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
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