# How do you simplify #sqrt65#?

##### 1 Answer

#### Explanation:

If a radicand (the part under the root sign) of a square root has a square factor, then it can be simplified:

#sqrt(a^2b) = abs(a) sqrt(b)#

or if you know that

#sqrt(a^2b) = a sqrt(b)#

For example,

In our example, we find

If you like, you can reexpress it:

#sqrt(65) = sqrt(5)sqrt(13)#

but that is not (as far as I know) considered 'simpler'.

Note that

**Bonus**

#65 = 64 + 1 = 8^2 + 1#

is of the form

As a result, the square root can be expressed as a very simple *continued fraction* ...

#sqrt(65) = [8;bar(16)] = 8+1/(16+1/(16+1/(16+1/(16+...))))#

You can use this to give you good *approximations* for

For example,

#[8; 16] = 8+1/16 = 129/16 = 8.0625#

#[8; 16, 16] = 8+1/(16+1/16) = 8+16/257 = 2072/257 ~~ 8.0622568#

Actually