SECTION 20.1: WHAT A SOLUTION
IS AND THE GENERAL RULES FOR MANIPULATING AN EQUATION
A "solution" is a number, when substituted into the
variables of an equation, makes the equation TRUE.
ex 1) Is 7 a solution for 3x + 4 = 2x + 11?
I'll test by putting 7 into the variable x in 3x + 4 = 2x + 11 to see if I
get a true statement.
3(7) + 4 ?=? 2(7) + 11
21 + 4 ?=? 14 + 11
25 = 25
It's a true statement, therefore 7 is a solution!
ex 2) Is 6 a solution for 2x + 4 = 5x  7?
I'll test by putting 6 into the variable x in 2x + 4 = 5x  7 to see if I
get a true statement.
2(6) + 4 ?=? 5(6)  7
12 + 4 ?=? 30 + 11
16 ≠ 19
Since 16 is not equal to 19, 6 is not a solution for 2x + 4 = 5x  7
GENERAL RULE FOR SYMBOL MANIPULATING AN EQUATION 
IF you are given an equation that is true
AND
IF you perform the same operation on both sides of the given equation
THEN the resulting equation will also be true. 
Here are some simple numerical examples of the above idea:
ex. 1) Given: 6 + 20 = 10 + 16 is an equation that is true.
2(6 + 20) = 2(10 + 16) I multiplied each side of the original
equation, 6 + 20 = 10 + 16, by 2.
2(26) = 2(26) I added the results inside each set of parentheses
52 = 52 The result is still true. I've proven my point!
ex 2) Given: 6 + 20 = 10 + 16 is an equation that is true.
= I divided each side of the original equation, 6 + 20 = 10 + 16, by 2.
= I added the numerators on each side of the equation.
13 = 13 The result is still true. I've proven my point!
ex 3) Given: 6 + 20 = 10 + 16 is an equation that is true.
6 + 20 + 7 = 10 + 16 +7 I added 7 to each side of the
original equation.
33 = 33 I added all the numbers on each side of the equation. The result still true.
ex 4) Given: 6 + 20 = 10 + 16 is an equation that is true.
6 + 20  3 = 10 + 16  3 I subtracted 3 from each side
of equation.
23 = 23 I added all the numbers on each side of the equation. The result still true.
SUMMING UP: I HAVE SHOWN YOU THAT I CAN PERFORM THE SAME OPERATION ON BOTH SIDES OF A TRUE EQUATION AND THE EQUATIONS REMAINS TRUE!
A MOMENTARY ASIDE: SOME IMPORTANT ARITHMETIC FACTS 
 1 times any number is equal to that number.
 Any number multiplied by the reciprocal of the number is equal to 1.
 Any fraction can be rewritten as an integer times a fraction.
 Any fraction containing a variable in the numerator can be rewritten as a fraction times the variable.

SECTION 20.2: SOLVING LINEAR EQUATIONS
"Solving an equation" is a symbol manipulation
procedure for finding a numerical value, called a "solution."
The solution, when substituted into the variables
of the given equation, will result in a "true statement."
Your goal, when "Solving an Equation," is to isolate a variable on one side of the equation and
isolate a numerical value on the other. For example, you will end up with
equations like x = 2 or x = 7 or x =
The simplest equation to solve is called a "linear equation."
The definition of a "linear equation" is
 an
equation that contains the same lettered variable throughout the equation,
 the power of that variable is 1 and
 all the terms are either constants or constants times a variable of a power of 1.
EXAMPLES OF LINEAR EQUATIONS
ex. 1) x + 5 = 20 
ex. 2) 2x  3 = 12 
ex. 3) 2x + 5 = 7x + 4 
ex. 4) 2(3x + 5) = 5x  2 
ex 1. Solve 5x + 13 = 3x  7
I will use the rules in section 20.1 to help me.
step1) 5x + 13  13 = 3x  7 13
I subtracted 13 from both sides of the equation
step 2) 5x = 3x  20
I simplified each side of the equation.
step 3) 5x  3x = 3x  20  3x
I subtracted 3x from both sides of the equation.
step 4) 2x = 20
I simplified each side of the equation.
step 5)
I divided both sides of the equation by 2.
step 6) 1
I cancelled like factors on each side of the equation.
step 7) x = 10
I multiplied 1x to get x and 10 is the solution.
CHECK: If I place 10 in the original equation, the equation will come out
true! Watch!
5x + 13 = 3x  7
5(10) + 13 = 3(10)  7
50 + 13 = 30  7
37 = 37 A true statement results!
ex 2. Solve 7x  10 = 18x  32
step1)
I added 10 to both sides of the equation
step 2)
I simplified each side of the equation
step 3)
I added 18x to both sides of the equation.
step 4)
I simplified both sides of the equation
step 5)
I divided both sides of the equation by 11.
step 6) 1x = 2
I simplified by canceling like factors.
step 7) x = 2
I multiplied 1x to get x and 2 is the solution.
CHECK: If I place this solution in the original equation, the equation will
come out true! Watch!
4 = 4 The result is true!
PARENTHESES AND THE DISTRIBUTIVE PROPERTY 
When you have parentheses in an equation make sure you apply the distributive property before you perform any other step in your solving process.
This concept is demonstrated in the next example: 
ex 3) Solve
To humor some students I will change all subtraction to addition before I
distribute. Remember all subtraction is really addition!
s1)
Now I will apply the Distributive Property
s2)
The result of distributing
s3)
I simplified step 2 to get s3) above
s4)
I subtracted 15 from both sides of the equation
s5)
I simplified s4) to get s5) above
s6)
I subtracted 7x from both sides of the equation
s7)
s6)
was simplified to get s7)
s8)
I divided
both sides by 17
s9) 1
I cancelled like factors in s8)
s10)
1x simplifies to x and the solution is 3/17.
SECTION 20.3: LITERAL EQUATIONS
Equations that have several variables (letters) are called "literal
equations." You are usually asked to solve for one of those
variables. We will deal with literal equations whose variables have exponents
of 1.
HOW TO SOLVE LITERAL EQUATIONS 
Step 1) Bring all terms with the variable that you are solving for
to one side of the equation and all other terms without that variable to the other side of the equation.
Step 2) Isolate the variable that you are solving for using the general rules for solving equations. 
ex 1. Solve for a in 6a + 7b = 3a  7d
1a =
 GENERAL COMMENT 
As you solve an equation, I suggest that you talk to yourself, noting your steps.
Ask yourself why you are performing your steps. Your ability to verbalize
what you are doing in mathematics is an important skill.

ex 2. Solve for m in 6m + 4c  3x = 5x  7(4c  3m)
1
SECTION 20.4: QUADRATIC EQUATIONS
A "quadratic equation" is an equation with the following form ax^{2} + bx + c = 0 where a,b and c are real numbers and a ≠ 0.
The number "c" is called the "numerical part." Notice that all the
terms on the lefthand side of the equation are added!
ex. 1) Determine a, b and c in 3x^{2} + 2x  5 = 0.
Rewrite as .
a = 3; b = 2; c = 5
ex. 2) Determine a, b and c in 5x^{2 }= 0.
Rewrite as
a = 5; b = 0; c = 0
ex. 3) Determine a, b and c in 7x^{2}  4 = 0.
Rewrite as
a = 7; b = 4; c = 0
ex. 4) Determine a, b and c in 9x^{2}  11 = 0.
Rewrite as
a = 9; b = 0; c = 11
USEFUL DEFINITIONS FOR NEXT SECTION 
ALGEBRAIC EXPRESSION_{definition} : a math statement made up of numbers, variables, numbers times one or more variables, operations( +,  , x, ÷ ) and exponents.
ex. 1) 5  6^{2} ex. 2) 5x  3 ex. 3) 3x^{2} + 2y  7
FACTOR_{definition}: any number, variable or algebraic expression that is multiplied.
ex. 1) (5)(7)(8) 5, 7, and 8 are factors ex. 2) 3xy^{2} 3, x and y^{2 }are factors
ex. 3) (2x+3)(5x7) (2x+3) and (5x7) are factors 
SECTION 20.5: THE LOGIC OF
FACTORS SET EQUAL TO 0
Let's imagine you have two boxes marked A and B, and there is a number in each box;
What if I told you that those numbers hiding in the boxes, when multiplied,
produce a result of 0. What can you absolutely conclude about the numbers
inside those boxes?
SECTION 20.6: SOLVING
QUADRATIC EQUATIONS BY FACTORING
HOW TO SOLVE QUADRATIC EQUATIONS BY FACTORING 
Step 1) Make sure all terms of the quadratic are
on one side of the equation and set equal to 0.
Step 2) Factor the quadratic in the given equation.
Comment: If there is a greatest common factor
then factor that first.
Step 3) Set each factor equal to 0.
Step 4) Solve each equation produced from your
factors.
Step 5) Test your results in your original equation as one or more of your answers MAY NOT be solutions. A RESULT THAT IS NOT A SOLUTION IS CALLED AN "EXTRANEOUS SOLUTION." 
ex 1) Solve x^{2} + 5x + 6 = 0 by factoring
s1) The equation is already prepared for us
s2) (x + 3)(x + 2) = 0
s3) x + 3 = 0 and x + 2 = 0
s4) x + 3 = 0 ; x = 3 and x + 2 = 0 ; x = 2
x = 3 and x = 2 may be the solutions of the quadratic equation x^{2} + 5x + 6 = 0
s5) If you substitute x = 3 OR x = 2 into the equation x^{2} + 5x + 6 = 0 we must get 0 as a result.
x = 3: .
3 is a solution!
You try 2. You will find that 2 is a solution as well.
Therefore the solutions are 3 and 2.
DO NOT ASSUME THAT YOUR RESULTS ARE SOLUTIONS!
ex. 2) Solve by factoring
2x + 1 = 0 and 3x  2 = 0
2x + 1 = 0
and are the solutions! How did I know? I checked!
I'll check for you:
6(1/2)^{2 } (1/2)  2 ?=? 0 → 6(1/4) + 1/2  2 ?=? 0 → 6/4 + 1/2  2 ? =? 0
→3/2 + 1/2  2 ?=? 0 → 4/2 2 ?=? = 0 → 2  2 = 0
You can try and see if is also a solution.
ex. 3) Solve by factoring
s1)
s2)(3x + 2)(x  5) = 0
s3) and
s4)
s5) Solutions are and x = 5.
I'll test
This is true: therefore, is a solution.
You'll find that x = 5 is also a solution.
example 4) 3x² + 6x  45 = 0
s1) 3(x² + 2x  15) = 0
Please note that I factored the GCF(greatest common factor first)
s2) 3(x+5)(x3)=0
s3) 3≠0 and x + 5 = 0 and x  3 = 0
Obviously the factor 3 is not equal to 0.
s4) 3≠0 and x=5 and x=3
You can check and see that 5 and 3 are solutions to 3x² + 6x  45 = 0
WHAT ARE "EXTRANEOUS SOLUTIONS"? 
You must check your socalled solutions because not all socalled solutions are guaranteed to make your equation true.
If the socalled "solution" does NOT MAKE YOUR EQUATION EQUAL TO 0, that "solution" is called an "extraneous solution."

SECTION 20.7: SOLVING A
QUADRATIC EQUATION USING THE QUADRATIC FORMULA
There is a formula that you can use to get the solutions of a quadratic
equation
ax + bx + c = 0. This useful formula is called the "Quadratic
Formula."
THE QUADRATIC FORMULA 
If ax^{2} + bx + c = 0
Then
The ± above means that you will get two formulas that will produce two possible solutions.
or
When would you use this formula? When you have difficulty
factoring or the quadratic cannot be factored.
Please note that the quadratic equation must be in the form of addition if you want to get the correct values of a, b and c which are the coefficients of each term of the given quadratic equation.

ex. 1) Solve using the Quadratic formula.
s1) Determine a, b and c: I will get in the form of addition: 5x^{2} + 3x + 4 = 0
Therefore, a = 5; b = 3; c = 4
s2) Fill in the quadratic formula:
and
ex. 2) Solve using the Quadratic formula.
note: The above quadratic can be factored, but I will show
you that you can still use the quadratic formula.
s1) Get 6x^{2}  5x  4 = 0 in the form of addition: 6x^{2} + 5x + 4 = 0
Therefore, a = 6; b = 5 ; c = 4
Now solving by factoring and prove to yourself that the quadratic formula in
fact is accurate!
SECTION 20.8: TWO PREREQUISITE
CONCEPTS NEEDED FOR THE UPCOMING TOPIC " COMPLETING THE SQUARE"
Before I teach you a symbol manipulation technique called "Completing
the Square," I need to teach you a few prerequisite ideas. I will
construct two general math rules that you will find quite useful in "completing the square."
Let's consider what motivates the first rule that I will introduce:
Consider x^{2} = 4. What if I asked you for the solutions to this, i.e. What numbers make
x = 4 true?
I think you can see 2 and 2 will make this equation true. Test and see: (2)^{2}= 4 and (2)^{2}= 4.
Of course if one could not do
the above "in their head" there would be an issue over getting the solutions.
There is a rule that will allow you to quickly get solutions to equations in
the form of "a variable squared set equal to a number."
THE VARIABLE SQUARED RULE 
If x^{2} = b where b is a real number then x = ± √ b
COMMENT: Any variable raised to an even integer power will have two solutions! That is, if x^{n} = b where n is an even integer power then.

ex. 1) Solve x^{2} = 25
→x = ± √ 25
The solutions are x = 5 and x = 5
ex. 2) Solve 3x^{2} = 48
Comment: Notice this equation IS NOT a variable squared set
equal to a math expression.
We must get the equation in this form before we apply the above rule.
THE GENERALIZED VARIABLE SQUARED RULE 
If math expression A^{2} = math expression B
Then math expression A = ± √ math expression B
Comment: This rule will hold for any even integer power not just a power of
2.

ex. 1) Solve (x + 2)^{2} = 81
We have a math expression, x + 2, that is being squared. We can use the above
rule.
So there are two solutions: x = 9  2 = 11 and x = 9  2 = 7
ex. 2) Solve7(3x  5)^{2} = 70
Comment: We do not have a math expression squared equal to
another math expression. Note the coefficient, 7, in front of the math expression being squared. Let's prepare the equation so we can use the rule.
(3x  5) = ± √ 10
Solution 1: 3x  5 = √ 10 

Solution 2: 3x  5 =  √ 10 
3x = 5 + √ 10 

3x = 5  √ 10 
x =( 5 + √ 10 )/3 

x = (5  √ 10 )/3 
WHAT IS A PERFECT SQUARE TRINOMIAL 
A trinomial that can be factored into a binomial squared.
See examples below
Teacher Comment: You will need to understand this concept when you learn completing the square, a topic coming up soon.

ex 1) Factor x^{2} + 4x + 4
x^{2} + 4x + 4 = (x + 2)(x + 2) = (x + 2)^{2}
x^{2} + 4x + 4 is a "perfect square trinomial" since it can be factored into
a binomial squared.
ex. 2) Factor 9x^{2}  12x + 4
9x^{2}  12x + 4 = (3x  2)(3x  2) = (3x 2)^{2}
9x^{2}  12x + 4 is a "perfect square trinomial" since it can be factored into
a binomial squared.
Now lets use the idea of a perfect square trinomial in the next section.
SECTION 20.9: COMPLETING THE SQUARE
I will introduce you to "Completing The Square," the last
technique for solving a quadratic equation. Using factoring and the quadratic
formula are the two techniques you currently know which enable you to solve a
quadratic equation.
Consider the following quadratic equation x^{2} + 4x  11 = 0.
The quadratic x^{2} + 4x  11 cannot be factored. So I can't use factoring to solve this quadratic
equation. I could use the quadratic formula but I will show you how to use "Completing The Square." I will model the "Completing The
Square" procedure with a step by step explanation.
ex 1) Use Completing the Square to solve x^{2} + 4x  11 = 0.
s1) Isolate the numerical part to the right hand side of the
equation.
x^{2} + 4x  11 = 0
x^{2} + 4x = 11
s2) Take the coefficient of x (with the operation in front of
it as its sign), divide it by 2, square the result, and add that result to
both sides of the equation in step 1.
x^{2} + 4x + 4 = 11 + 4 is still an equivalent to the
original equation x^{2} + 4x  11 = 0.
s3) The quadratic on the left hand side of the equation is
now a "perfect square trinomial." Factor the perfect square trinomial on the left hand side of the equation and
add the numbers on the right hand side of the equation.
x^{2} + 4x +4 = 11 + 4
(x + 2)^{2} = 15
s4) Now use the "generalized variable squared
rule" to solve the resulting equation in step 3
So the solutions to x^{2} + 4x  11 = 0 are:
and
ex. 2) Solve x^{2} 12x  28 = 0 using completing the square.
s1) Isolate the numerical part to the right hand side of the
equation.
x^{2} 12x  28 = 0
x^{2} 12x = 28
s2) Take the coefficient of x(with the operation in front of
it as its sign), divide it by 2, square the result, and add that result to
both sides of the equation in step 1.
s3) The quadratic on the left hand side of the equation is a
"perfect square trinomial." Factor the perfect square trinomial on the left hand side of the equation and
add the numbers on the right hand side of the equation.
x^{2} 12x + 36 = 28 + 36
s4) Now use the "generalized variable squared rule" to solve
the result of step 3.
and
So the solutions to x^{2}  12x  28 = 0 are x_{1} = 14 and x_{2} = 2
Based on the above important comment, 4x^{2} + 8x  40 = 0 would not be a candidate for completing the square as it stands because x^{2} has a coefficient of 4 and not 1. But let's force the equation, 4x^{2} + 8x  40 = 0 ,
to have its squared variable with a coefficient of 1. How? Divide
both sides of the equation by 4, the coefficient of
x. Let me show you.
Now the above is set for completing the square. I'll leave
this to you to finish.
SECTION 20.10: SOLVING BASIC
RATIONAL EQUATIONS
A "rational equation" is an equation that has a fraction set
equal to a fraction and one of those fractions (or both) will contain
variables. Here are 4 examples of rational equations.
ex. 1) 

ex. 2) 



ex. 3) 

ex. 4) 
Solving equations such as these is not difficult if you understand the "rule of extrememeans" which states the following:
In effect, we multiply numerators and denominators together when we use the
above rule. A practical number example follows to clarify the above rule. We
know The above rule states that if then (1)(14) = (2)(7) which in fact is true.
So I will use the above rule in order to solve rational equations.
ex. 1) Solve
ex. 2) Solve
ex. 3) Solve
Since I know how to solve a quadratic equation, I'll bring all the terms to the
left hand side of the equation to create a quadratic set equal to 0.
Now I'll solve this with the quadratic formula.
a = 3; b = 42; c = 6
and
The following symbol
" ≈ "
means "almost equal to." Why do you think I associated this symbol for
the above solutions?
SECTION 20.11: A REVIEW OF
RADICALS, INDEXES AND RADICANDS
I might suggest that you review the section on roots, i.e. Chapter 9. Let's do a review of radical expressions like etc. The following symbol is called a "radical symbol." Numbers are placed in various
positions of the radical symbol and the position of those numbers determines
what we call them. In a "radical symbol" there is an "index" and a "radicand."
See the image below.
When you are given a radical symbol with an index and radicand you are being
asked to find a specific number. You are being asked "Find the number when raised
to the power of the index that will produce the radicand."
So when I am given I am being asked to "Find the number when raised to the power of 3 that will
produce 125." The answer is 5 because So you would write .
There is one radical that does not have an index, for example √ 64 .
If you come across this type of radicand, which I'm sure you have, it is understood that the
index is 2. √ 64 is asking you "What positive number when raised to the power of 2
will produce 64?" 8 is the answer since 8 raised to the power of 2 will
produce 64, i.e.
8^{2} = 64. Therefore, √ 64 = 8.
MEANING OF THE RADICAL 
IF a is a positive even number
THEN is asking you
"What POSITIVE real number raised to the power of a will produce b?"
IF a is an ODD positive number
THEN THEN is asking you
"What real number raised to the power of a will produce b?"
MAKE SURE YOU UNDERSTAND THE DISTINCTION IN THESE TWO RULES! 
EXPLAIN THE MEANING OF THE FOLLOWING RADICANDS
ex. 1)
ans: is asking "What POSITIVE integer raised to the power of 2 will produce 100?"
The
answer is 10 since
Therefore, .
ex. 2)
ans: is asking "What REAL number raised to the power of 3 will produce 64?"
The answer is
4 since .
Therefore,
ex. 3)
ans: is asking "What POSITIVE real number raised to the power of 4 will produce 16?"
The answer
is 2 since .
Therefore, .
ex. 4)
ans: is asking "What REAL number raised to the power of 5 will produce 32?"
The answer is
2 since (2)^{5} = 32.
Therefore, = 2.
ex. 5)
ans: is asking "What POSITIVE number raised to the power of 2 will produce 16?"
There is
NO REAL NUMBER when raised to a power of 2 that will produce a
negative 16. Can you think of one?
Therefore, has no answer real number answer.
SECTION
20.12: SOLVING EQUATIONS IN THE FORM x^{a} = b
Here are some examples of equations that you must be able to solve
ex. 1)
ex. 2)
ex. 3)
You will need the following rules to solve the above equations.
ex. 1) Solve
x^{3} = 27
(x^{3})= 27 What
you do to one side of an equation you must do to the other.
I used rule 2 on the lefthand side of this equation.
x = 27 or I used rule 1 to write this as a radical expression.
is asking you what real number raised to the power of 3 will produce 27. The
answer is 3.
Therefore,
ex. 2) Solve
x
SOME COMMENTS ON EXAMPLE
2: When you have a fractional power with an even integer in the numerator you will always generate two potential solutions. Notice that Rule 2 and 3 were used in the above example. It is smart to "remove" the denominator of the fractional power first so you won't lose touch with the fact that there will be two solutions.
ex. 3) Solve
x= 8
This is asking you, "what real number raised to a power of 3 will produce 64?"
The answer is 4.
USING YOUR CALCULATOR TO EVALUATE FRACTIONAL POWERS 
To solve numerical bases with fractional powers use the caret symbol(^).
example 1) Solve 8^{1/3} using
your calculator.
ans: Below is the order for pressing keys, and the home screen of my TI 84 PLUS.
example 2) Solve (16)^{3/4 }using
your calculator.
ans: Below is the order for pressing keys, and the home screen of my TI 84 PLUS.

