Notesale: Turn your study into money

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Preface
Here are my online notes for my Algebra course that I teach here at Lamar University, although I
have to admit that it’s been years since I last taught this course
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Despite the fact that these are my “class notes”, they should be accessible to anyone wanting to
learn Algebra or needing a refresher for algebra
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However, they do assume that you’ve had some
exposure to the basics of algebra at some point prior to this
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Here are a couple of warnings to my students who may be here to get a copy of what happened on
a day that you missed
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Because I wanted to make this a fairly complete set of notes for anyone wanting to learn
algebra I have included some material that I do not usually have time to cover in class
and because this changes from semester to semester it is not noted here
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2
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Likewise, even
if I do work some of the problems in here I may work fewer problems in class than are
presented here
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Sometimes questions in class will lead down paths that are not covered here
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Sometimes a very good question gets asked in class
that leads to insights that I’ve not included here
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4
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THESE NOTES ARE NOT A SUBSTITUTE FOR ATTENDING CLASS!!
Using these notes as a substitute for class is liable to get you in trouble
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Radicals
We’ll open this section with the definition of the radical
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The left side of this equation is often called the radical form and the right side is often called the
exponent form
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Note as well that the index is required in these to make sure that we correctly evaluate the radical
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For square roots we have,
2

a= a

In other words, for square roots we typically drop the index
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Example 1 Write each of the following radicals in exponent form
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When we
convert to exponent form and the radicand consists of more than one term then we need to
enclose the whole radicand in parenthesis as we did with these two parts
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Therefore, the radical form of
this is,
1
10

= 8 10 x ≠ 10 8 x
8x
So, we once again see that parenthesis are very important in this class
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Since we know how to evaluate rational exponents we also know how to evaluate radicals as the
following set of examples shows
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(a) 16 and 4 16 [Solution]
(b) 5 243 [Solution]
(c) 4 1296 [Solution]
(d) 3 −125 [Solution]
(e) 4 −16 [Solution]
Solution
To evaluate these we will first convert them to exponent form and then evaluate that since we
already know how to do that
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Let’s
take a look at both of these
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Different indexes will give different evaluations so make sure that you
don’t drop the index unless it is a 2 (and hence we’re using square roots)
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Note however that we can evaluate
the radical of a negative number if the index is odd as the previous part shows
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In this part we made the
claim that

16 = 4 because 42 = 16
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We also have ( −4 ) = So, why didn’t we use -4 instead? There is a general rule
16
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When evaluating
square roots we ALWAYS take the positive answer
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− 16 =
−4
This may not seem to be all that important, but in later topics this can be very important
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Note that we don’t have a similar rule for radicals with odd indexes such as the cube root in part
(d) above
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We can also write the general rational exponent in terms of radicals as follows
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Properties
If n is a positive integer greater than 1 and both a and b are positive real numbers then,
1
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n

3
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When we run across those situations we will acknowledge them
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Also note that while we can “break up” products and quotients under a radical we can’t do the
same thing for sums or differences
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5 = 25 = 9 + 16 ≠ 9 + 16 = 3 + 4 = 7
If we “break up” the root into the sum of the two pieces we clearly get different answers! So, be
careful to not make this very common mistake!
We are going to be simplifying radicals shortly so we should next define simplified radical
form
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1
...

3
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All exponents in the radicand must be less than the index
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No fractions appear under a radical
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In our first set of simplification examples we will only look at the first two
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Example 3 Simplify each of the following
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y7

(a)
(b)

9

[Solution]

x6

[Solution]

18x 6 y11 [Solution]

(c)
(d)

4

32x 9 y 5 z12

(e)

5

x12 y 4 z 24

(f)

3

9 x2 3 6 x2

[Solution]
[Solution]
[Solution]

Solution
(a) y 7
In this case the exponent (7) is larger than the index (2) and so the first rule for simplification is
violated
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So, let’s note
that we can write the radicand as follows
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The radical then becomes,

(y )

3 2

y7 =

y

Now use the second property of radicals to break up the radical and then use the first property of
radicals on the first term
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Before moving on let’s briefly discuss how we figured out how to break up the exponent as we
did
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We then determined the largest multiple of 2 that
is less than 7, the exponent on the radicand
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Next, we noticed that 7=6+1
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[Return to Problems]

(b) 9 x 6
This radical violates the second simplification rule since both the index and the exponent have a
common factor of 3
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9

6
x=

6

1
6 9

( x )=

2

1
2 3

( x )=

9
3
x= x=

3

x2
[Return to Problems]

(c) 18x 6 y11
Now that we’ve got a couple of basic problems out of the way let’s work some harder ones
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There is more than one term here but everything works in exactly the same fashion
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e
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= 9= 9 ( x3 )
18 x 6 y11
x 6 y10 ( 2 y )

2

( y ) (2y)
5 2

Don’t forget to look for perfect squares in the number as well
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9 ( x3 ) ( y 5 ) ( 2 y )
=
2

18 x 6 y11
=

2

9 = 3x3 y 5 2 y
( x3 ) ( y 5 ) 2 y
2

2

Note that we used the fact that the second property can be expanded out to as many terms as we
have in the product under the radical
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This will happen on occasion
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So, instead of get perfect
squares we want powers of 4
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4
32 x 9 y 5 z12
=

16 x8 y 4 z12 ( 2 xy )
=

4

4

16 4 ( x 2 ) 4= 2 x 2 y z 3 4 2 xy
y 4 4 ( z 3 ) 4 2 xy
4

4

[Return to Problems]

(e) 5 x12 y 4 z 24
Again this one is similar to the previous two parts
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That will
happen on occasion
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Individually both of the radicals are in simplified form
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The unspoken rule is that we should
have as few radicals in the problem as possible
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=
( 9 x 2 )( 6 x 2 )

3
=
9x2 3 6x2

3

3

54 x 4

Now that it’s in this form we can do some simplification
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Performing these operations with radicals
is much the same as performing these operations with polynomials
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Recall that to add/subtract terms with x in them all we need to do is add/subtract the coefficients
of the x
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For instance,

4 x + 9 x = 9 ) x =x
13
(4 +

3 10 5 − 11 10 5 =11) 10 5 = 5
−8 10
(3 −

We’ve already seen some multiplication of radicals in the last part of the previous example
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What we need to look at now are
problems like the following set of examples
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Assume that x is positive
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Recall,

( 3x − 5)( x + 2 ) =3x ( x ) + 3x ( 2 ) − 5 ( x ) − 5 ( 2 ) =3x 2 + 6 x − 5 x − 10 =3x 2 + x − 10

With radicals we multiply in exactly the same manner
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[Return to Problems]

(

(b) 3 x −

y

)( 2

x −5 y

)

Don’t get excited about the fact that there are two variables here
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[Return to Problems]

(

)(

(c) 5 x + 2 5 x − 2

)

Not much to do with this one
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That will happen on occasion so don’t get
excited about it when it happens
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Okay, we are now ready to take a look at some simplification examples illustrating the final two
rules
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To get rid of them we will use some of the multiplication ideas that we looked at
above and the process of getting rid of the radicals in the denominator is called rationalizing the
denominator
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They are really more
examples of rationalizing the denominator rather than simplification examples
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Assume that x is positive
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The first two parts
illustrate the first type of problem and the final two parts illustrate the second type of problem
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4
x

(a)

In this case we are going to make use of the fact that n a n = a
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Once we figure this
out we will multiply the numerator and denominator by this term
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4
=
x

4 x 4 x 4 x
= =
x
x x
x2

Remember that if we multiply the denominator by a term we must also multiply the numerator by
the same term
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[Return to Problems]

(b)

5

2
x3

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
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5

This means that we need to multiply by

2
5
=
x3

5
5

x 2 so let’s do that
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To do this one we
will need to instead to make use of the fact that

( a + b )( a − b ) = a 2 − 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
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By doing this we were able to eliminate the radical
in the denominator when we then multiplied out
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The only difference is that both terms
in the denominator now have radicals
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5
=
4 x+ 3

(

5(4 x − 3)
) = 5(4 x − 3)
) ( 4 x + 3 )( 4 x − 3 ) 16 x − 3

(
)(

4 x− 3
5
=
4 x+ 3 4 x− 3

[Return to Problems]

Rationalizing the denominator may seem to have no real uses and to be honest we won’t see
many uses in an Algebra class
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We will close out this section with a more general version of the first property of radicals
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This was done to make the work in this section a little easier
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Here is the property for a general a (i
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positive or negative)
n

a
an = 
a

if n is even
if n is odd

where a is the absolute value of a
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All that you need to do is know at this point is that absolute value
always makes a a positive number
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