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Part III:
TEACHING-LEARNING ACTIVITIES (TLA)
The following are review materials needed to be read first before accomplishing the tasks
required from the student. The topics are arranged in such a way that you need to accomplish
the tasks in chronological order in order to determine the solution of linear equations, quadratic
equations and higher degree polynomials. Moreover, after mastering the operations involved in
finding the solutions of these equations, the students may proceed to determining the solutions
of systems of equations.
1. Linear Equations
The general form of the linear equation is ax + c = 0 where a and c are real numbers and
a is not zero. It is an equation which involves only one variable whose highest power is 1 in that
variable.
Solution of linear equation or Root of linear equation
The value of the variable which makes left hand side equal to right hand side in the given
equation is called the solution or the root of the equation. A linear equation has only one root.
Rules for solving a linear equation in one variable:
The equation remains unchanged if –
(a)The same number is added to both sides of the equation.
Example:
1. x - 4 = 7
⇒ x - 4 + 4 = 7 + 4 (Add 4 to both sides)
⇒ x = 11
2. x - 2 = 10
⇒ x - 2 + 2 = 10 + 2 (Add 2 to both sides)
⇒ x = 12
b) The same number is subtracted from both sides of the equation.
Example:
1. x + 5 = 9
⇒ x + 5 - 5 = 9 - 5 (Subtract 5 from both sides)
⇒x+0=4
⇒x=4
2. x + 1/2 = 3
x + 1/2 - 1/2 = 3 - 1/2 (Subtract 1/2 from both sides)
⇒ x = 3 - 1/2
⇒ x = (6 - 1)/2
⇒ x = 5/2
(c) The same number is multiplied to both sides of the equation.
Example:
1. x/2 = 5
⇒ x/2 × 2 = 5 × 2 (Multiply 2 to both the sides)
⇒ x = 10
2. x/5 = 15
⇒ x/5 × 5 = 15/5 (Multiply 5 to both the sides)
⇒x=3
(d) The same non-zero number divides both sides of the equation.
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Example:
1. 0.2x = 0.24
⇒ 0.2x/0.2 = 0.24/0.2 (Divide both sides by 0.2)
⇒ x = 0.12
2. 5x = 10
⇒ 5x/5 = 10/5 (Divide both sides by 2)
⇒x=2
Source: https://www.math-only-math.com/linear-equation.html
2. Quadratic Equations
A quadratic equation is an equation of the form, ax2 + bx + c = 0, where a, b and c are real
numbers with a ≠ 𝟎.
Solution of linear equation or Root of linear equation
The value of the variable which makes left hand side equal to right hand side in the given
equation is called the solution or the root of the equation. A quadratic equation may have one
or two roots.
Methods for Solving Quadratic Equations
2.1. Factoring
To solve a quadratic equation by factoring,
1. Put all terms on one side of the equal sign, leaving zero on the other side.
2. Factor.
3. Set each factor equal to zero.
4. Solve each of these equations.
5. Check by inserting your answer in the original equation.
Example:
1. Solve x 2 – 6 x = 16.
Following the steps,
x 2 – 6 x = 16 becomes x 2 – 6 x – 16 = 0
Factor.
( x – 8)( x + 2) = 0
Setting each factor to zero,
Then to check,
Both values, 8 and –2, are solutions to the original equation.
A quadratic with a term missing is called an incomplete quadratic (as long as the ax 2 term isn't
missing).
Example:
2. Solve x 2 – 16 = 0.
Factor.
To check, x 2 – 16 = 0
Both values, 4 and –4, are solutions to the original equation.
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2.2. The Quadratic Formula
Many quadratic equations cannot be solved by factoring. This is generally true when the roots, or
answers, are not rational numbers. A second method of solving quadratic equations involves the
use of the following formula:
a, b, and c are taken from the quadratic equation written in its general form of ax 2 + bx + c = 0
where a is the numeral that goes in front of x 2, b is the numeral that goes in front of x, and c is
the numeral with no variable next to it (a.k.a., “the constant”).
When using the quadratic formula, you should be aware of three possibilities. These three
possibilities are distinguished by a part of the formula called the discriminant. The discriminant is
the value under the radical sign, b 2 – 4 ac. A quadratic equation with real numbers as coefficients
can have the following:
1. Two different real roots if the discriminant b 2 – 4 ac is a positive number.
2. One real root if the discriminant b 2 – 4 ac is equal to 0.
3. No real root if the discriminant b 2 – 4 ac is a negative number.
Example:
3. Solve for x: x 2 – 5 x = –6.
Setting all terms equal to 0,
x2–5x+6=0
Then substitute 1 (which is understood to be in front of the x 2), –5, and 6 for a, b,
and c, respectively, in the quadratic formula and simplify.
Because the discriminant b 2 – 4 ac is
Example produces rational roots.
positive,
you
get
two
different
real
roots.
Example:
4. Solve for y: y 2 = –2y + 2.
Setting all terms equal to 0,
y2+2y–2=0
Then substitute 1, 2, and –2 for a, b, and c, respectively, in the quadratic formula and simplify.
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Note that the two roots are irrational.
Example:
5. Solve for x: x 2 + 2 x + 1 = 0.
Substituting in the quadratic formula,
Since the discriminant b 2 – 4 ac is 0, the equation has one root.
The quadratic formula can also be used to solve quadratic equations whose roots are imaginary
numbers, that is, they have no solution in the real number system.
Example:
6. Solve for x: x( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0.
Substituting in the quadratic formula,
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Since the discriminant b 2 – 4 ac is negative, this equation has no solution in the real number
system.
But if you were to express the solution using imaginary numbers, the solutions would
be
.
2.3. Completing the Square
A third method of solving quadratic equations that works with both real and imaginary roots is
called completing the square.
1. Put the equation into the form ax 2 + bx = – c.
2. Make sure that a = 1 (if a ≠ 1, multiply through the equation by
3. Using the value of b from this new equation, add
form a perfect square on the left side of the equation.
4. Find the square root of both sides of the equation.
5. Solve the resulting equation.
before proceeding).
to both sides of the equation to
Example:
7. Solve for x: x 2 – 6 x + 5 = 0.
Arrange in the form of
Because a = 1, add
, or 9, to both sides to complete the square.
Take the square root of both sides.
x – 3 = ±2
Solve.
Example:
8. Solve for y: y 2+ 2 y – 4 = 0.
Arrange in the form of
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Because a = 1, add
, or 1, to both sides to complete the square.
Take the square root of both sides.
Solve.
Source: https://www.cliffsnotes.com/study-guides/algebra/algebra-i/quadratic-equations
3. Higher Degree Polynomials
A polynomial of degree n has at least one root, real or complex.
A polynomial P(x) of degree n has exactly n roots, real or complex.
If the leading coefficient of P(x) is 1, then the Factor Theorem allows us to conclude:
P(x) = (x − rn)(x − rn − 1). . . (x − r2)(x − r1)
Hence a polynomial of the third degree, for example, will have three roots.
Example:
Find the roots of the polynomial x3 − 2x2 − x + 2 = 0 using the Factor Theorem.
x3 − 2x2 − x + 2 = 0
(x − 2x2) – ( x – 2 ) = 0
Group the common terms.
2
x (x–2) –(x–2)=0
Factor out the common terms.
( x – 2 ) ( x2 – 1 ) = 0
(x − 2) (x + 1)(x − 1) = 0
The factors are (x − 2) (x + 1)(x − 1). Therefore the roots are −1, 1, 2.
3
Using Synthetic Division
Find the three roots of P(x) = x3 − 2x2 − 9x + 18, given that one root is 3.
Solution. Since 3 is a root of P(x), then according to the factor theorem, x − 3 is a
factor. Therefore, on dividing P(x) by x − 3, we can find the other, quadratic factor.
We have
x3 − 2x2 − 9x + 18 = (x2 + x − 6)(x − 3)
= (x − 2)(x + 3)(x − 3)
The three roots are: 2, −3, 3.
Again, since x − 3 is a factor of P(x), the remainder is 0.
Integer Root Theorem
If an integer is a root of a polynomial whose coefficients are integers and whose leading
coefficient is ±1, then that integer is a factor of the constant term.
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This Integer Root Theorem is an instance of the more general Rational Root Theorem:
If the rational number r/s is a root of a polynomial whose coefficients are integers, then the
integer r is a factor of the constant term, and the integer s is a factor of the leading coefficient.
Example:
What are the possible integer roots of x3 − 4x2 + 2x + 4?
Answer. If there are integer roots, they will be factors of the constant term 4;
namely: ±1, ±2, ±4.
By trial and error using the roots until the remainder becomes zero:
Now, is 1 a root? To answer, we will divide the polynomial by x − 1, and hope for remainder 0.
1
−4
+2
+4
|1
+1
−3
−1
------------------------------------------------------------------------------------------------------1
−3
−1
+3
The remainder is not 0. 1 is not a root. Let's try −1:
1
−4
+2
+4
|−1
−1
+5
−7
-----------------------------------------------------------------------------------------------------1
−5
+7
−3
The remainder again is not 0. Let's try 2:
1
−4
+2
+4
|2
+2
−4
−4
------------------------------------------------------------------------------------------------------1
−2
−2
+0
Yes! 2 is a root. We have
x3 − 4x2 + 2x + 4 = (x2 − 2x − 2)(x − 2)
We can now find the roots of the quadratic by completing the square.
x=1±
Therefore, the three roots are:
1+
, 1−
, 2.
Source: https://themathpage.com/aPreCalc/factor-theorem.htm
4. Systems of Equations
A system of equations is a set of equations that involve the same variables. A system of
linear equations is a system of equations in which each equation is linear. A solution of a
system is an assignment of values for the variables that makes each equation in the system
true. To solve a system means to find all the solutions of the system.
Methods of Solving Systems of Equations
4.1. Substitution Method
4.1.1. Solve for one variable. Choose one equation and solve for one variable in terms of
the other variable.
4..1.2. Substitute. Substitute the expression you found in step 1 into the other equation to
get an equation in one variable, then solve for that variable.
4.1.3. Back-Substitute. Substitute the value you found in step 2 back into the expression
found in step 1 to solve for the remaining variable.
Example:
Find all solutions of the system. 2x + y = 1
3x + 4y = 14
(1)
(2)
4.1.1.
y = 1 – 2x
(3)
from (1)
4.1.2. 3x + 4(1 – 2x) = 14
substitute (3) into (2)
3x + 4 – 8x
= 14
expand
- 5x + 4 = 14
simplify
x=-2
solve for x
4.1.3.
y=1–2(-2)
y=5
The solution is the ordered pair ( - 2, 5).
4.2. Elimination Method
4.2.1. Adjust the coefficients. Multiply one or more of the equations by appropriate
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numbers so that the coefficient of one variable in one equation is the negative of its
coefficient in the other equation.
4.2.2. Add the equations. Add the two equations to eliminate one variable, then solve for
the remaining variable.
4.2.3. Back-substitute. Substitute the value that you found in step 2 back into one of the
original equations, and solve for the remaining variable.
Example:
Find all solutions of the system:
3x + 2y = 14
x – 2y = 2
(1)
(2)
Since the coefficients of the y-terms are negatives of each other, we can add the equations to
eliminate y.
4.2.2.
3x + 2y = 14
x – 2y = 2
4x
= 16
x= 4
4.2.3.
4 – 2y = 2
y =1
The solution is the ordered pair ( 4, 1).
(1)
(2)
add
solve for x
substitute x to (2) or (1)
solve for y
5. Applications of Linear and Quadratic Equations and Systems of Equations
The following guide steps will help you solve worded problems. There are no hard and fast
rules but you must be critical and analytical when solving worded problems. When you cannot
get the problem after reading it for the first time, please read it again until such time you
understand what the problem is stating.
Principles of Problem Solving
1. Understand the problem. Read the problem and make sure that you understand it clearly.
Be guided by the following questions: What is the problem?
What are the given quantities?
What are the given conditions?
Illustrate with a diagram and identify the given and required quantities on the diagram.
2.Think of a plan. Find the connection between the given information and the unknown that
will enable you to calculate the unknown.
2.1. Try to recognize something familiar by relating it to a previous knowledge.
2.2. Try to recognize patterns by seeing repetitions in the problem.
2.3. Use analogy by relating it to a simpler problem compared to the original problem.
2.4. Introduce something extra that can help in making the connection between the given and
the unknown.
2.5 Take cases by splitting the problem into several cases and give a different argument for
each of the cases.
2.6. Work backward, step by step, until you arrive at the given data. Working in reverse can
aid in constructing a solution to the original problem.
3. Carry out the problem. In carrying out the plan you have to check each stage of the plan
and write details that prove that each stage is correct.
4. Look back. This is to check for errors in the solution and partly to see if you can think of an
easier way to solve the problem. Another reason for looking back is to be familiar with the
method of solution which may be useful for solving future problems.
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Examples:
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6. Partial Fraction Decomposition
The partial fraction decomposition or partial fraction expansion of a rational fraction is an
operation that consists of expressing the fraction as a sum of a polynomial and one or several
fractions with a simpler denominator.
𝑃(𝑥)
𝐻(𝑥) =
𝑄(𝑥)
A partial fraction is called a proper fraction if the degree of P(x) is less than the degree of
Q(x). Otherwise, if the degree of P(x) is greater than the degree of Q(x), then it is called an
improper fraction. Partial fraction decomposition is important especially in integral calculus
where many integrals involving rational expressions can be done if we first do partial
fractions on the integrand.
Steps in Partial Fraction Decomposition
1.
2.
3.
4.
5.
6.
Start with a Proper Rational Expressions (if not, do division first)
Factor the bottom into:
a. linear factors
b. or "irreducible" quadratic factors
Write out a partial fraction for each factor (and every exponent of each)
Multiply the whole equation by the bottom
Solve for the coefficients by
a. substituting zeros of the bottom
b. making a system of linear equations (of each power) and solving
Substitute the values of the coefficients back into the expression.
The following are different cases for which partial fraction decomposition can be used.
Case 1. Factors of Q(x) are all linear and none is repeated.
Example.
𝟓𝒙 − 𝟔
𝑨
𝑩
=
+
𝒙(𝒙 + 𝟑)
𝒙
𝒙+𝟑
𝟓𝒙 − 𝟔 = 𝑨 (𝒙 + 𝟑) + 𝑩𝒙
𝟓𝒙 − 𝟔 = 𝑨𝒙 + 𝟑𝑨 + 𝑩𝒙
𝟓𝒙 − 𝟔 = 𝒙 (𝑨 + 𝑩) + 𝟑𝑨
𝒙𝟎 :
(𝟏)
−𝟔 = 𝟑𝑨
(𝟐) since there is only one variable, solve for
−𝟔
𝑨=
the value of A
𝟑
𝑨 = −𝟐
Substitute the value of A into equation (1) to solve for the value of B:
𝟓 = −𝟐 + 𝑩
𝑩=𝟕
Therefore,
𝟓𝒙 − 𝟔
−𝟐
𝟕
=
+
𝒙(𝒙 + 𝟑)
𝒙
𝒙+𝟑
𝒙:
𝟓 = 𝑨+𝑩
Case 2. Factors of Q(x) are all linear and some are repeated.
Example.
𝒙𝟐 :
𝟎 = 𝑨+𝑩
𝑨 = −𝑩 (𝟏)
𝟐𝒙 − 𝟕
𝑨
𝑩
𝑪
=
+
+
𝟐
(𝒙 + 𝟏)(𝒙 − 𝟐)
𝒙+𝟏 𝒙−𝟐
(𝒙 − 𝟐)𝟐
𝟐
𝟐𝒙 − 𝟕 = 𝑨(𝒙 − 𝟐) + 𝑩 ( 𝒙 + 𝟏)(𝒙 − 𝟐) + 𝑪(𝒙 + 𝟏)
𝟐𝒙 − 𝟕 = 𝑨 (𝒙𝟐 − 𝟒𝒙 + 𝟒) + 𝑩(𝒙𝟐 − 𝒙 − 𝟐) + 𝑪𝒙 + 𝑪
𝟐 𝒙 − 𝟕 = 𝑨𝒙𝟐 − 𝟒𝑨𝒙 + 𝟒𝑨 + 𝑩𝒙𝟐 − 𝑩𝒙 − 𝟐𝑩 + 𝑪𝒙 + 𝑪
𝟐𝒙 − 𝟕 = 𝒙𝟐 (𝑨 + 𝑩) + 𝒙( −𝟒𝑨 − 𝑩 + 𝑪) + (𝟒𝑨 − 𝟐𝑩 + 𝑪
𝒙:
𝒙𝟎 ∶
𝟐 = −𝟒𝑨 − 𝑩 + 𝑪
− 𝟕 = 𝟒𝑨 − 𝟐𝑩 + 𝑪
𝟐 = −𝟒(−𝑩) − 𝑩 + 𝑪
− 𝟕 = −𝟔𝑩 + 𝑪
𝟐 = 𝟑𝑩 + 𝑪 (𝟐)
𝑪 = 𝟔𝑩 − 𝟕
(𝟑)
substitute equation (3) into (2) and solve for B:
𝟐 = 𝟑𝑩 + 𝟔𝑩 − 𝟕
𝟗𝑩 = 𝟗
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𝑩=𝟏
substitute B into equations (1) & (3):
𝑨 = −𝟏 ; 𝑩 = 𝟏; 𝑪 = −𝟏
𝟐𝒙 − 𝟕
−𝟏
𝟏
−𝟏
=
+
+
𝟐
(𝒙 + 𝟏)(𝒙 − 𝟐)
𝒙+𝟏 𝒙−𝟐
(𝒙 − 𝟐)𝟐
Case 3. All factors of Q(x) are quadratic and repeating.
Example.
𝒙𝟑 :
𝟑=𝑨
𝟑𝒙𝟑 + 𝒙𝟐 − 𝒙 − 𝟓
𝑨𝒙 + 𝑩
𝑪𝒙 + 𝑫
= 𝟐
+ 𝟐
𝟐
𝟐
(𝒙 + 𝟐𝒙 + 𝟓)
𝒙 + 𝟐𝒙 + 𝟓 (𝒙 + 𝟐𝒙 + 𝟓)𝟐
𝟑
𝟐
𝟑𝒙 + 𝒙 − 𝒙 − 𝟓 = (𝑨𝒙 + 𝑩)(𝒙𝟐 + 𝟐𝒙 + 𝟓) + 𝑪𝒙 + 𝑫
= 𝑨𝒙𝟑 + 𝟐𝑨𝒙𝟐 + 𝟓𝑨𝒙 + 𝑩𝒙𝟐 + 𝟐𝑩𝒙 + 𝟓𝑩 + 𝑪𝒙 + 𝑫
= 𝑨𝒙𝟑 + 𝒙𝟐 (𝟐𝑨 + 𝑩) + 𝒙(𝟓𝑨 + 𝟐𝑩 + 𝑪) + 𝟓𝑩 + 𝑫
𝒙𝟐 :
𝒙:
𝒙𝟎 ∶
𝟏 = 𝟐𝑨 + 𝑩
−𝟏 = 𝟓𝑨 + 𝟐𝑩 + 𝑪
− 𝟓 = 𝟓𝑩 + 𝑫
𝟏 = 𝟐(𝟑) + 𝑩
− 𝟏 = 𝟓(𝟑) + 𝟐(−𝟓) + 𝑪
− 𝟓 = 𝟓(−𝟓) + 𝑫
𝑩 = −𝟓
𝑪 = −𝟔
𝑫 = 𝟏𝟓
𝟑𝒙𝟑 + 𝒙𝟐 − 𝒙 − 𝟓
𝟑𝒙 − 𝟓
−𝟔𝒙 + 𝟏𝟓
= 𝟐
+ 𝟐
𝟐
𝟐
(𝒙 + 𝟐𝒙 + 𝟓)
𝒙 + 𝟐𝒙 + 𝟓 (𝒙 + 𝟐𝒙 + 𝟓)𝟐
Case 4. All factors of Q(x) are quadratic and non-repeating.
Example.
𝒙𝟐 + 𝒙 + 𝟑
𝑨𝒙 + 𝑩
𝑪
= 𝟐
+
𝟐
(𝒙 + 𝟕)( 𝒙 − 𝟑)
𝒙 +𝟕
𝒙−𝟑
𝒙𝟐 + 𝒙 + 𝟑 = (𝑨𝒙 + 𝑩)( 𝒙 − 𝟑) + 𝑪(𝒙𝟐 + 𝟕)
= 𝑨𝒙𝟐 + 𝑩𝒙 − 𝟑𝑨𝒙 − 𝟑𝑩 + 𝑪𝒙𝟐 + 𝟕𝑪
= 𝒙𝟐 (𝑨 + 𝑪) + 𝒙 (𝑩 − 𝟑𝑨) − 𝟑𝑩 + 𝟕𝑪
𝒙𝟐 :
𝟏 = 𝑨+𝑪
𝑨 = 𝟏 − 𝑪 (𝟏)
𝒙:
𝟏 = 𝑩 − 𝟑𝑨
𝑩 = 𝟑𝑨 + 𝟏 (𝟐)
𝒙𝟎 ∶
−𝟑 = −𝟑𝑩 + 𝟕𝑪
(𝟑)
Solving by substitution equations (1), (2) & (3):
𝟏
𝟏𝟗
𝟏𝟓
𝑨=
;𝑩 =
;𝑪 =
𝟏𝟔
𝟏𝟔
𝟏𝟔
𝟏
𝟏𝟗
𝟏𝟓
𝒙+
𝒙𝟐 + 𝒙 + 𝟑
𝟏𝟔
𝟏𝟔
=
+ 𝟏𝟔
(𝒙𝟐 + 𝟕)( 𝒙 − 𝟑)
𝒙𝟐 + 𝟕
𝒙−𝟑
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Module 3 | Page 14