white)(.)26color(white)(.)00`
`color(white)(xxxxxxxx)ul(1color(white)(.)05color(white)(.)97color(white)(.)41)`
`color(white)(xxxxxxxxx)color(white)(.)52color(white)(.)85color(white)(.)9`

Hence `sqrt34.667895~~5.8879`

If you want more accuracy, you can bring down next pair of `00` and continue.

Answer 2

`sqrt34.667895 ~~5.888`

Explanation:

The manual calculation of square roots is a long and complicated process, especially in grade 6 and with such an uncomfortable number.!!

My guess is that it is the idea of what a square root is, that is more important here....

The square root of a number is another number, which, when multiplied by itself gives the original number.

`5xx5 = 5^2 = 25" "`If you do this the other way: `sqrt25 = 5`

`3xx3 = 3^2 = 9" "` this means that `sqrt9 = 3`

`8xx8 = 8^2 = 64" "` this means that `sqrt64 = 8`

`1.2 xx1.2 = 1.2^2 = 1.44" "` Doing it in reverse: `sqrt1.44 = 1.2`

Numbers which have an exact square root are called square numbers.

Most numbers do not have an exact square root. The simplest way to do those calculations is to use a calculator.

In your example, we need to find out:

`?? xx ?? = 34.667895`

We can see that it is going to be a number between `5 and 6`,

`5xx5 = 25" " and " "6xx6 = 36` and

`34.667895` is between `25 and 36.`

Using a calculator we find:

`sqrt34.667895 = 5.887944888......`

This answer goes on and on and on and will never make a pattern.

It would be enough to give the answer as `5.888`

Checking: `5.888 xx 5.888 = 34.668544` which is pretty close to the given number.