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Radical on the Bottom of a Fraction

Date: 12/10/2002 at 21:03:13
From: Joe
Subject: Rational Expressions

The question is:

Simplify:  4/3-root 5

I can't figure out how my teacher keeps getting the answer she does.

Thanks.

Date: 12/11/2002 at 11:41:32
From: Doctor Ian
Subject: Re: Rational Expressions

Hi Joe,

Do you mean

4
----------- ?
3 - sqrt(5)

The first thing you need to know is that

this   this   something
---- = ---- * ---------
that   that   something

^
|_______ this is equal to 1

so long as 'something' is not zero. This is one of the handiest tricks
in all of mathematics, so it's worth becoming familiar with it.

We use it, for instance, to make equivalent fractions:

1   2   2
- * - = -
2   2   4

1   3   3
- * - = -
2   3   6

and so on. The question is, what should 'something' be in order to

To answer that, we have to mention another pattern that is very handy,
called a 'difference of squares'.  It looks like this:

(this + that)(this - that) = this^2 - that^2

You can use the distributive property to verify this, if you want:

(a + b)(a - b) = a(a - b) + b(a - b)

= (a^2 - ab) + (ab - b^2)

= a^2 - b^2

Now, we don't actually _have_ this pattern in our fraction, but we
have _half_ of it, i.e.,

4
-------
(a - b)

So we can supply the other half using our something/something trick:

4      (a + b)   4(a + b)
------- * ------- = ---------
(a - b)   (a + b)   a^2 - b^2

Now, if a or b are square roots, the radicals disappear, which is what
we want to happen.  Here's how it looks with your fraction:

a     b
|     |
4         3 + sqrt(5)
----------- * -----------
3 - sqrt(5)   3 + sqrt(5)
|     |       |     |
a     b       a     b

a     b
|     |
4(3 + sqrt(5))
=  -----------------
3^2 - (sqrt(5))^2
|        |
a^2      b^2

4(3 + sqrt(5))
=  -----------------
9 - 5

4(3 + sqrt(5))
=  -----------------
4

= 3 + sqrt(5)

Does this make sense?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
Associated Topics:
Middle School Square Roots

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