Associated Topics || Dr. Math Home || Search Dr. Math

### Rationalizing the Denominator

```Date: 07/10/2003 at 18:52:04
From: Indy Bowe
Subject: Algebra

Question:
1 divided by (the square root of 3 plus the square root of 5 plus
the square root of 7).

The answer the book gives is: 9 times the sq root of 3, plus 5
times the sq root of 5, plus the sq root of 7, minus 2 times the sq
root of 105.
```

```
Date: 07/10/2003 at 21:35:17
From: Doctor Achilles
Subject: Re: Algebra

Hi,

Thanks for writing to Dr. Math.

The problem that you are doing isn't one about adding square roots,
it's about a process called "rationalizing the denominator."

One thing that comes in handy is knowing how to multiply square roots.
When you multiply square roots, just multiply the numbers inside. For
example:

sqrt(5) * sqrt(3) = sqrt(15)

Let me give you an easier example of rationalizing the denominator
first that only involves two square roots in the denominator instead
of three:

1
-------------------
sqrt(3) + sqrt(5)

To get rid of all the sqrts in the denominator, let's multiply by:

sqrt(3) - sqrt(5)
-------------------
sqrt(3) - sqrt(5)

We can do this because we are really just multiplying by 1 since the
numerator and denominator are the same.

So we have:

1               sqrt(3) - sqrt(5)
-------------------  *  -------------------
sqrt(3) + sqrt(5)       sqrt(3) - sqrt(5)

The numerator just becomes:

sqrt(3) - sqrt(5)

For the denomoniator, we have to multiply each term in the first part
by each term in the second part:

sqrt(3) * sqrt(3) equals 3

sqrt(3) * -sqrt(5) equals -sqrt(15)

sqrt(5) * sqrt(3) equals sqrt(15)

sqrt(5) * -sqrt(5) equals -5

So our denominator is:

3 - sqrt(15) + sqrt(15) - 5

Which simplifies to:

3 - 5

or

-2

So our fraction is:

sqrt(3) - sqrt(5)
-------------------
-2

We can multiply the fraction by:

-1
----
-1

Which will give us:

-sqrt(3) + sqrt(5)
--------------------
2

Or:

sqrt(5) - sqrt(3)
-------------------
2

Now let's use that method to answer your question, let's move on to
the tougher one:

1
-----------------------------
sqrt(3) + sqrt(5) + sqrt(7)

To answer that, let's multiply by:

sqrt(3) - sqrt(5) - sqrt(7)
-----------------------------
sqrt(3) - sqrt(5) - sqrt(7)

The numerator is:

sqrt(3) - sqrt(5) - sqrt(7)

For the denominator, we have to multiply each term in the first part
by each term in the second part.

sqrt(3) * sqrt(3) = 3

sqrt(3) * -sqrt(5) = -sqrt(15)

sqrt(3) * -sqrt(7) = -sqrt(21)

sqrt(5) * sqrt(3) = sqrt(15)

sqrt(5) * -sqrt(5) = -5

sqrt(5) * -sqrt(7) = -sqrt(35)

sqrt(7) * sqrt(3) = sqrt(21)

sqrt(7) * -sqrt(5) = -sqrt(35)

sqrt(7) * -sqrt(7) = -7

The -sqrt(15) and the sqrt(15) will cancel and so will the -sqrt(21)
and the sqrt(21), so our denominator will be:

3 - 5 - sqrt(35) - sqrt(35) - 7

Which will simplify to:

-9 - 2*sqrt(35)

So now we have:

sqrt(3) - sqrt(5) - sqrt(7)
-----------------------------
-9 - 2*sqrt(35)

So now we have a square root in the denominator. So what do we do with
it? We have to rationalize (again). So multiply by:

-9 + 2*sqrt(35)
-----------------
-9 + 2*sqrt(35)

Which when all is said and done should give you:

-9sqrt(3) + 9sqrt(5) + 9sqrt(7) + 2sqrt(105) - 10sqrt(7) - 14sqrt(5)
----------------------------------------------------------------------
-59

[Please write back and let me know if you get something different.]

This should all simplify to:

-9*sqrt(3) - 5*sqrt(5) - sqrt(7) + 2*sqrt(105)
------------------------------------------------
-59

Which you can multiply by:

-1
----
-1

To get:

9*sqrt(3) + 5*sqrt(5) + sqrt(7) - 2*sqrt(105)
------------------------------------------------
59

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

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search