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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

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