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Rationalizing the DenominatorDate: 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/
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