Thursday, September 1, 2011

Rationalizing Denominators

In my previous posts, about nth roots and the properties of nth roots, I mentioned that I would post next about rationalizing denominators, and how to rationalize the denominator.  This may be a new concept to some students, so I will explain it first and describe where it will come in handy, before I actually show you how to do it.

Rationalizing denominators is a way of simplifying radicals, specifically fractions that have a radical on the bottom (in the denominator).  You may come across expressions like these, for example, when dealing with property #3 of nth roots.  In fact, you will soon find that equations are often not considered 'completely simplified' if there is still a radical in the denominator.  To get rid of this, you must rationalize the denominator.

Here's how you do it.  I think you will find it much easier than its name makes it sound!

First, let's consider a couple of points.

The first one is an obvious one, and you will see why in a second... that is, x / x = 1.  Right?  Anything divided by itself is 1 (except zero).  3 / 3 is 1.  100 / 100 is 1.  57483 / 57483 is 1.  That's how fractions work: if you have all of the pieces of the pie, out of all the pieces of the whole pie, you have the whole pie.  Easy, probably doesn't even need to be said.  But keep it in mind.

The second point is a refresher of the properties of nth roots... specifically, property 1 which says (n√x)n = x.  So, for n = 2, we have (2√x)2 = (√x)=  (√x)(√x) = x

So then, with those two points in mind, here is the trick to rationalizing the denominator when there is only a radical down there.  Quite simply, you take the expression that you have, and whatever radical is in the denominator, you multiply both the top and the bottom by that.  Let me show you what I mean:

Take 2 / √5.  To rationalize the denominator, we multiply the top and the bottom by √5.  So, this is what we get:

Hopefully, you can easily see how we really haven't really changed anything at all.  We multiplied the whole thing by 1, and then simplified the top and bottom using simple multiplication and division rules that you already know.  And that would then be considered to be fully simplified.  However, I should point out that this is all that's necessary when there is ONLY a radical in the denominator and nothing else.

If we had √5 + 2 in the denominator, that needs a bit more work.  It's still rationalizing the denominator, because we 'make the denominator rational' by getting rid of the radical.  But, when there's a plus or minus down there as well, there is an extra twist.

We still multiply the whole thing by 1 (that is, by something over something), but now it's a bit more complicated.  If you try to multiply top and bottom by √5 + 2, YOU ARE INCORRECT.  You can try (just follow the typical FOIL rules for multiplying binomials), but you will find that you end up with more of a mess on the bottom... more terms than you start with!  It doesn't simplify anything at all!

To properly get rid of this radical, you multiply the top and the bottom by √5 - 2.  That's right, notice that the sign in the middle is flipped!  This term is called a conjugate, and the two terms with the opposite sign in the middle are conjugates of each other.  If you FOIL out (√5 + 2)(√5 - 2), you end up with 5 - 4 = 1 on the bottom!  (It won't always be 1, but you will always get rid of the radical and can simplify further!).

So, let's try another example to put it all together.

Rationalize the denominator and simplify:  5 / (2 + √6)

And that would be your simplified answer!  Note that I multiplied the top and bottom by -1 (still not changing the equation!) to switch the negative signs around so that I didn't have a negative on the bottom.  Not always necessary, but it looks much cleaner.  (Technically, some teachers would say the proper answer is (5√6 / 2) - 5.  That works too.)

So that is all there is to rationalizing the denominator, and using conjugates to do it.  Since this is now a fairly long post, I'll leave it there.  But if you'd like another example or two with variables instead of numbers, or nth roots higher than squares, leave me a note in the comments and I'll either update the post or do another one with more!

1 comment:

  1. Thanks I think you have explained it to me so that I so i understand why it works properly.
    Great post much appreciated.


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