1. Quadratic Equations.pptx

What this
module is about
Equations like 3x + 4 = 10 and
5x – 15 = 0 are called linear
equations or first-degree in
one variable. Recall that these
equations can be written in the
form 𝑨𝒙 + 𝑩 = 𝟎 where A and
B are real numbers and 𝑨 ≠ 𝟎.
In this chapter, you will learn
how to solve another type of
equation: quadratic equations.
The speed of a vehicle can be
modeled using a quadratic
function.

Here’s a simple guide for you in going
about the module:
1. Read and follow instruction very carefully.
2. Read the different lessons included in this module.
3. Perform ALL the activities, as these will help you have a
better understanding of the topic
4. Finally, take the summative test at the end of the module.
5. Kindly submit your worksheets only at the end of the quarter.
(For modules only)

In this lesson, you will be able to:
1. Illustrate quadratic equations
2. Solve quadratic equations by: (a) factoring; (b) completing
the square; and (c) using the quadratic formula
3. Characterize the roots of a quadratic equation using the
discriminant
4. Describe the relationship between the coefficients and the
roots of a quadratic equation
5. Solve equations transformable to quadratic equations
(including rational algebraic equations)
6. Solve problems involving quadratic equations and rational
algebraic equations.

Lesson 1.1
The
Standard
Form of
Quadratic
Equations

The quadratic equation of our love story
has only two roots: You and me.
A mathematical Lover

Lesson 1.1 The Standard Form of Quadratic
Equations
A quadratic equation is an equation that can be
written in the form
𝒂𝒙𝟐
+ 𝒃𝒙 + 𝒄 = 𝟎
where a, b, and c are real numbers and 𝑎 ≠ 0.
Quadratic equations of this form are said to be in
standard form. A quadratic equation is also called
a second-degree equation in one variable.

Lesson 1.1 The Standard Form of Quadratic
Equations
Example.
Quadratic Equation Standard Form Values of a, b, and c
𝑥2
− 5𝑥 + 6 = 0
2𝑥2 + 𝑥 = 7
𝑥2
− 5𝑥 + 6 = 0
2𝑥2 + 𝑥 − 7 = 0
𝑎 = 1, 𝑏 = −5, 𝑐 = 6
𝑎 = 2, 𝑏 = 1, 𝑐 = −7

Lesson 1.2
Solutions of
Quadratic
Equations by
Factoring,
Completing the
Square and
Quadratic Formula

Life is a math equation. In order to gain the
most, you have to know how to convert
negatives into positives.
Anonymous

Lesson 1.2 Solutions of Quadratic Equations by Factoring,
Completing the Square and Quadratic Formula
Factoring.
A quadratic equation can be solve
by factoring.
Example.
Solve 𝒙𝟐
+ 𝟑𝒙 = 𝟒 by factoring

Solution.
Write the equation in standard
form. Then factor the left-hand
side of the equation.

Solution.
𝑥2
+ 3𝑥 − 4 = 0
𝑥 + 4 𝑥 − 1 = 0

Solution.
Equate each factor to zero
then solve each equation.

Solution.
𝑥 + 4 = 0 𝑥 − 1 = 0
𝑥 = −4 𝑥 = 1

Therefore, the solution set is
−4, 1

Example.
Solve 𝟐𝒙𝟐
= 𝟏𝟎 + 𝒙 by
factoring.

Solution.
Write the equation in standard
form. Then factor its left-hand
side.

Given Reason Result
2𝑥2
= 10 + 𝑥 Transform into standard
form
𝟐𝒙𝟐
− 𝒙 − 𝟏𝟎 = 𝟎
2𝑥2
− 𝑥 − 10 = 0 Remove and multiply the
leading coefficient (2) to
the constant term (-10)
𝒙𝟐
− 𝒙 − 𝟐𝟎 = 𝟎
𝑥2 − 𝑥 − 20 = 0 Factor (find roots
wherein their sum is -1
and their product is -20 =
-5 and 4)
𝒙 − 𝟓 𝒙 + 𝟒 = 𝟎

𝑥 − 5 𝑥 + 4 = 0 Divide the factors of the
constant term (-5 and 4) by
the leading coefficient (2)
𝒙 −
𝟓
𝟐
𝒙 +
𝟒
𝟐
= 𝟎
𝑥 −
5
2
𝑥 +
4
2
= 0 Simplify 𝟐𝒙 − 𝟓 𝒙 + 𝟐 = 𝟎
2𝑥 − 5 𝑥 + 2 = 0
Equate each
factor to zero
𝟐𝒙 − 𝟓 = 𝟎
𝒙 =
𝟓
𝟐
𝒙 + 𝟐 = 𝟎
𝒙 = −𝟐
Therefore, the solution set is
−𝟐,
𝟓
𝟐

Completing-the-Square Method.
The expression 𝒙𝟐
+ 𝒃𝒙 becomes a perfect square when
𝒃
𝟐
𝟐
is added to it. Thus,
𝒙𝟐 + 𝒃𝒙 +
𝒃
𝟐
𝟐
= 𝒙 +
𝒃
𝟐
𝟐
To solve a quadratic equation using this approach is called
completing the square.

Completing the Square
In an expression of the form
𝒙𝟐
+ 𝒃𝒙 or 𝒙𝟐
− 𝒃𝒙, add the constant
term
𝒃
𝟐
𝟐
to complete the square.

Example.
Complete the square for the
expression 𝒙𝟐
+ 𝟏𝟒𝒙.

Solution.
Given Reason Result
𝑥2
+ 14𝑥 Divide the linear term
(𝑏 = 14) by 2 and then
square the quotient.
𝟏𝟒
𝟐
𝟐
= 𝟒𝟗
𝑥2 + 14𝑥 + 49 The number to be added
to make it a perfect
square trinomial is 49
𝒙𝟐
+ 𝟏𝟒𝒙 + 𝟒𝟗
= 𝒙 + 𝟕 𝟐

Example.
Solve 𝒙𝟐
+ 𝟒𝒙 = 𝟏𝟐 by completing the
square.

Solution.
𝑥2 + 4𝑥 = 12
𝑥2
+ 4𝑥 + 4 = 12 + 4
(Addition Property of Equality)

Given Reason Result
𝑥2
+ 4𝑥 = 12 Complete the
square (left-
hand side)
𝒙𝟐
+ 𝟒𝒙 + 𝟒
𝑥2
+ 4𝑥 + 4 Addition
Property of
Equality
𝒙𝟐
+ 𝟒𝒙 + 𝟒
= 𝟏𝟐 + 𝟒

𝑥2
+ 4𝑥 + 4 = 12 + 4 Factor the left-hand side
of the equation and
simplify
𝒙 + 𝟐 𝟐 = 𝟏𝟔
𝑥 + 2 2 = 16 Square Root Property 𝒙 + 𝟐 = ±𝟒
𝑥 + 2 = ±4 Equate 𝑥 + 2 to 4 and −4.
Then solve each equation.
𝒙 + 𝟐 = 𝟒
𝒙 = 𝟐
𝒙 + 𝟐 = −𝟒
𝒙 = −𝟔
Therefore, the solution set is −𝟔, 𝟐

Quadratic Formula
The solutions to the quadratic equation 𝒂𝒙𝟐
+ 𝒃𝒙 +
𝒄 = 𝟎 are given by the quadratic formula
𝒙 =
−𝒃 + 𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
where 𝒂, 𝒃, and 𝒄 are real numbers and 𝒂 ≠ 𝟎

Example.
Solve 𝑥2
+ 11𝑥 = −28 using the
quadratic formula.

Solution.
Write the given equation in standard
form 𝑥2 + 11𝑥 + 28 = 0. Thus, 𝑎 = 1,
𝑏 = 11 and 𝑐 = 28. Then substitute
these values in the quadratic formula.

𝒙 =
−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
𝒙 =
−𝟏𝟏 ± 𝟏𝟏𝟐 − 𝟒 𝟏 𝟐𝟖
𝟐 𝟏

𝒙 =
−𝟏𝟏 ± 𝟏𝟐𝟏𝟐 − 𝟏𝟏𝟐
𝟐
x =
−𝟏𝟏 ± 𝟗
𝟐
=
−𝟏𝟏 + 𝟑
𝟐
𝑎𝑛𝑑
−𝟏𝟏 − 𝟑
𝟐

Equate 𝑥 to
−𝟏𝟏+𝟑
𝟐
and
−𝟏𝟏−𝟑
𝟐
. Then solve
each resulting equation.
Therefore, the solution set is −𝟕, −𝟒
𝑥 =
−11 + 3
2
=
−8
2
= −4 𝑥 =
−11 − 3
2
=
−14
2
= 7

Lesson 1.3
Relationship
Between the
Coefficients and
the Solutions of
Quadratic
Equations

“Politics is for the present, but an equation is
for eternity.”
Albert Einstein

Lesson 1.3 Relationship Between the Coefficients and
the Solutions of Quadratic Equations
You can determine the sum and the product of
the solutions of a given quadratic equation
without solving it. Consider
𝒂𝒙𝟐
+ 𝒃𝒙 + 𝒄 = 𝟎 and denote its solutions by
𝒓𝟏 and 𝒓𝟐.
Sum of roots: 𝑟1 + 𝑟2 = −
𝑏
𝑎
Product of roots: 𝑟1𝑟2 =
𝑐
𝑎

Example.
Find the sum and product of
the roots of the quadratic
equation 𝟏𝟎𝒙𝟐
+ 𝟓𝒙 − 𝟐 = 𝟎

Solution.
In the equation, 𝑎 = 10, 𝑏 = 5,
𝑐 = −2. Hence, the sum of the roots is
−
𝒃
𝒂
= −
𝟓
𝟏𝟎
= −
𝟏
𝟐
.
The product of the roots is
𝒄
𝒂
=
−𝟐
𝟏𝟎
= −
𝟏
𝟓
.

Lesson 1.4
The
Discriminant
of Quadratic
Equations

“me plus you. Multiply by your smile. Minus the
drama. A fraction of your heart. I’ll solve your
problems. We make the perfect equation.
anonymous

Lesson 1.4 The Discriminant of
Quadratic Equations
Discriminant of Quadratic Equations
In the quadratic formula, 𝒙 =
−𝒃+ 𝒃𝟐−𝟒𝒂𝒄
𝟐𝒂
, the
expressions 𝒃𝟐
− 𝟒𝒂𝒄 found inside the radical sign is
called the discriminant of the quadratic equation
𝒂𝒙𝟐
+ 𝒃𝒙 + 𝒄 = 𝟎. The discriminant gives you an
idea of the kind of solutions a quadratic equation
has, even without solving it.

Quadratic Equations
Nature of the Solutions to a Quadratic Equation
Value of the Discriminant Nature of the Discriminant
Positive and a perfect square Two different rational numbers
Positive but not a perfect square Two different irrational numbers
Zero Two equal rational numbers
Negative Two different imaginary numbers

Quadratic Equations
Example.
Determine the nature of the solutions to each
quadratic equation by computing its discriminant.
1. 𝒙𝟐
− 𝟓𝒙 + 𝟔 = 𝟎 3. 𝒙𝟐
− 𝟒𝒙 + 𝟒 = 𝟎
2. 𝒙𝟐
+ 𝒙 − 𝟓 = 𝟎 4. 𝒙𝟐
+ 𝒙 + 𝟒 = 𝟎

Quadratic Equations
Solution.
1. In 𝑥2 − 5𝑥 + 6 = 0, 𝑎 = 1, 𝑏 = −5, and 𝑐 = 6
𝑏2 − 4𝑎𝑐 = −5 2 − 4 1 6 = 25 − 24 = 1
(1 is a positive number and a perfect square)
There are two different rational solutions: 2 and 3.
2. In 𝑥2 + 𝑥 − 5 = 0, 𝑎 = 1, 𝑏 = 1, and 𝑐 = −5
𝑏2 − 4𝑎𝑐 = 1 2 − 4 1 −5 = 1 + 20 = 21
(21 is a positive number but not a perfect square)
There are two different irrational solutions:
−𝟏+ 𝟐𝟏
𝟐
and
−𝟏− 𝟐𝟏
𝟐

Quadratic Equations
Solution.
3. In 𝑥2 − 4𝑥 + 4 = 0, 𝑎 = 1, 𝑏 = −4, 𝑐 = 4
𝑏2 − 4𝑎𝑐 = −4 2 − 4 1 4 = 16 − 16 = 0
There are two rational solutions both equal to 2.
4. In 𝑥2 + 𝑥 + 4 = 0, 𝑎 = 1, 𝑏 = 1, and 𝑐 = 4
𝑏2 − 4𝑎𝑐 = 1 2 − 4 1 4 = 1 − 16 = −15
(-15 is a negative)
There are two imaginary numbers as solutions:
−𝟏+𝒊 𝟏𝟓
𝟐
, and
−𝟏−𝒊 𝟏𝟓
𝟐

Lesson 1.5
Real-life
Application
of Quadratic
Equations

“The best mathematical equation I have ever seen:
1 cross + 3 nails = 4 given.”
Anonymous

Lesson 1.5 Real-life Application of
Quadratic Equations
Example.
Motion Problem. Two hikers, Michael and
Bongbong, start from the same place at the
same time and walk toward Novaliches, which
is 90km away from their place. Bongbong walks
1kph faster than Michael and reaches
Novaliches 1hr before him. Find the rate of each
hiker.

Quadratic Equations
Solution.
This is a motion problem and therefore,
we are going to use the concept:
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝒓𝒂𝒕𝒆 × 𝒕𝒊𝒎𝒆 ⟶ 𝒕𝒊𝒎𝒆 =
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝒓𝒂𝒕𝒆

Quadratic Equations
Let 𝒙 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑀𝑖𝑐ℎ𝑎𝑒𝑙 𝑖𝑛 𝑘𝑝ℎ
𝒙 + 𝟏 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝐵𝑜𝑛𝑔𝑏𝑜𝑛𝑔 𝑖𝑛 𝑘𝑝ℎ
Representing these on the table, we have:
r t d
Michael 𝑥 90
𝑥
90
Bongbong 𝑥 + 1 90
𝑥 + 1
90

Quadratic Equations
The working equation is:
𝟗𝟎
𝒙
−
𝟗𝟎
𝒙+𝟏
= 𝟏, 𝑳𝑪𝑫 = 𝒙 𝒙 + 𝟏
𝟗𝟎𝒙 𝒙 + 𝟏
𝒙
−
𝟗𝟎𝒙 𝒙 + 𝟏
𝒙 + 𝟏
= 𝟏𝒙 𝒙 + 𝟏
(The difference between their time traveled is
1hr)

Quadratic Equations
Manipulating the equation, we have:
𝟗𝟎𝒙 + 𝟗𝟎 − 𝟗𝟎𝒙 = 𝒙𝟐 + 𝒙
(Multiply both sides by the LCD: 𝒙 𝒙 + 𝟏 )
𝒙𝟐
+ 𝒙 − 𝟗𝟎 = 𝟎
(Quadratic equation in Standard Form)

Quadratic Equations
𝒙 + 𝟏𝟎 𝒙 − 𝟗 = 𝟎
(Factor the equation)
𝒙 = −𝟏𝟎, 𝒙 = 𝟗
(Equate each factors to zero to find the roots.)

Quadratic Equations
We take only the positive value
for the rate. Therefore, the
walking rate of Michael is 9kph
and Bongbong’s rate is 10kph.

Quadratic Equations
Example.
Work Problem. Elaine and Gigi can finish
cleaning the house in 2hrs. If it takes Elaine
working alone 3hrs longer than it takes Gigi
working alone, how many hours will each girl
finish the work alone?

Quadratic Equations
Solution.
This is a work problem and therefore,
we are going to use the concept:
𝒘𝒐𝒓𝒌 = 𝒓𝒂𝒕𝒆 × 𝒕𝒊𝒎𝒆 ⟶ 𝒓𝒂𝒕𝒆 =
𝒘𝒐𝒓𝒌
𝒕𝒊𝒎𝒆

Quadratic Equations
Let 𝑥 = ℎ𝑜𝑢𝑟𝑠 𝑖𝑡 𝑡𝑎𝑘𝑒𝑠 𝐺𝑖𝑔𝑖 𝑡𝑜 𝑑𝑜 𝑡ℎ𝑒 𝑗𝑜𝑏
𝑥 + 3 = ℎ𝑜𝑢𝑟𝑠 𝑖𝑡 𝑡𝑎𝑘𝑒𝑠 𝐸𝑙𝑎𝑖𝑛𝑒 𝑡𝑜 𝑑𝑜 𝑡ℎ𝑒 𝑗𝑜𝑏
Representing these on the table, we have:
Time to finish the job Rate in doing the job
Gigi 𝑥 1
𝑥
Elaine 𝑥 + 3 1
𝑥 + 3
Gigi and Elaine 2 1
2

Quadratic Equations
The working equation is:
1
𝑥
+
1
𝑥 + 3
=
1
2
, 𝐿𝐶𝐷 = 2𝑥 𝑥 + 3
2𝑥 𝑥 + 3
𝑥
+
2𝑥 𝑥 + 3
𝑥 + 3
=
2𝑥 𝑥 + 3
2

Quadratic Equations
Manipulating the equation:
𝟐 𝒙 + 𝟑 + 𝟐𝒙 = 𝒙 𝒙 + 𝟑
𝟐𝒙 + 𝟔 + 𝟐𝒙 = 𝒙𝟐
+ 𝟑𝒙 ⟶ 𝒙𝟐
+ 𝟑𝒙 − 𝟒𝒙 − 𝟔 = 𝟎
(Multiply both sides by the LCD: 𝟐𝒙 𝒙 + 𝟑 )
𝒙𝟐
− 𝒙 − 𝟔 = 𝟎
(Quadratic equation in standard form)

Quadratic Equations
𝒙 − 𝟑 𝒙 + 𝟐 = 𝟎
(Factor the equation)
𝒙 = 𝟑, 𝒙 = −𝟐
(Equate each factors to zero to find the roots.)

Quadratic Equations
We take only the positive value
for the rate. Hence, it will take
Gigi 3hrs and Elaine 6hrs to clean
the house if each girl works
alone.

The Standard Form of Quadratic Equations
A quadratic equation is an equation that can be
written in the form
•𝒂𝒙𝟐
+ 𝒃𝒙 + 𝒄 = 𝟎
where a, b, and c are real numbers and 𝑎 ≠ 0.
Quadratic equations of this form are said to be
in standard form. A quadratic equation is also
called a second-degree equation in one
variable.

Solutions of Quadratic Equations by Factoring,
Completing the Square
In an expression of the form 𝒙𝟐
+ 𝒃𝒙 or 𝒙𝟐
−
𝒃𝒙, add the constant term
𝒃
𝟐
𝟐
to complete the
square.

Solutions of Quadratic Equations by Factoring,
Quadratic Formula
The solutions to the quadratic equation 𝒂𝒙𝟐
+
𝒃𝒙 + 𝒄 = 𝟎 are given by the quadratic formula
𝒙 =
−𝒃 + 𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
Where 𝒂, 𝒃, and 𝒄 are real numbers and 𝒂 ≠ 𝟎.

Relationship Between the Coefficients and
Sum of roots: 𝑟1 + 𝑟2 = −
𝑏
𝑎
Product of roots: 𝑟1𝑟2 =
𝑐
𝑎

Real-life Application of Quadratic
Equations
Motion problem
𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝒓𝒂𝒕𝒆 × 𝒕𝒊𝒎𝒆
Work problem
𝒘𝒐𝒓𝒌 = 𝒓𝒂𝒕𝒆 × 𝒕𝒊𝒎𝒆

What does the Bible say about finding
solutions to your problems?
• The LORD is my strength and my shield; my heart
trusted in him, and I am helped: therefore, my heart
greatly rejoices; and with my song will I praise him.
Psalm 28:7
• Trust in the LORD, and do good; so, shall you dwell
in the land, and truly you shall be fed.
Psalm 37:3

What does the Bible say about finding
solutions to your problems?
• Rest in the LORD, and wait patiently for him: fret
not yourself because of him who prospers in his way,
because of the man who brings wicked devices to
pass.
Psalm 37:7
• Trust in the LORD with all your heart; and lean not
to your own understanding.
Proverbs 3:5

1. Quadratic Equations.pptx

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