Example 1 Multiply each
of the following.
Solution
In all of these
problems all we need to do is recall how to FOIL binomials. Recall,
(3x
− 5)( x + 2) = 3x
( x) + 3x
(2) − 5( x) − 5(2) = 3x2
+ 6x
− 5x
−10 = 3x2
+ x −10
With radicals
we multiply in exactly the same manner. The main difference is that on occasionwe’ll need to do some simplification
after doing the multiplication.
As noted above
we did need to do a little simplification on the first term after doing the multiplication.
Don’t get
excited about the fact that there are two variables here. It works the same
way!
Again, notice
that we combined up the terms with two radicals in them.
Not much to do
with this one.
Notice that, in
this case, the answer has no radicals. That will happen on occasion so don’t
get excited about
it when it happens.
The last part
of the previous example really used the fact that
(a + b)(a − b) = a2 − b2
If you don’t
recall this formula we will look at it in a little more detail in the next
section.
Example 2 Rationalize the denominator for each of the following.
Solution
There are
really two different types of problems that we’ll be seeing here. The first two
parts illustrate the
first type of problem and the final two parts illustrate the second type of
problem.
Both types are
worked differently.
In this case we
are going to make use of the fact that. We need to determine what to multiply the
denominator by so that this will show up in the denominator. Once we figure this out we will
multiply the numerator and denominator by this term.
Here is the
work for this part.
Remember that
if we multiply the denominator by a term we must also multiply the numerator by
the same term. In this way we are really multiplying the term by 1 (since a/a= 1)
and so aren’t changing its value in any way.
We’ll need to
start this one off with first using the third property of radicals to eliminate
the fraction from
underneath the radical as is required for simplification.
Now, in order
to get rid of the radical in the denominator we need the exponent on the x to
be a 5.
In this case we
can’t do the same thing that we did in the previous two parts. To do this one
we will need to instead to make use of the fact that.
(a + b)(a − b) = a2 − b2
When the
denominator consists of two terms with at least one of the terms involving a
radical we will do the following to get rid of the radical.
So, we took the
original denominator and changed the sign on the second term and multiplied the numerator and denominator by this new term. By doing this we were able to
eliminate the radical in the denominator when we then multiplied out.
This one works
exactly the same as the previous example. The only difference is that both
terms in the denominator now have radicals. The process is the same however.
Rationalizing
the denominator may seem to have no real uses and to be honest we won’t see many uses in an Algebra class. However, if you are on a track that will take you
into a Calculus class you will find that rationalizing is useful on occasion at
that level.
We will close
out this section with a more general version of the first property of radicals.
Recall that when we first wrote down the properties of radicals we required
that a be a positive number. This was done to make the work in this
section a little easier. However, with the first property that doesn’t
necessarily need to be the case.
Here is the
property for a general a (i.e. positive or negative)
where
│a│ is the absolute value of a. If you don’t recall
absolute value we will cover that in detail in a section in the next chapter. All
that you need to do is know at this point is that absolute value always makes a
a positive number.
So,
as a quick example this means that,
For square
roots this is,
This will not
be something we need to worry all that much about here, but again there are
topics in courses after an Algebra course for which this is an important idea
so we needed to at least acknowledge it.
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