Transcript of "Module in solving quadratic equation "
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A premier university in CALABARZON, offering academic
programs and related services designed to respond to the requirements
of the Philippines and the global economy, particularly in Asian
Countries.
VISIO
N
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The University shall primarily provide advanced education, professional,
technological and vocational instruction in agriculture, fisheries, forestry, science,
engineering, industrial technologies, teacher education, medicine, law, arts and
sciences, information technologies and other related fields. It shall also undertake
research and extension services and provide progressive leadership in its areas of
specialization.
MISSIO
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In pursuit of the college vision/mission the College of Education is
committed to develop the full potentials of the individuals and equip them with
knowledge, skills and attitudes in Teacher Education allied fields to effectively
respond to the increasing demands, challenges and opportunities of changing
time for global competitiveness.
GOALS
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OBJECTIVES OF
BSED
Produce graduate who can demonstrate and practice the
professional and ethical requirement for the Bachelor of Secondary Education
such as:
1.To serve as positive and powerful role models in the pursuit of the learning thereby
maintaining high regards to professional growth.
2. Focus on the significance of providing wholesome and desirable learning environment.
3. Facilitate learning process in diverse type of learners.
4. Used varied learning approaches and activities, instructional materials and learning
resources.
5. Used assessment data, plan and revise teaching – learning plans.
6. Direct and strengthen the links between school and community activities.
7. Conduct research and development in Teacher Education and other related activities.
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This Teacher’s “MODULE IN SOLVING QUADRATIC EQUATION” is part of the requirements in
Educational Technology 2 under the revised curriculum based on CHED Memorandum Order (CMO)-30, Series
of 2004. Educational Technology 2 is a three (3)-unit course designed to introduce both traditional and
innovative technologies to facilitate and foster meaningful and effective learning where students are expected to
demonstrate a sound understanding of the nature, application and production of the various types of
educational technologies.
Students are provided with guidance and assistance of selected faculty The members of the College on
the selection, production and utilization of appropriate technology tools in developing technology-based teacher
support materials. Through the role and functions of computers especially the Internet, the student researchers
and the advisers are able to design and develop various types of alternative delivery systems. These kinds of
activities offer a remarkable learning experience for the education students as future mentors especially in the
preparation and utilization of instructional materials.
The output of the group’s effort may serve as a contribution to the existing body instructional materials
that the institution may utilize in order to provide effective and quality education. The lessons and evaluations
presented in this module may also function as a supplementary reference for secondary teachers and students.
reference for secondary teachers and students.
Aleli M. Ariola
Module Developer
Shane Maureen D. Atendido
Module Developer
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7.
This Teacher’s “MODULE IN SOLVING QUADRATIC EQUATION” is part of the requirements in Educational
Technology 2 under the revised curriculum based on CHED Memorandum Order (CMO)-30, Series of 2004.
Educational Technology 2 is a three (3)-unit course designed to introduce both traditional and innovative technologies
to facilitate and foster meaningful and effective learning where students are expected to demonstrate a sound
understanding of the nature, application and production of the various types of educational technologies.
Students are provided with guidance and assistance of selected faculty The members of the College through
the selection, production and utilization of appropriate technology tools in developing technology-based teacher
support materials. Through the role and functions of computers especially the Internet, the student researchers and
the advisers are able to design and develop various types of alternative delivery systems. These kinds of activities
offer a remarkable learning experience for the education students as future mentors especially in the preparation and
utilization of instructional materials.
The output of the group’s effort may serve as a contribution to the existing body instructional materials that the
institution may utilize in order to provide effective and quality education. The lessons and evaluations presented in
this module may also function as a supplementary reference for secondary teachers and students. reference for
secondary teachers and students.
FOR-IAN V. SANDOVAL
Computer Instructor / Adviser
Educational Technology 2
DELIA F. MERCADO
Module Consultant / Instructor 3
Principal of Laboratory High School
LYDIA R. CHAVEZ
Dean College of Education
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The authors would like to acknowledge with deep appreciation and gratitude the invaluable help of the following
persons:
Mr. For-Ian V. Sandoval our module adviser, Computer Instructor / Adviser Educational Technology 2 for giving
us opportunity to participate on this project, and for guiding us and pursue us to finish this module.
Mrs. Delia F. Mercado, Instructor III and Director of Laboratory High School, for being our Teacher Consultant for
the completion of this modular workbook.
Mrs. Corazon San Agustin, our instructor in Educational Technology I, for giving us guidance and
encouragement us in completing the requirement.
Mrs. Lydia Chavez, Dean of Education for the support and guidance.
We also wish to thank our family and friends as an inspiration and understand us they were robbed of many
precious moments as we looked ourselves in our rooms when our minds went prolific and our hands itched to write.
And finally, we thank Almighty God, the source of all knowledge, understanding and wisdom. From him we owe
all that we have and all that we are!
Once again, we thank all those who have encourage and helped us in preparing this module for publication and
who have extended us much understanding, patience, and support.
THE AUTHORS
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A quadratic equation is a second-order polynomial equation in a single variable x.
The general form is where x represents a variable, and a, b, and c, represent coefficients
and constants, with a ≠ 0. (If a = 0, the equation becomes a linear equation.)
Among his many other talents, Major General Stanley in Gilbert and Sullivan's operetta the Pirates of Penzance
impresses the pirates with his knowledge of quadratic equations in "The Major General's Song" as follows: "I am the very
model of a modern Major-General, I've information vegetable, animal, and mineral, I know the kings of England, and I
quote the fights historical, From Marathon to Waterloo, in order categorical; I'm very well acquainted too with matters
mathematical, I understand equations, both the simple and quadratic, About binomial theorem I'm teeming with a lot o'
news-- With many cheerful facts about the square of the hypotenuse."
The constants a, b, and c, are called respectively, the quadratic coefficient, the linear coefficient and the constant
term or free term. Quadratic comes from quadratus, which is the Latin word for "square." Quadratic equations can be
solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula. One common
use of quadratic equations is computing trajectories in projectile motion.
This module centers on the different ways of solving the quadratic equation by factoring, by finding square
roots, by completing the square, and by using the quadratic formula. Students are given guides to determine the most
appropriate method to use. Identifying the disciminant of the quadratic equation and finding the relationship between the
coefficient and the root of the quadratic equation are also discussed.
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At the end of this module, students are expected to:
•distinguish what is quadratic equation and complex numbers;
•recognize property of the number I and solutions in solving the four arithmetic operations;
•solve quadratic equation by factoring, completing the square, quadratic formula and solving by graphing;
•learn the technique on how to use any method for solving quadratic equation, taking the square root and
transforming quadratic equations by appropriate substitution;
•determine discriminant, and relations between roots and coefficient; and
•use calculator in solving quadratic equation.
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VMGO’s of BSEd
Foreword
Acknowledgement
Introduction
General Objective’s
Table of Contents
Chapter I. Identify the Quadratic Equation
Lesson 1. Quadratic Equation
Chapter II. Complex Number
Lesson 2. Defining Complex Number
Lesson 3. Number i
Lesson 4. Complex Plane
Lesson 5. Complex Arithmetic
Chapter III. Solving Quadratic Equation
Lesson 6. Factoring
Lesson 7. Completing the Square
Lesson 8. Quadratic Formula
Lesson 9. Solving by Graphing
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Chapter IV. Solving Equation on Quadratic
Lesson 10. Equation in Quadratic Form
Lesson 11. Equation Containing Radicals
Lesson 12. Equation Reducible to Quadratic Equation
Chapter V. The Discriminant, Roots and Coefficient
Lesson 13. Discriminant and the Roots of a Quadratic Equation
Lesson 14. Relation between Roots and Coefficient
Chapter VI. Solving Quadratic Equation on a Calculator
Lesson 15. Equation on a Calculator
References
Demo (aleli)
Demo (shane)
Slide share (aleli)
Slide share (shane)
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13.
Chapter I
This chapter deals with equations which are classified according to the highest power of its variable. An
equation in the variable x whose highest power is 2 is called a quadratic equation. It will be observed here that
variable a, b and c are real numbers and a cannot be 0.
TARGET SKILLS:
At the end of this chapter, students are expected to:
• identify quadratic equation;
• discuss real numbers and standard form of the quadratic equation;
• express quadratic equation in standard form; and
• apply distributive property in solving quadratic equation.
TARGET SKILLS:
At the end of this chapter, students are expected to:
• identify quadratic equation;
• discuss real numbers and standard form of the quadratic equation;
• express quadratic equation in standard form; and
• apply distributive property in solving quadratic equation.
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Lesson
1 Identifying the quadratic
equation
OBJECTIVES:
At the end of this lesson, students are expected to:
•define the quadratic equation;
•discuss real numbers in quadratic equation; and
•improve writing the standard form of the quadratic equation.
Polynomials are classified according to the highest power of its variable. A first degree polynomial, like 2x +
5 is linear; a second degree polynomial, like x2
+ 2 – 3 is quadratic; a third degree polynomial, like x3
+ 4x2
– 3x + 12
is cubic.
Similarly, equation and inequalities are classified according to the highest power of its variable. An equation
in the variable x whose highest power is two is called a quadratic equation. Some examples are x2
– 64, 4n2
= 25,
3x2
– 4x + 1 = 0.
An equation of the form ax2
+ bx + c = 0, where a, b and c are constant and a not equal to 0, a id a
quadratic equation. to 0, a id a quadratic equation.
Any quadratic equation can be written in the form ax2
+ bx + c = 0. This is also called the standard
form of the quadratic equation. Here, a, b and c are real numbers and a cannot be 0.
Example A. Express x2
= 8x in standard form
x2
= 8x can be written as x2
- 8x = 0
where a=1, b= ˉ8, and c=0.
Example B. Express x2
= 64 in standard form
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x2
= 64 can be written as x2
– 64 = 0
where a=1. B=0, and c=ˉ64.
Example C. Express the fractional equation x = 1/x-3 as a quadratic equation.
x = 1/x-3
x (x-3) = 1 multiply both sides by x-3
x² - 3x = 1 using the distributive property
x² - 3x - 1 = 0 a=1, b=ˉ3, c=ˉ1
Exercises:
Which of the following equations are quadratic?
1.3x = x² - 5
2. 2x =1
3. x² = 25
4. 2x - 3 = x + 5
5. 5x – 2y = 0
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Name: ___________________ Section: _______
Instructor: ________________ Date: _______ Rating: _______
Instruction: Write the following equations in the form ax2
+ bx + c = 0, and give the value of a, b, and c.
1. x2
= 6x
_____________________________________________
2. 2x2
= 32
_____________________________________________
3. 3x2
= 5x – 1
_____________________________________________
4. 10 = 3x – x2
_____________________________________________
5. (x + 2)2
= 9
_____________________________________________
6. 4x2
= 64
_____________________________________________
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18.
A. Define each of the following terms.
1. Quadratic equation
2. Standard form of a quadratic equation
3. Real numbers
B. Which of the following equations are quadratic?
1. 4x = 2x2
– 6
2. 3x = 1
3. 5x2
= 30
4. 3x – 2 = 2x + 6
5. 2x – 5y = 0
6. 4x + 2x2
– 3x3
= 0
7. 12x2
– x = 11
C. Write the following equations in the form ax2
+ bx + c = 0, and give the values of a, b and c.
1. 3x2
= 6x
2. 3x2
= 32
3. 2x2
= 5x – 1
4. 12 = 4x – x2
5. (x + 3)2
= 8
6. 4x2
= 56
7. 1/x + x = 6
8. x(x – 4) – 1 = 0
9. 9x = x2
10.(1/x)2
= 10
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19.
Chapter II
This chapter centers on the complex numbers which is a number comprising a real number and an imaginary
number. Under this, we have the number i, the complex plane where the points are plotted and the 4 arithmetic
operations such as addition and subtraction, multiplication and division of complex numbers. To round up the chapter,
simple equation involving complex numbers will be studied and solved.
TARGET SKILLS:
At the end of this chapter, students are expected to:
• identify complex numbers;
• differentiate the real pat and imaginary part of complex
numbers; and
• explore solving of the 4 arithmetic operations on the
complex numbers.
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20.
Less
on 2Defining Complex
NumbersOBJECTIVES:
At the end of this lesson, students are expected to:
•identify complex numbers;
•differentiate the real number and standard imaginary unit; and
•extend the ordinary real number.
A complex number, in mathematics, is a number comprising a real number and an imaginary number; it can
be written in the form a + bi, where a and b are real numbers, and i is the standard imaginary unit, having the
property that i2
= −1. The complex numbers contain the ordinary real numbers, but extend them by adding in extra
numbers and correspondingly expanding the understanding of addition and multiplication.
Equation 1: x2
- 1 = 0.
Equation 1 has two solutions, x = -1 and x = 1. We know that solving an equation in x is equivalent to finding
the x-intercepts of a graph; and, the graph of y = x2
- 1 crosses the x-axis at (-1,0) and (1,0).
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Equation 2: x2
+ 1 = 0
Equation 2 has no solutions, and we can see this by looking at the graph of y = x2
+
1.
Since the graph has no x-intercepts, the equation has no solutions. When we define complex numbers,
equation 2 will have two solutions.
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Less
on 3The
Number i
OBJECTIVES:
At the end of this lesson, students are expected to:
•recognize the property of the number i;
•discuss the powers of i; and
•solve the high powers of imaginary unit.
Equation 1 Equation 2
x2
- 1 = 0. x2
+ 1 = 0.
x2
= 1. x2
= -1.
Consider Equations 1 and 2 again.
Equation 1 has solutions because the number 1 has two square roots, 1 and -1. Equation 2 has no solutions
because -1 does not have a square root. In other words, there is no number such that if we multiply it by itself we get
-1. If Equation 2 is to be given solutions, then we must create a square root of -1.
The imaginary unit i is defined by
The definition of i tells us that i2
= -1. We can use this fact to find other powers of i.
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25.
Example
i3
= i2
* i = -1*i = -i.
i4
= i2
* i2
= (-1) * (-1) = 1.
Exercise:
Simplify i8
and i11
.
We treat i like other numbers in that we can multiply it by numbers, we can add it to other numbers, etc. The
difference is that many of these quantities cannot be simplified to a pure real number.
For example, 3i just means 3 times i, but we cannot rewrite this product in a simpler form, because it is not a real
number. The quantity 5 + 3i also cannot be simplified to a real number.
However, (-i)2
can be simplified. (-i)2
= (-1*i)2
= (-1)2
* i2
= 1 * (-1) = -1.
Because i2
and (-i)2
are both equal to -1, they are both solutions for Equation 2 above.
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26.
Name: ___________________ Section: _______
Instructor: ________________ Date: _______ Rating: ______
Instruction: Express each number in terms of i and simplify.
1.
2.
3.
4.
5.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
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28.
Lesson 4
The Complex Plane
OBJECTIVES:
At the end of this lesson, students are expected to:
•distinguish the points on the plane;
•differentiate the real and imaginary part; and
•draw from memory the figure form by the plot points on the complex plane.
A complex number is one of the form a + bi, where a and b are real numbers. a is called the real part of the
complex number, and b is called the imaginary part.
Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. I.e.,
a+bi = c+di if and only if a = c, and b = d.
Example.
2 - 5i.
6 + 4i.
0 + 2i = 2i.
4 + 0i = 4.
The last example above illustrates the fact that every real number is a complex number (with imaginary part 0).
Another example: the real number -3.87 is equal to the complex number -3.87 + 0i.
It is often useful to think of real numbers as points on a number line. For example, you can define the order
relation c < d, where c and d are real numbers, by saying that it means c is to the left of d on the number line.
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29.
We can visualize complex numbers by associating them with points in the plane. We do this by letting the
number a + bi correspond to the point (a,b), we use x for a and y for b.
Exercises: Represent each of the following complex number by a point in the plane.
1. 3 + 2i
2. 1 – 4i
3. 4 + 3i
4. 2 – 5i
5. 4 – 3i
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30.
Name: ___________________ Section: _______
Instructor: ________________ Date: _______ Rating: ______
Instruction: Represent each of the following Complex Numbers by a point in the plane.
1.
2.
3.
4. 0
5. 3
6.
7. 1/2
8.
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
______________________________________________________
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9.
10.
11.
12.
13.
14.
15.
_____________________________________________________
_____________________________________________________
_____________________________________________________
____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
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32.
Lesson 5
Complex Arithmetic
OBJECTIVES:
At the end of this lesson, students are expected to:
•define the four arithmetic operations on complex numbers;
•comply with the steps in solving the different operations; and
•solve the four arithmetic operations.
When a number system is extended the arithmetic operations must be defined for the new numbers, and the
important properties of the operations should still hold. For example, addition of whole numbers is commutative. This
means that we can change the order in which two whole numbers are added and the sum is the same: 3 + 5 = 8 and 5
+ 3 = 8.
We need to define the four arithmetic operations on complex numbers.
Addition and
Subtraction To add or subtract two complex numbers, you add or subtract the real parts and the
imaginary parts.
(a + bi) + (c + di) = (a + c) + (b + d)i.
(a + bi) - (c + di) = (a - c) + (b - d)i.
Example
(3 - 5i) + (6 + 7i) = (3 + 6) + (-5 + 7)i = 9 + 2i.
(3 - 5i) - (6 + 7i) = (3 - 6) + (-5 - 7)i = -3 - 12i.
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33.
Note
These operations are the same as combining similar terms in expressions that have a variable. For example, if
we were to simplify the expression (3 - 5x) + (6 + 7x) by combining similar terms, then the constants 3 and 6 would be
combined, and the terms -5x and 7x would be combined to yield 9 + 2x.
The Complex Arithmetic applet below demonstrates complex addition in the plane. You can also select the other
arithmetic operations from the pull down list. The applet displays two complex numbers U and V, and shows their sum.
You can drag either U or V to see the result of adding other complex numbers. As with other graphs in these pages,
dragging a point other than U or V changes the viewing rectangle.
Multiplication
The formula for multiplying two complex numbers is
(a + bi) * (c + di) = (ac - bd) + (ad + bc)i.
You do not have to memorize this formula, because you can arrive at the same result by treating the complex
numbers like expressions with a variable, multiply them as usual, then simplify. The only difference is that powers of i
do simplify, while powers of x do not.
Example
(2 + 3i)(4 + 7i) = 2*4 + 2*7i + 4*3i + 3*7*i2
= 8 + 14i + 12i + 21*(-1)
= (8 - 21) + (14 + 12)i
= -13 + 26i.
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Notice that in the second line of the example, the i2
has been replaced by -1.
Using the formula for multiplication, we would have gone directly to the third line.
Exercise
Perform the following operations.
(a) (-3 + 4i) + (2 - 5i)
(b) 3i - (2 - 4i)
(c) (2 - 7i)(3 + 4i)
(d) (1 + i)(2 - 3i)
Division
The conjugate (or complex conjugate) of the complex number a + bi is a - bi.
Conjugates are important because of the fact that a complex number times its conjugate is real; i.e., its
imaginary part is zero.
(a + bi)(a - bi) = (a2
+ b2
) + 0i = a2
+ b2
.
Example
Number Conjugate Product
2 + 3i 2 - 3i 4 + 9 = 13
3 - 5i 3 + 5i 9 + 25 = 34
4i -4i 16
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35.
Suppose we want to do the division problem (3 + 2i) ÷ (2 + 5i). First, we want to rewrite this as a fractional
expression .
Even though we have not defined division, it must satisfy the properties of ordinary division. So, a number
divided by itself will be 1, where 1 is the multiplicative identity; i.e., 1 times any number is that number.
So, when we multiply by, , we are multiplying by 1 and the number is not changed.
Notice that the quotient on the right consists of the conjugate of the denominator over itself. This choice was
made so that when we multiply the two denominators, the result is a real number. Here is the complete division
problem, with the result written in standard form.
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36.
Exercise:
Write (2 - i) ÷ (3 + 2i) in standard form.
We began this section by claiming that we were defining complex numbers so that some equations would have solutions.
So far we have shown only one equation that has no real solutions but two complex solutions. In the next section we will
see that complex numbers provide solutions for many equations. In fact, all polynomial equations have solutions in the set
of complex numbers. This is an important fact that is used in many mathematical applications. Unfortunately, most of these
applications are beyond the scope of this course. See your text (p. 195) for a discussion of the use of complex numbers in
fractal geometry.
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37.
Name: ___________________ Section: _______
Instructor: ________________ Date: _______ Rating: ______
Instruction: Perform the indicated operations and express the result in the form .
1.
2.
3.
4.
5.
6.
7.
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
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39.
A. Define and/or describe each of the following terms.
• Imaginary part
• Real number
• Complex number
• Complex plane
• Imaginary unit
• Commutative property
• Complex conjugate
B. 1. Simplify:
1. i15
2. i25
3. i106
4. i207
5. i21
2. Perform the indicated operation and express each answer.
a. +
b. +
c. +
d. +
e. +
f. +
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40.
3. Represent each complex numbers by a point in the plane.
a.3 – i
b.-2 + 4i
c.-3 + 3i
d.4 + 5i
e.-3 + 5i
4. Give the real part and the imaginary part of each complex numbers in #3.
5. Perform the indicated operations.
a. (3 – 2i) + (-7 + 3i)
b. (-4 + 7i) + (9 – 2i)
c. (14 – 9i) + (7 – 6i)
d. (5 + i) – (3 + 2i)
e. (7 – 2i) – (4 – 6i)
f. (8 + 3i) – (-4 – 2i)
g. (3 – 2i) (3 +2i)
h. (5 + 3i) (4 – i)
i. (11 + 2i)2
(5 – 2i)
j. (5 + 4i) / (3 – 2i)
k. (4 + i) (3 – 5i) / (2 – 3i)
l. (7 + 3i) / (3 – 3i / 4)
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41.
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.
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42.
Lesson 6
Solving by
factoring
OBJECTIVES:
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
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43.
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
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44.
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.
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45.
(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:
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46.
•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:
•(x – 3)(x – 5) = 0.
•x2
+ 6x + 7 = 0.
•x2
– 4 = 2x.
•x2
– 6x = 0.
•x2
– 8 = 0.
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49.
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".
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50.
The answer can also be written in rounded form as
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51.
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.)
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52.
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
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55.
Lesson 8
Quadratic Formula
OBJECTIVES:
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
+ =
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) = ± ( /2a
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56.
Step 4. Solve for x.
x = or
Let a, b and c be real constant, where a ≠ 0. Then the roots of ax2
+ bx + c = 0 are
x =
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 =
=
=
=
=
The roots are and .
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
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57.
Here a=2, b=⁻5 c=2. By the quadratic formula
x = =
The roots are 2 and .
.
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 = 1/2 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
4. x2
+ x – 30 = 0
5. x2
+ 3x = 40
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58.
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
_____________________________________________________
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60.
Lesson 9
Solving "by
Graphing
OBJECTIVES:
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
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61.
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.
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62.
• 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:
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63.
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.
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67.
Chapter IV
Much of the study in quadratic equation consist of different solving equation, we have equation in
quadratic form, equation containing radicals and equation reducible to quadratic equation. They have their own
steps and procedures to be followed in order to solve the given equation.
TARGET SKILLS:
At the end of this chapter, students are expected to:
• discuss solving equation on quadratic;
• determine the index and its radicals;
• interpret the solution of the original equation; and
• select appropriate method in solving quadratic equation.
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68.
Lesson 10
Equation in Quadratic Form
OBJECTIVES:
At the end of this lesson, students are expected to:
•identify equation in quadratic form;
•select appropriate method in solving quadratic equation; and
•change the equation in standard form.
Quadratic in Form
An equation is quadratic in form when it can be written in this standard form
where the same expression is inside both ( )'s.
In other words, if you have a times the square of the expression following b plus b times that same
expression not squared plus c equal to 0, you have an equation that is quadratic in form.
If we substitute what is in the ( ) with a variable like t, then the original equation will become a quadratic
equation.
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69.
Solving Equations that are
Quadratic in Form
If it is not in standard form, move any term(s) to the appropriate side by using the
addition/subtraction property of equality.
Also, make sure that the squared term is written first left to right, the expression not squared is
second and the constant is third and it is set equal to 0.
Step 1: Write in Standard Form, , if needed.
In other words, substitute your variable for what is in the ( ) when it is in standard form, .
I’m going to use t for my substitution, but really you can use any variable as long as it is not the
variable that is used in the original equation.
Step 2: Substitute a variable in for the expression that follows b in the second term.
You can use any method you want to solve the quadratic equation: factoring, completing the square
or quadratic formula.
Step 3: Solve the quadratic equation created in step 2.
Keep in mind that you are finding a solution to the original equation and that the variable you
substituted in for in step 2 is not your original variable.
Use the substitution that was used to set up step 2 and then solve for the original variable.
Step 4: Find the value of the variable from the original equation.
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70.
In some cases, you will be working with rational exponents and square roots in your problems. Those
types of equations can cause extraneous solutions. Recall that an extraneous solution is one that is a
solution to an equation after doing something like raising both sides of an equation by an even power,
but is not a solution to the original problem.
Even though not all of the quadratic in form equations can cause extraneous solutions, it is better to
be safe than sorry and just check them all.
Step 5: Check your solutions.
Example 1: Solve the equation that is quadratic in form:
.
Standard Form,
*Rewriting original equation to show it is quadratic in form
*Note that (y squared) squared = y to the fourth
*When in stand. form, let t = the expression following b.
Next, we need to substitute t in for y squared in the original equation.
*Original equation
*Substitute t in for y squared
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71.
Note how we ended up with a quadratic equation when we did our substitution. From here, we need
to solve the quadratic equation that we have created.
Solve the quadratic equation: factoring, completing the square or quadratic formula.
*Factor the trinomial
*Use Zero-Product Principle
*Set 1st factor = 0 and solve
*Set 2nd factor = 0 and solve
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72.
Let's find the value(s) of y when t = -4:
*Plug in - 4 for t
*Use square root method to solve for y
*First solution
*Second solution
Let's find the value(s) of y when t = 1:
*Plug in 1 for t
*Use square root method to solve for y
*First solution
*Second solution
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73.
Example 2: Solve the equation that is quadratic in form:
.
Standard Form,
*Inverse of add. 3 is sub. 3
*Equation in standard form
Note how when you square x to the 1/3 power you get x to the 2/3 power, which is what you have
in the first term.
* Rewriting original equation to show it is quadratic in
form
*Note that (x to the 1/3 power) squared = x to the 2/3
power
*When in stand. form, let t = the expression following b.
Next, we need to substitute t in for x to the 1/3 power in the original equation.
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74.
*Original equation
*Substitute t in for x to the 1/3 power
You can use any method you want to solve the quadratic equation: factoring, completing the
square or quadratic formula.
*Factor the trinomial
*Use Zero-Product Principle
*Set 1st factor = 0 and solve
*Set 2nd factor = 0 and solve
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75.
Let's find the value(s) of x when t = 3:
*Plug in 3 for t
*Solve the rational exponent equation
*Inverse of taking it to the 1/3 power is
raising it to the 3rd power
Let's find the value(s) of x when t = -1:
*Plug in -1 for t
*Solve the rational exponent equation
*Inverse of taking it to the 1/3 power is
raising it to the 3rd power
Let's double check to see if x = 27 is a solution to the original equation.
*Plugging in 27 for x
*True statement
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76.
Since we got a true statement, x = 27 is a solution.
Let's double check to see if x = -1 is a solution to the original equation.
*Plugging in -1 for x
*True statement
Since we got a true statement, x = -1 is a solution.
There are two solutions to this equation: x = 27 and x = -1.
Exercises:
1. a4
+ 2a2 –
5 = 0
2. x2
– 3x + 2 = 0
3. s6
+ 8s3
– 6 = 0
4. n2
– 6n + 10 = 0
5. g8
+ 2g4
– g = 0
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79.
Lesson 11
Equation Containing Radicals
OBJECTIVES:
At the end of this lesson, students are expected to:
•determine the index and its radicals;
•positively respond to the note to be remembered; and
•perform isolation of one radical if there are two radicals in the equation.
In the radicals which is read the “nth root of b,” the positive integer n is called the index or
order of the radical, and b is called its radicand. When n is 2, 2 is no longer written, just simply write
instead of to indicate the square root of b, thus is read as “cube root of b”; as “4th
of b”.
Note:
•In order to solve for x, you must isolate x.
•In order to isolate x, you must remove it from under the radial.
•If there are two radicals in the equation, isolate one of the radicals.
•Then raise both sides of the equation to a power equal to the index of the isolated radical.
•Isolate the the remaining radical.
•Raise both sides of the equation to a power equal to the index of the isolated radical.
•You should now have a polynomial equation. Solve it.
•Remember that you did not start out with a polynomial; therefore, there may be extraneous solutions.
Therefore, you must check your answers.
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80.
Example 1:
First make a note of the fact that you cannot take the square root of a negative number. Therefore,
the term is valid only if and the second term is valid if
Isolate the term
Square both sides of the equation.
Isolate the term
Square both sides of the equation.
Check the solution by substituting 9 in the original equation for x. If the left side of the equation
equals the right side of the equation after the substitution, you have found the correct answer.
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81.
• Left side:
•Right Side:1
•
Since the left side of the original equation does not equal the right side of the original equation
after we substituted our solution for x, then there is no solution.
You can also check the answer by graphing the equation:
.The graph represents the right side of the original equation minus the left side of the original equation.. The x-
intercept(s) of this graph is (are) the solution(s). Since there are no x-intercepts, there are no solutions.
Exercises:
Solve each of the following equation.
1. = 3
2. = x + 1
3. =
4. = 5
5. = 10
= 5
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82.
Name: ___________________ Section: _______
Instructor: ________________ Date: _______ Rating: ______
Instruction: Solve each of the following equation.
1.
2.
3.
4.
5.
6.
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
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84.
Lesson 12
Equations Reducible to Quadratic Equations
OBJECTIVES:
At the end of this lesson, students are expected to:
•interpret the solution of the original equation;
•organize the equation if it is quadratic equation ; and
•solve the equation by factoring or quadratic formula.
A variety of equations can be transformed into quadratic equations and solved by methods that we have
discussed in the previous section. We will consider fractional equations, equations involving radicals and
equation that can be transformed into quadratic equations by appropriate substitutions. Since the transformation
process may introduce extraneous roots which are not solutions of the original equation, we must always check
the solution in the original equation.
Example: Solve 1 1 7
x+2 + x+3 = 12
Solution: First note that neither -2 nor -3 can be a solution since at either of these points the equation is
meaningless.
Multiplying by the LCD, 12(x+2) (x+3), we get
12(x+3) + 12(x+2) = 7(x+2) (x+3)
24x + 60 = 7(x2
+ 5x + 6)
or
7x2
+ 11x – 18 = 0
Factoring, we get, (7x + 18)(x – 1) = 0
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85.
x = 1 or -18
7
If x = 1, _1_ _1_ _1_ _1_ _7_
1+2 1+3 = 3 + 4 = 12
Therefore x = 1 is a solution.
If x = -18, __1__ + __1__
7 -18/7 + 2 -18/7 + 3
= __7__ + __7__
-18 + 14 -18 + 21
= _-7_ + _7_ = _7_
4 3 12
Therefore, x = _-18_ is a solution.
7
Example 2. = - 2
Solution: squaring both sides of the equation, we obtain,
= + 4
2x – 16 = 4
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86.
Dividing both sides by 2 gives, x – 8 = -2
Squaring both sides of the equation we get
(x - 8) =
2
x2
- 16x + 64 = 4 (x +16)
x2
– 20x = 0
x(x – 20) =0
x = 0 or x = 20
Check: if x = 0, =
=
8 = 6 – 2
8 ≠ 4
Therefore x = 20 is not a solution of the original equation.
Thus the only root of
Many equations are not quadratics equations. However, we can transform them by means of appropriate
substitutions into quadratics equations and then solve these by techniques that we know.
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87.
Example: Solve:
1. 2x-2
– 7x-1
+ 3 = 0
2. x4
– 2x2
– 2 = 0
3.
2
–
Solutions
a.Let u = x-1
. Then u2
= (x-1
)2
= x-2
and our equation becomes
2u2 – 7u + 3 = 0, a quadratic equation in u.
To solve the equation, we factor the left-hand side.
(2u – 1)( u – 3) = 0
U = ½ or u = 3
Since u = x-1
, x-1
=
or x-1
= 3, from which
x = 2 or x =
Check: if x = 2, 2(2-2
) – 7(2-1
) + 3 =
Thus x = 2 is solution
If x = 1/3, 2(1/3)-2
– 7(1/3)-1
+ 3 = 2(3)2
– 7(3) + 3
So, x = 1/3 is a solution.
b. Let u = x2
. Then u2
= x4
and the given equation becomes a quadratic equation in u.
u2
– 2u – 2 = 0
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88.
u = =
u = 1 + or u =
Since u = x2
and u = < 0, we have to discard this solution.
u = x2
= implies
x = ±
It is simple to verify that both values of x satisfy the original equation. The roots of x4
– x2
– 2 = 0 are
and -
.
c. Let u = .This substitution yields a quadratic equation in u.
u2
– u – 2 = 0
(u – 2)(u + 1) = 0
u = 2 0r u = ˉ1
u = = 2 implies x = 2(4x + 1)
or x =
u = = 1 implies x = ˉ4x – 1
or x =
Again, it can easily be verified that both solutions check in the original equation.
The roots are and
.
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89.
Name: ___________________ Section: _______
Instructor: ________________ Date: _______ Rating: ______
Instruction: Solve the following equation.
1.
2.
3.
4.
5.
6.
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
_____________________________________________________
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92.
Chapter V
The discriminant gives additional information on the nature of the roots beyond simply whether there
are any repeated roots: it also gives information on whether the roots are real or complex, and rational or
irrational. More formally, it gives information on whether the roots are in the field over which the polynomial is
defined, or are in an extension field, and hence whether the polynomial factors over the field of coefficients.
This is most transparent and easily stated for quadratic and cubic polynomials; for polynomials of degree 4 or
higher this is more difficult to state.
TARGET SKILLS:
At the end of this chapter, students are expected to:
•determine discriminant, roots and coefficient;
• discuss the relation the roots and coefficient;
• find the sum and product of the roots; and
• change quadratic equation to discriminant formula.
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93.
Lesson 13
The Discriminant and the roots of a
Quadratic Equation
OBJECTIVES:
At the end of this lesson, students are expected to:
•determine discriminant and the roots;
•compare discriminant and the nature of the roots; and
•change quadratic equation to discriminant using the nature of the roots.
Example
1. Find the x-intercept of y = 3x² - 6x + 4.
Solution: As already mentioned, the values of x for which 3x² - 6x + 4 = 0 give the x-intercepts of the function.
We apply the quadratic formula in solving the equation.
3x² - 6x + 4 = 0
x = =
Since is not a real number, the equation 3x² - 6x + 4 = 0 has no real root. This means that the
parabola y = 3x² - 6x + 4 does not intersect the x-axis.
Let us write the equation in the form y = a(x – h)²+ k.
y = 3(x² – 2x)² + 4
= 3(x – 1)² + 1
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94.
∆ = b² - 4ac Roots of ax² + bx + c = 0
Positive Real and distinct
r =
s =
Zero Real and equal
r = s =
Negative No real roots
Example 2. Use the disciminant to determine the nature of the roots of the
following quadratic equation.
a. x² - x + ¼ = 0
a = 1, b = ˉ1, c = ¼
b² - 4ac = (ˉ1)² - 4 (1)(¼)
= 1 – 1
= 0 There is only one solution, that is, a double root.
Note that x² - x = ½ = (x - ½), so that double root is ˉb/2a = ½.
b. 5x² - 4x + 1 = 0
a = 5, b = ˉ4, c = 1
b² - 4ac = (ˉ4)² - 4 (5)(1)
= 16 – 20
= ˉ4 < 0 There are no real roots since a negative number has no real square root.
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95.
Name: ___________________ Section: _______
Instructor: ________________ Date: _______ Rating: ______
Instruction: Use the Discriminant to determine the nature of the root of the following Quadratic Equations.
1.x2
- 2x – 3=0
_____________________________________________________
2. 6x2
– x – 1 = 0
_____________________________________________________
3. 2x2
– 50 = 0
_____________________________________________________
4. x2
– 8x + 12 = 0
_____________________________________________________
5. x2
+ 5x – 14 = 0
_____________________________________________________
6. -4x2
– 4x + 1 = 0
_____________________________________________________
7. 7x2
+ 2x – 1 = 0
_____________________________________________________
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97.
Lesson 14
Relation between roots and coefficient
OBJECTIVES:
At the end of this lesson, students are expected to:
•classify roots and coefficient;
•discuss relations between the roots and coefficient of the quadratic equation; and
•find the sum and product of the roots of a given quadratic equation
There are some interesting relations between the sum and the product of the roots of a quadratic
equation. To discover these, consider the quadratic equation ax2
+ bx + c = 0, where a ≠ 0.
Multiply both sides of this equation by 1/a so that the coefficient of x2
is 1.
(ax2
+ bx + c) =
We obtain an equivalent quadratic equation in the form
x2
+ + = 0
If r and s are the roots of the quadratic equation ax2
+ bx + c = 0, then from the quadratic formula
r = and s =
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98.
Adding the roots, we obtain
r + s = +
= =
Multiplying the roots, we obtain
rs =
= c / a
Observe the coefficient in the quadratic equation x2
+ bx/a + c/a = 0. How do they compare with the
sum and the product of the roots? Did you observe the following?
1. The sum of the roots is equal to the negative of the coefficient of x.
r + s = -b / a
2. The product of the roots is equal to the constant term
rs = c / a
An alternate way of arriving at these relations is as follows
Let r and s be the roots of x2
+ bx/a + c/a = 0. Then
x - r)(x – s) = 0
Expanding gives, x2
– rx – sx + rs = 0
or x2
– (r + s)x + rs = 0
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99.
Comparing the coefficients of the corresponding terms, we obtain
r + s = -b / a and rs c / a
The above relations between the roots and the coefficients provide a fast and convenient means of
checking the solutions of a quadratic equation.
Example: Solve and check. 2x2
+ x – 6 = 0
Solutions: 2x2
+ x – 6 = (2x – 3)(x + 2) = 0
x = 3/2 or x = 2
The roots are 3/2 and 2.
To check, we add the roots, 3/2 = (-2) = -1/2 = -b/a.
and multiply them 3/2 = (-2) = -3 = c/a
Example: Find the sum and the product of the roots of 3x2
– 6x + 8 = 0 without having to first determine the roots.
Solution: The sum of the roots is r + s = -c/a = -(-6)/3 = 2
and their product is rs = c/a = 8/3
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100.
Name: ___________________ Section: _______
Instructor: ________________ Date: _______ Rating: ______
Instruction: Without solving the roots, find the sum and product of the roots of the following.
1. 6x2
– 5x + 2 = 0
_____________________________________________________
2. x2
+ x – 182 = 0
_____________________________________________________
3. x2
– 5x – 14 = 0
_____________________________________________________
4. 2x2
– 9x + 8 = 0
_____________________________________________________
5. 3x2
- 5x – 2 = 0
_____________________________________________________
6. x2
– 8x – 9 = 0
_____________________________________________________
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102.
A. Use the discriminant to determine which of the following quadratic equations have two, one or no real
roots. Give reasons.
1. x2
– 5x – 5 = 0
2. x2
– 3x – 2 = 3x – 11
3. 5x2 –
9x = 2x – 7
4. 2x2
– 7x + 8 = 0
5. 3 – 4x – 2x2
= 0
6. 4x2
– 9x + 5 = 3x – 7
B. By using the relations between roots and coefficients, determine if the given #s are roots of the
corresponding given equation.
1. 6x2
– 5x + 3 = 0 (1/6, -1)
2. x2
+ x _ 182 = 0 (13, -4)
3. x2
– 5x – 14 =0 (2, -7)
4. 2x – 9x + 8 = 0 (3, 4/3)
5. 3x2
– 5x – 2 = 0 (2, -1/3)
C. Given one roots of the equation, find the other.
A. x2
– 8x – 9 = 0; r= 1
B. 2x2
– 3x – 9 = 0; r= 3
C. x2
+ x – 2 - = 0 r =
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103.
Chapter VI
In this chapter, students will be taught how to find solutions to quadratic equations. This lesson
assumes students are already familiar with solving simple quadratic equations by hand, and that they have
become relatively comfortable using their graphing calculator for solving arithmetic problems and simple
algebra problems. Students will also be shown strategies on how to use the keys on the graphing calculator to
show a complete graph.
TARGET SKILLS:
At the end of this chapter, students are expected to:
• use calculator in solving quadratic equation;
• solve equation on a calculator; and.
• improve skills using calculator by solving quadratic equation.
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104.
Lesson 15
Equation on a Calculator
OBJECTIVES:
At the end of this lesson, students are expected to:
• acquire knowledge using calculator in solving quadratic equation;
• resolve equations on a calculator; and
• improve skills on solving quadratic equation using a calculator.
The simplest way to solve a quadratic equation on a calculator is to use the
quadratic formula.
x=-b ± where d=b² - 4ac
2a
As we have seen, if d < 0, there are no real solutions. But f d ≥ 0, then we can use calculator to get the
solutions.
To solve 3x² + 5x – 7 = 0, first compute the discriminant.
d = b² - 4ac = 5² - 4(3)(-7).
On an arithmetic calculator, the keystroke sequence for d is,
[AC][MC] 4 [x] 3 [x] 7 [=][M+] 5 [x][=][+][MR][=]
The display will show the value of the discriminant to be 109, and so the quadratic equation has two distinct
roots. To compute the roots, proceed as follows:
[AC][MC] 109 [√][M+][MR][-] 5 [+] 2 [x] 3 [=] the first root and
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105.
0 [-][MR][-] 5 [+] 2 [ x] 3 [=] the second root.
On an algebraic calculator, the keystroke sequence is easier. Recall that the actual roots are:
x= -5 +
2(3)
x= -5 -
2(3)
First, we compute the square root of the discriminant and store it is memory.
[AC] 5 [2] – 4 [x] 3 [x] 7 [+/-][=][√][Min]
Then, compute the first root as:
x1 =
(5 [+/-][+][MR])[÷] (2 [x] 3) [=];
and then compute the second root as:
x2 =
(5[+/-][-][MR])[+] (2 [x] 3)[=].
Exercise:
Find the roots of the following equations.
• 3.5x2
+ 1.2x – 3.2 = 0
• 7.6 -2.2x – 1.7x2
= 0
• 2.5x2
+ 5.6x – 13.5 = 0
• x – 77.3 + 2.3x2
= 0
• x2
- 1000.5 + 32.3 = 0
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106.
Name: ___________________ Section: _______
Instructor: ________________ Date: _______ Rating: ______
Instruction: Find the roots of the following equations.
1. 5.3x2
+ 2.1v – 2.3 = 0
_____________________________________________________
2. 6.7 v - 2.2x – 7.1x2
= 0
_____________________________________________________
3. 5.2x2
+ 6.5x – 5.13 = 0
_____________________________________________________
4. x2
– 50.001 + 33.2 = 0
_____________________________________________________
5. x – 7.73 + 2.3x2
= 0
_____________________________________________________
6. 3.3x2
– 1.9x – 7.10 = 0
_____________________________________________________
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109.
Salamat, Lorina G., College Algebra, National Book Store 1988 pg. 151 – 159.
Coronel, Iluminada C. F.M.M, Mathematics 3 An Integrated Approach, Bookmark Inc. 1991 pg. 77 –
79, 134 – 158.
Coronel, Iluminada C. F.M.M, Mathematics 4 An Integrated Approach, Bookmark Inc. 1992 pg. 276 –
297.
Borwein, P. and Erdélyi, T. "Quadratic Equations." §1.1.E.1a in Polynomials and Polynomial
Inequalities. New York: Springer-Verlag, p. 4, 1995.
URL – Images
http://www.mathwarehouse.com/algebra/complex-number/absolute-value-complex-number.php
http://commons.wikimedia.org/wiki/File:Quadratic_equation_coefficients.png
http://mathworld.wolfram.com/PolynomialRoots.html
http://www.livephysics.com/shop/tools-and-gadgets.html
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