Question

How do you find the square root of 270?

Answer 1

See the solution process below:

Explanation:

We can use this rule of radicals to rewrite this expression:

`sqrt(a * b) = sqrt(a) * sqrt(b)`

`sqrt(270) = sqrt(9 * 30) = sqrt(9) * sqrt(30) = +-3sqrt(30)`

If necessary, the `sqrt(30) = +-5.477`

And therefore:

`+-3sqrt(30) = +-3 *+- 5.477 = +-16.432` rounded to the nearest thousandth.

Answer 2

A = `3 * sqrt30`

Explanation:

Find a `sqrt` which can be divided by the whole number:
= `sqrt30`

Then find the quotient of the number and the divisor (`sqrt`):
= 270 / `sqrt30`
= 3
= 3 * `sqrt30`

Answer 3

`sqrt(270) = 3sqrt(30) ~~ 15873/966 ~~ 16.431677`

Explanation:

First note that `270`, like any positive number, has two square roots. We denote the positive square root by `sqrt(270)` and the negative one by `-sqrt(270)`. However, the expression "the square root" is often used to refer to the principal, positive square root.

Next note that if `a, b >= 0` then:

`sqrt(ab) = sqrt(a)sqrt(b)`

[[ The same is not true if both `a < 0` and `b < 0` ]]

Also, if `a >= 0` then:

`sqrt(a^2) = a`

So we find:

`sqrt(270) = sqrt(9*30) = sqrt(9)sqrt(30) = 3sqrt(30)`

This is the simplest form of the principal square root.

`3sqrt(30)`, like `sqrt(30)` is an irrational number.

We can calculate approximations to `sqrt(30)` using its continued fraction. Note that `30=5*6` is of the form `n(n+1)`. Hence its square root has a short regular repeating pattern:

`sqrt(30) = [5;bar(2,10)] = 5+1/(2+1/(10+1/(2+1/(10+1/(2+1/(10+...))))))`

[[ In general `sqrt(n(n+1)) = [n;bar(2,2n)]` ]]

We can get decent approximations for `sqrt(30)` by truncating just before one of the `10`'s as follows:

`sqrt(30) ~~ 5+1/2 = 11/2`

`sqrt(30) ~~ 5+1/(2+1/(10+1/2)) = 241/44`

`sqrt(30) ~~ 5+1/(2+1/(10+1/(2+1/(10+1/2)))) = 5291/966`

Let's stop there and use this to give us an approximation for `sqrt(270)`...

`sqrt(270) = 3sqrt(30) ~~ 3*5291/966 = 15873/966 ~~ 16.431677`