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```
Date: 02/15/99 at 17:05:38
From: holly

What is the process for doing solving problems like

7 sqrt32
--------
5 sqrt63  ?

```

```
Date: 02/15/99 at 17:34:33
From: Doctor Pat

To divide by a simple radical like the one in your problem, simply
multiply both the numerator and denominator by the radical. This is
called "rationalizing the denominator" and will leave a whole number in
the denominator. Here is what it looks like:

7 sqrt32     7 sqrt(32)*sqrt(63)    7 sqrt(32*63)
--------  =  ------------------- =  -------------- =
5 sqrt63     5 sqrt(63)*sqrt(63)    5 * 63

Now all you have to do is simplify. The 7 in the numerator will divide
into the 63 in the denominator, and the sqrt (32*63) can be simplified
also (HINT: look for perfect square factors).

- Doctor Pat, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 02/17/99 at 10:50:19
From: Doctor Bonnie

Let us first break down all the stuff that we are taking the square
root of into primes:

7 sqrt(2*2*2*2*2)
-----------------
5 sqrt (3*3*7)

So now we can do some actual work. Group the 2's in the numerator into
groups and the 3's in the denominator:

7 sqrt{(2*2)*(2*2)*2}
---------------------
5 sqrt{(3*3)*7}

Because all the stuff under the sqrt sign is multiplied, we can take
the sqrt of them separately:

7 sqrt(2*2) sqrt(2*2) sqrt(2)
-----------------------------
5 sqrt(3*3) sqrt(7)

So now we can take some sqrts:

7*2*2 sqrt 2
------------
5*3 sqrt 7

All we can do now is rationalize the denominator. This is not
necessary, but it is sometimes fun to do, and it might lead to a bit

(7*2*2 sqrt2)*(sqrt7)
---------------------
(5*3 sqrt7)*(sqrt7)

So we get:

7*2*2 sqrt 14
-------------
5*3*7

And doing the multiplication and canceling:

4 sqrt 14
---------
15

So I think this is about as pretty as you are going to get it.

3 sqrt 8
----------
11 sqrt 44

and:

2 sqrt 18
---------
3 sqrt 12

- Doctor Bonnie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Square & Cube Roots