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Using Conjugates to Simplify Radicals


Date: 08/04/98 at 20:40:54
From: varun
Subject: Conjugates of square roots

When using conjugates, how do we know when to change the sign? Would 
the sign have to be changed if we had to rationalize a denominator 
like sqrt(a+b)? How can we solve this problem: 1/[3+sqrt(3)-sqrt(6)]?


Date: 08/05/98 at 12:05:30
From: Doctor Peterson
Subject: Re: Conjugates of square roots

Hi, Varun. 

Conjugates are a neat trick, and the best way to see when and how to 
use them is to think about how they work, rather than just treat them 
as magic. The basic idea is that we have some expression in the form 
of a sum:

   a + b

and we want to multiply it by:

   a - b

so that the result will be simpler:

  a^2 - b^2

The question is, when will this be simpler? It will if both a and b are 
either radicals or mere numbers. So if you have expressions like:

   2 + 4sqrt(3)    or    5sqrt(3) - 3sqrt(5)

you can simplify them by multiplying them by their conjugates:

   2 - 4sqrt(3)    or    5sqrt(3) + 3sqrt(5)

respectively, so that there will be no radicals left.

If you have more than two terms in an expression, you have to decide 
how to break it up into "a" and "b"; I'm not sure that this is 
technically a conjugate, but it is the same idea extended. In your 
example:

    3 + sqrt(3) - sqrt(6)

there would be no benefit in breaking it up as:

   [ 3 ] + [ sqrt(3) - sqrt(6) ]

since squaring [ sqrt(3) - sqrt(6) ] doesn't simplify it; but we can 
take a step in the right direction by writing it as either:

    [ 3 + sqrt(3) ] - [ sqrt(6) ]

or as:

    [ 3 - sqrt(6) ] + [ sqrt(3) ]

In each case, the left part ("a") will not simplify when you square it, 
but the right side ("b") will. You can then gather terms and use a 
second conjugate to finish the job. Try both ways and see what happens.

Now, if your denominator is sqrt(a+b), it just doesn't have the form 
of a sum, so there is no way to form a conjugate. You could try 
multiplying by sqrt(a-b), of course, but all you will get is 
sqrt(a^2 - b^2), which doesn't help at all. So the best you can do to 
simplify:

       1
   ---------
   sqrt(a+b)

is to multiply numerator and denominator by sqrt(a+b) itself:

       1       sqrt(a+b)
   --------- = ---------
   sqrt(a+b)     (a+b)

I hope that helps. Let me know if you have trouble finishing the 
solution to your last problem.

- Doctor Peterson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   


Date: 08/05/98 at 20:18:41
From: Anonymous
Subject: Re: Conjugates of square roots

Dr. Math,

Thank you for helping me out with conjugates. I got that problem from
one of my friend's books. But in the book the solution is given as:

   sqrt(18)[3-sqrt(3)+sqrt(6)] 
   ---------------------------
              36

Could you please tell me how I could get this solution?
                                                                              
Varun


Date: 08/06/98 at 09:16:40
From: Doctor Peterson
Subject: Re: Conjugates of square roots

Hi, Varun. 

Let's work through the problem starting with my suggestion that you 
think of:

   3 + sqrt(3) - sqrt(6)

as:

   [ 3 + sqrt(3) ] - [ sqrt(6) ]

Its conjugate is then:

   [ 3 + sqrt(3) ] + [ sqrt(6) ]

so we will multiply like this:

                 1                 [ 3 + sqrt(3) ] + [ sqrt(6) ]
   ----------------------------- * ----------------------------- =
   [ 3 + sqrt(3) ] - [ sqrt(6) ]   [ 3 + sqrt(3) ] + [ sqrt(6) ]

     [ 3 + sqrt(3) ] + [ sqrt(6) ]
   --------------------------------- =
   [ 3 + sqrt(3) ]^2 - [ sqrt(6) ]^2

   [ 3 + sqrt(3) ] + [ sqrt(6) ]
   ----------------------------- =
   [ 9 + 6 sqrt(3) + 3 ] - [ 6 ]

   3 + sqrt(3) + sqrt(6)
   --------------------- =
       6 + 6 sqrt(3)

   3 + sqrt(3) + sqrt(6)   6 - 6 sqrt(3)
   --------------------- * ------------- =
       6 + 6 sqrt(3)       6 - 6 sqrt(3)

   18 - 18 sqrt(3) + 6 sqrt(3) - 6*3 + 6 sqrt(6) - 6 sqrt(6) sqrt(3)
   ----------------------------------------------------------------- =
                         [6]^2 - [6 sqrt(3)]^2

   18 - 18 sqrt(3) + 6 sqrt(3) - 18 + 6 sqrt(6) - 18 sqrt(2)
   --------------------------------------------------------- =
                         36 - 36 * 3

   6 sqrt(6) - 12 sqrt(3) - 18 sqrt(2)
   ----------------------------------- =
                 -72

   3 sqrt(2) + 2 sqrt(3) - sqrt(6)
   -------------------------------
                 12

This is the answer I would give. As I predicted, I had to use 
conjugates twice, and do a lot of simplifying of radicals, but it all 
came out without too much trouble. I hope this is what you got too, or 
something close.

The book's answer is really not quite in simplest form. Let's see if 
we can show that it is equal to this answer:

   sqrt(18) * [3 - sqrt(3) + sqrt(6)] 
   ---------------------------------- =
                  36

   3 sqrt(2) * [3 - sqrt(3) + sqrt(6)]
   ----------------------------------- =
                  36

   3 sqrt(2) - sqrt(6) + 2 sqrt(3)
   ------------------------------- 
                 12

which does in fact match my answer.

So I'd say the book's answer isn't the best, but it's in agreement with 
what I got, so I'm satisfied. Sometimes in complex problems like this 
people can disagree on what is really simplest. We don't argue about 
it, but just recognize that both answers are equal and both are simpler 
than the original.

- Doctor Peterson, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Square & Cube Roots
Middle School Square Roots

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