Question

How do you find the square root of 529?

Answer 1

Check for divisibility by a perfect square to simplify. You will find that `23^2 = 529`

Explanation:

When we try to simplify a square root, we look for perfect square factors.

Do this by testing perfect squares until you get to a number whose square if greater that `529` (for example `25^2 = 625`)

So we test `2^2 = 4`. Clearly `4` is not a factor of `529` (nor is any other even number.

Test `3^2 = 9` which is not a factor. Skip `4^2` because it is even.

Obviously `5^2=25` is not a factor. Skip `6^2`.
Keep going . . .

`23^2 = 529` so we're finished.

`sqrt529 = 23`

Answer 2

Use a mixture of methods to find `sqrt(529) = 23`

Explanation:

There are quite a few different ways to find square roots.

Here's a bit of a mish-mash for this particular example...

Given `529`, first split off pairs of digits starting from the right hand side to get:

`5|29`

Next, note that `5` lies between two square numbers `2^2 = 4` and `3^2 = 9`

We can approximate where `sqrt(5)` lies between `2` and `3` by linearly interpolating. What's that? We approximate the part of the graph of `x^2` between `x=2` and `x=3` using a straight line to get our approximation.

Since `5` is `1/5` of the way between `2^2=4` and `3^2=5`, we can approximate `sqrt(5)` by `2+1/5 = 2.2`

Hence a good first approximation for `sqrt(500)` is `22`.

How about `529`?

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

Given a positive number `n` and an approximation `a_i` to its square root, a better approximation is given by the formula:

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

So putting `n=529` and `a_0=22`, we find:

`a_1 = (a_0^2+n)/(2a_0) = (22^2+529)/(2*22) = (484+529)/44 = 1013/44 = 23.02bar(27)`

That looks suspiciously close to `23` so try:

`23^2 = 529`

Thus `sqrt(529) = 23`

Answer 3

`sqrt529 =23`

Explanation:

Do a really rough estimate first using squares of multiples of `10`

`20^2 = 400 and 30^2 = 900`

`529` lies between `400 and 900`

so `sqrt529` will lie between `20 and 30`

Now look at the last digit ...`9`
There are only two numbers whose squares end with a `9`,

`3 and 7 rarr " "3^2 =9 and 7^2 =49`

So the possibilities are `23 and 27`

However, `25^2 = 625 and 529` is less than `625`

My first guess would therefore be `23`

Multiplying confirms that `23 xx 23 = 529`

Hence `sqrt529 =23`