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Skill 30 page 154 Solving Quadratic Equations Part One of Three Lessons
This lesson will be the first of three lessons that will allow you to be able to
solve equations that have a squared term, x2, in it. These kinds of equations
must be solved in one of two different ways.
Look at this problem from page 155. It is in a form that lets us name the solutions
by just looking at the problem.
Think about a problem like 5x = 0 X must be 0 because the only way to get 0 when
multiplying two numbers is for one of them to be 0.
Set each factor equal to zero and solve. X – 2 = 0 or x – 3 = 0
That means either x = 2 or x = 3 Both answers are required.
Look at this problem. It is set up to be solved by setting each factor equal to 0
There are two answers. -5 and -3.
X + 5 = 0 or x + 3 = 0
So x = -5 or x = -3
The two answers are -5 and -3.
Try this one ( x – 2 ) ( x + 4) = 0
x – 2 = 0 or x + 4 = 0
So x = 2 or x = -4
Look at this one x ( x -5) = 0
x = 0 or x – 5 = 0
So x = 0 or x = 5
The two answers are 0 and 5.
Try these problems: x ( x + 7) = 0 Answers: 0 and -7
(x – 2) ( x + 8) = 0 Answers; 2 and -8
Now, the only thing about these problems is that they don’t start out the way we have listed them
The problems start out like this: x2 + 6 x + 8 = 0 or
or
So, we have to learn two new ways to solve this problem. One is factoring and the
other is using the quadratic formula. This is one of the most challenging things you can
do in the GED math course. Sometimes, just factoring is a question all by itself.
We will deal with factoring first. There are handouts in your book that explain this process.
We will go to them now. Turn to pages 137, 138, and 138a.
Factoring
Factors in arithmetic are multipliers. Factors in algebra are also multipliers.
Factor: 24 This means we want to write 24 as a multiplication.
24 = 2(12) or 24 = 4(6) or 24 = 3(8) or 24 = 24 (1)
Factor: 3x - 12 We want to write the problem as a multiplication.
Think about the Greatest Common Factor of 3x and -12. This is 3.
Write as a multiplication of 3 and (x - 4). Factored 3x - 12 is 3(x - 4).
This is called factoring out a common factor.
Some numbers cannot be factored with factors other than themselves and one. These are
called prime numbers. Prime numbers are 2, 3, 5, 7, 11, etc.
Some polynomials are also prime and cannot be factored.
Factoring Trinomial Polynomials
I am teaching a method for factoring trinomials I learned from my Algebra I teacher, Mrs.
Sarah Marley. I have used it to teach thousands of students to factor. There are other
ways, but this one is systematic and efficient which appeals to me as a mathematician.
Other ways are either trial and error where you erase a hole in your paper or are so
complicated that I doubt you will remember them. You will need to look at the page in the
book to really see this.
Marley's Method for factoring trinomials:
Factor: x2 + 6x + 8
You will just need to read the book text pages 137, 138 closely to follow
this. I am sorry I can’t do this in person yet.

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Skill30 solvingquadraticequations

  • 1. Skill 30 page 154 Solving Quadratic Equations Part One of Three Lessons This lesson will be the first of three lessons that will allow you to be able to solve equations that have a squared term, x2, in it. These kinds of equations must be solved in one of two different ways. Look at this problem from page 155. It is in a form that lets us name the solutions by just looking at the problem. Think about a problem like 5x = 0 X must be 0 because the only way to get 0 when multiplying two numbers is for one of them to be 0. Set each factor equal to zero and solve. X – 2 = 0 or x – 3 = 0 That means either x = 2 or x = 3 Both answers are required.
  • 2. Look at this problem. It is set up to be solved by setting each factor equal to 0 There are two answers. -5 and -3. X + 5 = 0 or x + 3 = 0 So x = -5 or x = -3 The two answers are -5 and -3. Try this one ( x – 2 ) ( x + 4) = 0 x – 2 = 0 or x + 4 = 0 So x = 2 or x = -4 Look at this one x ( x -5) = 0 x = 0 or x – 5 = 0 So x = 0 or x = 5 The two answers are 0 and 5.
  • 3. Try these problems: x ( x + 7) = 0 Answers: 0 and -7 (x – 2) ( x + 8) = 0 Answers; 2 and -8 Now, the only thing about these problems is that they don’t start out the way we have listed them The problems start out like this: x2 + 6 x + 8 = 0 or or So, we have to learn two new ways to solve this problem. One is factoring and the other is using the quadratic formula. This is one of the most challenging things you can do in the GED math course. Sometimes, just factoring is a question all by itself. We will deal with factoring first. There are handouts in your book that explain this process. We will go to them now. Turn to pages 137, 138, and 138a.
  • 4. Factoring Factors in arithmetic are multipliers. Factors in algebra are also multipliers. Factor: 24 This means we want to write 24 as a multiplication. 24 = 2(12) or 24 = 4(6) or 24 = 3(8) or 24 = 24 (1) Factor: 3x - 12 We want to write the problem as a multiplication. Think about the Greatest Common Factor of 3x and -12. This is 3. Write as a multiplication of 3 and (x - 4). Factored 3x - 12 is 3(x - 4). This is called factoring out a common factor. Some numbers cannot be factored with factors other than themselves and one. These are called prime numbers. Prime numbers are 2, 3, 5, 7, 11, etc. Some polynomials are also prime and cannot be factored. Factoring Trinomial Polynomials I am teaching a method for factoring trinomials I learned from my Algebra I teacher, Mrs. Sarah Marley. I have used it to teach thousands of students to factor. There are other ways, but this one is systematic and efficient which appeals to me as a mathematician. Other ways are either trial and error where you erase a hole in your paper or are so complicated that I doubt you will remember them. You will need to look at the page in the book to really see this.
  • 5. Marley's Method for factoring trinomials: Factor: x2 + 6x + 8 You will just need to read the book text pages 137, 138 closely to follow this. I am sorry I can’t do this in person yet.