Chapter III

In this chapter, the different ways of solving the quadratic equation are recalled.

There are by using the factoring, completing the square, by quadratic formula and

solving by graphing. Students are given guides to determine the most appropriate

method to use.

TARGET SKILLS:

At the end of this chapter, students are expected to:

• distinguish appropriate method in solving quadratic equation;

• discuss and follow the steps in such different method; and

• resolve quadratic equation using any method you want.

Lesson 6

Solving by factoringOBJECTIVES:

At the end of this lesson, students are expected to:

define what is factoring;

discuss the Zero Factor Principle; and

solve equation by using the factoring method.

Factoring – rearrange the equation; factor the left member; equate each factor to zero

to obtain the two roots.

Solve (x – 3)(x – 4) = 0.

The Zero Factor Principle tells me that at least one of the factors must be

equal to zero. Since at least one of the factors must be zero, I'll set them each

equal to zero:

x – 3 = 0   or   x – 4 = 0 x = 3  or  x = 4

Solve: x = 3, 4

Note that "x = 3, 4" means the same thing as "x = 3  or  x = 4"; the only difference

is the formatting. The "x = 3, 4" format is more-typically used.

Checking x = 3 in (x – 3)(x – 4) = 0:

([3] – 3)([3] – 4) ?=? 0

     (3 – 3)(3 – 4) ?=? 0

               (0)(–1) ?=? 0

                       0   =   0

Checking x = 4 in (x – 3)(x – 4) = 0:

([4] – 3)([4] – 4) ?=? 0

     (4 – 3)(4 – 4) ?=? 0

                 (1)(0) ?=? 0

                       0   =   0

Solve x2 + 5x + 6 = 0.

This equation is already in the form "(quadratic) equals (zero)" but, unlike the

previous example, this isn't yet factored. The quadratic must first be factored, because it

is only when you MULTIPLY and get zero that you can say anything about the factors

and solutions. You can't conclude anything about the individual terms of the unfactored

quadratic (like the 5x or the 6), because you can add lots of stuff that totals zero.

So the first thing I have to do is factor:

x2 + 5x + 6 = (x + 2)(x + 3)

Set this equal to zero:

(x + 2)(x + 3) = 0

Solve each factor:

x + 2 = 0  or  x + 3 = 0

x = –2  or  x = – 3

The solution to x2 + 5x + 6 = 0 is x = –3, –2

Checking x = –3 and x = –2 in x2 + 5x + 6 = 0:

[–3]2 + 5[–3] + 6 ?=? 0

           9 – 15 + 6 ?=? 0

           9 + 6 – 15 ?=? 0

               15 – 15 ?=? 0

                        0   =   0

[–2]2 + 5[–2] + 6 ?=? 0

           4 – 10 + 6 ?=? 0

           4 + 6 – 10 ?=? 0

               10 – 10 ?=? 0

                        0   =   0

So both solutions "check".

Solve x2 – 3 = 2x.

This equation is not in "(quadratic) equals (zero)" form, so I can't try to

solve it yet. The first thing I need to do is get all the terms over on one side, with

zero on the other side. Only then can I factor and solve:

x2 – 3 = 2x

x2 – 2x – 3 = 0

(x – 3)(x + 1) = 0

x – 3 = 0  or  x + 1 = 0

x = 3  or  x = –1

Then the solution to x2 – 3 = 2x is x = –1, 3

Solve (x + 2)(x + 3) = 12.

The (10 + 2)(9 + 3) does not equal 12, you should never forget that you

must have "(quadratic) equals (zero)" before you can solve.

So, tempting though it may be, the factors above equal to the other side of

the equation and "solve". Instead, multiply out and simplify the left-hand side,

then subtract the 12 over to the left-hand side, and re-factor.

(x + 2)(x + 3) = 12

x2 + 5x + 6 = 12

x2 + 5x – 6 = 0

(x + 6)(x – 1) = 0

x + 6 = 0  or   x – 1 = 0

x = –6  or   x = 1

Then the solution to (x + 2)(x + 3) = 12 is x = –6, 1

Solve x(x + 5) = 0.

To "solve" the equation for "x + 5 = 0", divide it by x. But it can't divide by

zero; dividing off the x makes the implicit assumption that x is not zero. Used the

variable factors having variables and numbers (like the other factor, x + 5), a

factor can contain only a variable, so "x" is a perfectly valid factor. So set the

factors equal to zero, and solve:

x(x + 5) = 0

x = 0  or  x + 5 = 0

x = 0  or  x = –5

Then the solution to x(x + 5) = 0 is x = 0, –5

Solve x2 – 5x = 0.

Factor the x out of both terms, taking the x out front.

x(x – 5) = 0

x = 0  or  x – 5 = 0

x = 0  or  x = 5

Then the solution to x2 – 5x = 0 is x = 0, 5

There is one other case of two-term quadratics that you can factor:

Solve x2 – 4 = 0.

This equation is in "(quadratic) equals (zero)" form, it's ready to solve. The

quadratic itself is a difference of squares, then apply the difference-of-squares

formula:

x2 – 4 = 0

(x – 2)(x + 2) = 0

x – 2 = 0  or  x + 2 = 0

x = 2  or  x = –2

Then the solution is x = –2, 2

Note: This solution may also be formatted as "x = ± 2"

Exercises: Solve:

1. (x – 3)(x – 5) = 0.

2. x2 + 6x + 7 = 0.

3. x2 – 4 = 2x.

4. x2 – 6x = 0.

5. x2 – 8 = 0.

Name: ___________________ Section: _______

Instructor: ________________ Date: _______ Rating: ____

Instruction: Solve the following Quadratic Equation by Factoring Method.

1. x2 – 36 = 0

_____________________________________________________

2. x2= 25

_____________________________________________________

3. x2 – 12x + 35 = 0

_____________________________________________________

4. x2 – 3x – 40 = 0

_____________________________________________________

5. 2x2 – 5x = 3

_____________________________________________________

6. 3x2 + 25x = 18

_____________________________________________________

7. 15x2 – 2x – 8 = 0

_____________________________________________________

8. 3x2 – x = 10

_____________________________________________________

9. x2 + 6x – 27 = 0

_____________________________________________________

10.y2 – 2y – 3 = y – 3

_____________________________________________________

11.4y2 + 4y = 3

_____________________________________________________

12.3a2 + 10a = -3

_____________________________________________________

13.a2 – 2a – 15 = 0

_____________________________________________________

14.r2 + 6r – 27 = 0

_____________________________________________________

15.2z2 – 2 – 1 = 0

_____________________________________________________

Lesson 7

Solving by Completing the Square

OBJECTIVES:

At the end of this lesson, students are expected to:

analyze the techniques in completing the square;

comply with the techniques of completing the square; and

carefully change the exact signs for every equation.

Some quadratics is fairly simple to solve because they are of the form

"something-with-x squared equals some number", and then you take the square root of

both sides. An example would be:

(x – 4)2 = 5

x – 4 = ± sqrt(5)

x = 4 ± sqrt(5)

x = 4 – sqrt(5)  and  x = 4 + sqrt(5)

Unfortunately, most quadratics doesn’t come neatly squared like this. For your

average everyday quadratic, you first have to use the technique of "completing the

square" to rearrange the quadratic into the neat "(squared part) equals (a number)"

format demonstrated above. For example:

Find the x-intercepts of y = 4x2 – 2x – 5.

First off, remember that finding the x-intercepts means setting y equal to zero

and solving for the x-values, so this question is really asking you to "Solve 4x2 –

2x – 5 = 0".

This is the original problem. 4x2 – 2x – 5 = 0

Move the loose number over to the

other side.4x2 – 2x = 5

Divide through by whatever is

multiplied on the squared term.

Take half of the coefficient (don't

forget the sign!) of the x-term, and

square it. Add this square to both

sides of the equation.

Convert the left-hand side to

squared form, and simplify the

right-hand side. (This is where you

use that sign that you kept track of

earlier. You plug it into the middle

of the parenthetical part.)

Square-root both sides,

remembering the "±" on the right-

hand side.  Simplify as necessary.

Solve for "x =".

Remember that the "±" means that

you have two values for x.

The answer can also be written in rounded form as

You will need rounded form for "real life" answers to word problems, and for

graphing. But (warning!) in most other cases, you should assume that the answer

should be in "exact" form, complete with all the square roots.

When you complete the square, make sure that you are careful with the sign on the

x-term when you multiply by one-half. If you lose that sign, you can get the wrong

answer in the end, because you'll forget what goes inside the parentheses. Also, don't

be sloppy and wait to do the plus/minus sign until the very end. On your tests, you won't

have the answers in the back, and you will likely forget to put the plus/minus into the

answer. Besides, there's no reason to go ticking off your instructor by doing something

wrong when it's so simple to do it right. On the same note, make sure you draw in the

square root sign, as necessary, when you square root both sides. Don't wait until the

answer in the back of the book "reminds" you that you "meant" to put the square root

symbol in there. If you get in the habit of being sloppy, you'll only hurt yourself!

Solve x2 + 6x – 7 = 0 by completing the square.

Do the same procedure as above, in exactly the same order. (Study tip:

Always working these problems in exactly the same way will help you remember

the steps when you're taking your tests.)

This is the original equation. x2 + 6x – 7 = 0

Move the loose number over to the other

side.x2 + 6x      = 7

Take half of the x-term (that is, divide it by

two) (and don't forget the sign!), and

square it. Add this square to both sides of

the equation.

Convert the left-hand side to squared

form.  Simplify the right-hand side.(x + 3)2 = 16

Square-root both sides. Remember to do

"±" on the right-hand side.x + 3 = ± 4

Solve for "x =". Remember that the "±"

gives you two solutions. Simplify as

necessary.

   x = – 3 ± 4

      = – 3 – 4, –3 +

4

      = –7, +1

If you are not consistent with remembering to put your plus/minus in as soon as

you square-root both sides, then this is an example of the type of exercise where you'll

get yourself in trouble. You'll write your answer as "x = –3 + 4 = 1", and have no idea

how they got "x = –7", because you won't have a square root symbol "reminding" you

that you "meant" to put the plus/minus in. That is, if you're sloppy, these easier

problems will embarrass you!

Exercise:

1. 3x2 – 4x – 6 = 0

2. 2x2 -3x + 4 = 0

3. x2 – 8x + 16 = 0

4. x2 + 18x + 72 = 0

5. 2x2 – 6x + 1 = 0

Name: ___________________ Section: _______

Instructor: ________________ Date: _______ Rating: ____

Instruction: Solve the following Quadratic Equation by Completing the Square.

1. x2 + 3x = 4

_____________________________________________________

2. x2 – 2x = 24

_____________________________________________________

3. x2 + 4 = 4x

_____________________________________________________

4. 2x2 – 6 = x

_____________________________________________________

5. 4a2 + 12a + 9 = 0

_____________________________________________________

6. 3a2 – 5 = 14a

_____________________________________________________

7. 16b2 + 1 = 16b

_____________________________________________________

8. 9b2 – 6b – 1 = 0

_____________________________________________________

9. 9z2 + 30z + 20 = 0

_____________________________________________________

10.2a2 + a = 10a

_____________________________________________________

11.2x2 + 17 = 10x

_____________________________________________________

12.2a2 + 6a + 9 = 0

_____________________________________________________

13.5x2 – 2x + 1 = 0

_____________________________________________________

14.3x2 + 2x + 1= 0

_____________________________________________________

15.2y2 + 5y = 42

_____________________________________________________

Lesson 8

Quadratic FormulaOBJECTIVES:

At the end of this lesson, students are expected to:

follow the step in solving quadratic formula;

distinguish the roots of the quadratic equation; and

perform substituting the values in the quadratic formula.

The following steps will serve as guide in solving this method.

Step 1. First subtract c from both sides of the equation and then, divide both sides by

(a≠ 0 by hypothesis) to obtain the equivalent equation,

x2 + bxa

= ⁻ca

Step 2. Complete the left-hand side in to the perfect square.

x2 + bx/a + (b/2a)2 = (b/2a)2 – c/a

or (x+b/2a)2 = (b2-4ac)/4a2

Step 3. Take the square roots of both sides of the last equation.

(x+b/2a) = ± (√b2 – 4ac)/2a

Step 4. Solve for x.

x = −b+√b2−4 ac2a

or x = −b−√b2−4ac2a

Let a, b and c be real constant, where a ≠ 0. Then the roots of ax2 + bx + c = 0 are

x = −b±√b2−4 ac2a

The above formula is referred to as the quadratic formula.

Example: Solve a. 3x2 – x – 5/2 = 0

Solutions: Here a=3, b= 1, c= 5/2⁻ ⁻

Substituting these values in the quadratic formula

we obtain x = −¿¿¿

= 1±√1+30

6

= 1±√316

The roots are 1+√316

and 1−√316

.

a. 2x2 – 5 (x-2) = 8

To be able to apply the formula, we must first put the given equation in standard form.

2x2 – 5 (x-2) = 8

2x2 – 5x + 10 = 8

2x2 – 5x + 2 = 0

Here a=2, b= 5 c=2. By the quadratic formula ⁻

x = −¿¿¿ = 5±34

The roots are 2 and ½.

Note that the expression 2x2 – 5x + 2 can be factored as

2x2 – 5x + 2 = (2x – 1) (x – 2)

The roots of the quadratic equation x = ½ and x = 2. This example shown that if

we can see that the given equation in factorable, it will be quicker to solve it by

factoring.

Exercises: Solve each equation by quadratic formula.

1. x2 – 14x + 49 = 0

2. x2 – 4x – 21 = 0

3. x2 + 5x – 36 = 0

Name: ___________________ Section: _______

Instructor: ________________ Date: _______ Rating: ____

Instruction: Solve the following equations by the Quadratic Formula.

1. 2a2 – 10 = 9

_____________________________________________________

2. 6b2 – b = 12

_____________________________________________________

3. 3x2 + x = 14

_____________________________________________________

4. 10a2 + 3 = 11a

_____________________________________________________

5. 2x2 + 5x = 12

_____________________________________________________

6. 4x2 + 5x = 21

_____________________________________________________

7. 2x2 – 7x + 3 = 0

_____________________________________________________

8. 3a2 – 6a + 2 = 0

_____________________________________________________

9. 3b2 – 2b – 4 = 0

_____________________________________________________

10.a2 – 3a – 40 = 0

_____________________________________________________

11.3y2 – 11y + 10 = 0

_____________________________________________________

12.3w2 = 9 + 2w

_____________________________________________________

13.15z2 + 22z = 48

_____________________________________________________

14.9a2 + 14 = 24a

_____________________________________________________

15.16m2 = 24m + 19

_____________________________________________________

Lesson 9

Solving "by GraphingOBJECTIVES:

At the end of this lesson, students are expected to:

define graphing;

resolve the equation by graphing; and

draw the points from the equations given.

To be honest, solving "by graphing" is an achingly trendy but somewhat bogus

topic. The basic idea behind solving by graphing is that, since the "solutions" to "ax2 +

bx + c = 0" are the x-intercepts of "y = ax2 + bx + c", you can look at the x-intercepts of

the graph to find the solutions to the equation. There are difficulties with "solving" this

way, though....

When you graph a straight line like "y = 2x + 3", you can find the x-intercept (to a

certain degree of accuracy) by drawing a really neat axis system, plotting a couple

points, grabbing your ruler and drawing a nice straight line, and reading the

(approximate) answer from the graph with a fair degree of confidence.On the other

hand, a quadratic graphs as a wiggly parabola. If you plot a few non-x-intercept points

and then draw a curvy line through them, how do you know if you got the x-intercepts

even close to being correct? You don't. The only way you can be sure of your x-

intercepts is to set the quadratic equal to zero and solve. But the whole point of this

topic is that they don't want you to do the (exact) algebraic solving; they want you to

guess from the pretty pictures.

So "solving by graphing" tends to be neither "solving" nor "graphing". That is, you

don't actually graph anything, and you don't actually do any of the "solving". Instead,

you are told to punch some buttons on your graphing calculator and look at the pretty

picture, and then you're told which other buttons to hit so the software can compute the

intercepts (or you're told to guess from the pretty picture in the book, hoping that the

printer lined up the different print runs for the different ink colors exactly right). I think the

educators are trying to "help" you "discover" the connection between x-intercepts and

solutions, but the concept tends to get lost in all the button-pushing. Okay, enough of

my ranting...

To "solve" by graphing, the book may give you a very neat graph, probably with at

least a few points labeled; the book will ask you to state the points on the graph that

represent solutions. Otherwise, it will give you a quadratic, and you will be using your

graphing calculator to find the answer. Since different calculator models have different

key-sequences, I cannot give instruction on how to "use technology" to find the

answers, so I will only give a couple examples of how to solve from a picture that is

given to you.

Solve x2 – 8x + 15 = 0 by using the following graph.

The graph is of the related quadratic, y = x2 – 8x + 15, with the x-intercepts

being where y = 0. The point here is to look at the picture (hoping that the points

really do cross at whole numbers, as it appears), and read the x-intercepts (and

hence the solutions) from the picture.

The solution is x = 3,

Since x2 – 8x + 15 factors as (x – 3)(x – 5), we know that our answer is correct.

Solve 0.3x2 – 0.5x –  5/3 = 0 by using the following graph.

For this picture, they labeled a bunch of points. Partly, this was to be

helpful, because the x-intercepts are messy (so I could not have guessed their

values without the labels), but mostly this was in hopes of confusing me, in case I

had forgotten that only the x-intercepts, not the vertices or y-intercepts,

correspond to "solutions".

The x-values of the two points where the graph crosses the x-axis are the

solutions to the equation.

The solution is x = –5/3, 10/3

Find the solutions to the following quadratic:

They haven't given me the quadratic equation, so I can't check my work

algebraically. (And, technically, they haven't even given me a quadratic to solve;

they have only given me the picture of a parabola from which I am supposed to

approximate the x-intercepts, which really is a different question....)

I ignore the vertex and the y-intercept, and pay attention only to the x-

intercepts. The "solutions" are the x-values of the points where the pictured line

crosses the x-axis:

The solution is x = –5.39, 2.76

"Solving" quadratics by graphing is silly in "real life", and requires that the

solutions be the simple factoring-type solutions such as "x = 3", rather than something

like "x = –4 + sqrt(7)". In other words, they either have to "give" you the answers (by

labeling the graph), or they have to ask you for solutions that you could have found

easily by factoring. About the only thing you can gain from this topic is reinforcing your

understanding of the connection between solutions and x-intercepts: the solutions to

"(some polynomial) equals (zero)" correspond to the x-intercepts of "y equals (that same

polynomial)". If you come away with an understanding of that concept, then you will

know when best to use your graphing calculator or other graphing software to help you

solve general polynomials; namely, when they aren't factorable.

Name: ___________________ Section: _______

Instructor: ________________ Date: _______ Rating: ____

Instruction: Solve each equation by graphing.

1. x2 – 6x + 9 = 0

_____________________________________________________

2. x2 – 5x + 10 = 0

_____________________________________________________

3. 2x2 – 6x + 8 = 0

_____________________________________________________

4. x2 – 7x + 12 = 0

_____________________________________________________

5. 2x2 – 8x + 10 = 0

_____________________________________________________

6. 3x2 + 6x – 9 = 0

_____________________________________________________

7. x2+ 8x – 12 = 0

_____________________________________________________

8. x2 + 4x – 3 = 0

_____________________________________________________

9. x2 – 2x – 2 = 0

_____________________________________________________

10.2x2 – 4x – 2 = 0

_____________________________________________________

11.4x2 – 8x – 16 = 0

_____________________________________________________

12.x2 – 9x + 21 = 0

_____________________________________________________

13.x2 + 10x + 18 = 0

_____________________________________________________

14.2x2 – 16x + 8 = 0

_____________________________________________________

15.3x2 – 12x – 9 = 0

_____________________________________________________

A. Solve by factoring.

1. x2 – 3x – 10 = 0

2. x2 + 2x = 8

3. x2 – x – 4 = 2

4. 2x2 – 6x – 36 = x2 – 15

5. 4x2 + 4x = 15

6. 6x2 + 11x – 2 = 8

7. 49x2 + 28x – 10 = 0

8. 6x4 – 4x3 – 10x2 = 0

9. 18 + 15x – 18x2 = 0

10. x4 – 4x2 + 3 = 0

B. Solve by completing the square.

11. x2 - 4x – 3 = 0

12. x2 + 3x – 6 = 0

13. x2 – 7x + 5 = 0

14. 2x2 + 5x + 1 = 0

15. 2x2 + 8x – 5 = 0

C. Solve for x by the quadratic formula.

1. x2.- 4x – 7 = 0

2. x2 – 3x + 4 = 0

3. 2x2 + 4x + 5 = 0

4. x2 + 7x – 3 = 0

5. x2 – 7x + 2 = 0

6. x2 + 5x – 7 = 0

7. x2 + 9x – 3 = 0

8. 4x2 – 6x + 2 = 0

9. 9x2 – 9x – 10 = 0

10.x2 + 5x + 8 = 0