Mathematics

Quarter 1-Module 5
Solving Quadratic Equation
Using Quadratic Formula
Week 1
Learning Code - M9AL-Ib-2.3

Mathematics – Grade
Alternative Delivery Mode
 GRADE 9
Learning Module for Junior High School Mathematics

Mathematics – Grade 9
Alternative Delivery Mode
Quarter 1 – Module 5 – New Normal Math for G9
First Edition 2020
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Published by the Department of Education
Secretary: Leonor Magtolis Briones
Undersecretary: Diosdado M. San Antonio

Development Team of the Module

Writers: Analynn M. Argel -MTII Rowena F. Reyes- T1

Editor: Sally C. Caleja - Head Teacher VI

Melody P. Rosales – Head Teacher VI
Annabelle R. Aguilar – Head Teacher VI
Validators: Remylinda T. Soriano, EPS, Math
Angelita Z. Modesto, PSDS
George B. Borromeo, PSDS

Illustrator: Writers
Layout Artist: Writers

Management Team: Malcolm S. Garma, Regional Director

Genia V. Santos, CLMD Chief
Dennis M. Mendoza, Regional EPS in Charge of LRMS and
Regional ADM Coordinator
Maria Magdalena M. Lim, CESO V, Schools Division
Superintendent
Aida H. Rondilla, Chief-CID
Lucky S. Carpio, Division EPS in Charge of LRMS and
Division ADM Coordinator
 GRADE 9
Learning Module for Junior High School Mathematics

MODULE SOLVING QUADRATIC EQUATION USING QUADRATIC

5 FORMULA

If you recall the previous lessons, the methods are just applicable for a specific
quadratic equation. For example, the process of “factoring” is appropriate only if the
quadratic expression is factorable. But, do not worry because there is one process that
is applicable for all forms of quadratic equations that will be tackled in this module.

WHAT I NEED TO KNOW

PPREPREVIER!
LEARNING COMPETENCY
The learners will be able to:
• solve quadratic equations by:(d) using the quadratic formula. M9AL-Ib-2.3

WHAT I KNOW
PPREPREVIER
! Find out how much you already know about the module. Write the letter that
you think is the best answer to each question on a sheet of paper. Answer all items.
After taking and checking this short test, take note of the items that you were not able
to answer correctly and look for the right answer as you go through this module.

1. Which of the following is the most appropriate method to solve for the roots of
quadratic equation 2.3𝑥 2 + 1.5𝑥 – 3.4 = 0?
A. Extracting the Square Root C. Completing the Squares
B. Factoring D. Quadratic Formula

2. The solutions to the quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 are ______________

−𝑏±√𝑏2 −4𝑎𝑐 −𝑏±√𝑏2 +4𝑎𝑐
A. x = C. x =
2𝑎 2𝑎
𝑏±√𝑏2 −4𝑎𝑐 −𝑏±√𝑏2 −4𝑎𝑐
B. x = D. x =
𝑎 𝑎

3. Which is the correct substitution of the values of a, b and c in the quadratic

formula of the equation 2(𝑥 2 − 2𝑥) = 5?
−2±√22 −4(1)(5) −(−4)±√(−4)2 −4(2)(−5)
A. 𝑥 = C. 𝑥 =
2(1) 2(2)
−(−2)±√(−2)2 −4(1)(5) −4±√42 −4(2)(−5)
B. 𝑥 = D. 𝑥 =
2(1) 2(2)
4. Solve − 6𝑥 + 2 = 0 using quadratic formula.
𝑥2
A. 2 ± √5 C. ±√5
B. 3 ± √7 D. ± √7

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5. If the roots of 2𝑥 2 = − 3(𝑥 + 2) is solve using quadratic formula, what is the

first thing to consider?
A. Identify the value of 𝑎, 𝑏, and 𝑐.
B. Rewrite the equation in general form.
C. Substitute the values of 𝑎, 𝑏, and 𝑐 in the quadratic formula
D. Divide both sides by a common factor
6. What are the real roots of the quadratic equation in #5?
3±√57 −3±√57 3±√39
A. B. C. D. None
4 4 4
7. Solve for the real roots of 𝑥 − 9𝑥 = −19 using quadratic formula.
2
9±√7 −9±√5 9±√5
A. B. C. D. None
4 2 2
8. Find the positive real root of the quadratic equation −3𝑥 2 = 8𝑥 – 12 using
quadratic formula.
−4+2√13 −4−2√13 4+2√13 4−2√13
A. B. C. D.
3 3 3 3
9. Which of the following quadratic equations does not have real roots when solved
using quadratic formula?
A. 𝑥 2 – 8𝑥 – 33 = 0 C. 𝑥 2 + 13𝑥 – 30 = 0
B. 𝑥 – 5𝑥 + 3 = 0
2
D. 𝑥 2 − 6𝑥 + 13 = 0
10. Erick throws the ball from the roof onto the ground 9m below. He tosses the
ball with an initial downward velocity of 3m per second. With H is the height of
an object after t seconds, v is the initial velocity, and h is the initial height, the
condition above is given by the equation 𝐻 = −16𝑡 2 + 𝑣𝑡 + ℎ, how long does it
take for the ball to hit the ground?
A. -0.66 sec B. 0.85 sec C. 1.2 sec D. 2.14 se

WHAT’S IN
PPREPREV
IER!
Before answering the questions in “WHAT’S NEW” part, let us have a quick
discussion on how to simplify radicals. Take note that a simplified radical has none of
the factors of the radicand can be written as powers greater than or equal to the index,
there are no fractions under the radical sign and no radicals in the denominator.
Study how the given radical expressions are simplified.

a. 4 + √500 = 4 + √(100)(5)
= 4 + 10√5
−6−√12 −6−√(4)(3)
b. =
4 4
−6−2√3
=
4
−3− √3
=
2
Try This! Simplify the following radicals.

1. √125
2. 27 - √49
3. 15 - √112

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Learning Module for Junior High School Mathematics

9+√108
4.
18
−7+ √72 −4(6)(−5)
5.
12

WHAT’S NEW Communication and Collaboration

Please Throw the Ball!

A certain group of boys play a ball outside
Erick’s house when one of them accidentally kicked it hard
and went up to Erick’s roof. They requested him to get the
ball and throw it back to them. Without hesitation, Erick
helped them and threw the ball from the roof onto the
ground 9.14m below. He tossed the ball with an initial
downward velocity of 3.05m per second.

With H as the height of an object after t seconds,

v as the initial velocity, and h as the initial height, the
condition above is given by the equation 𝐻 = −16𝑡 2 + 𝑣𝑡 + ℎ.

What will be the equation to find how long it takes the ball to hit the ground?
Do you think the methods of solving quadratic equation that you have learned in the
previous modules will help you solve the equation you made?

Communication, Critical
WHAT IS IT Thinking, and Collaboration

The quadratic equation that can be formed from the story model earlier is ----
-16t2 + 3.05t + 9.14 = 0. Do you think the methods discussed previously can be used
to solve for the roots of the obtained equation? Why?

There are quadratic equations that are difficult to solve by extracting the square
roots, factoring or even completing the squares. These quadratic equations need a
formula that will help you in solving for their roots.

By completing the square of the quadratic equation ax2 + bx + c = 0, a formula

can be developed that gives the solutions to any quadratic equation in standard form.
The formula is called Quadratic Formula.

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To have a better understanding of what the quadratic formula is, let us review
first the steps in completing the squares as previously discussed.

1. Place the constant term on the right side of the equation. All the terms with
unknowns are on the left side.
2. The numerical coefficient of x2 should be 1. Divide each term of the equation with
the numerical coefficient of x2 if necessary.
3. To get the constant term needed to complete the square, get the numerical
coefficient of x, divide it by 2 and square it. Add the result to both sides of the
equation.
4. Factor the perfect square trinomial.
5. Extract the square root from both sides. Two values will be obtained for the right
side of the equation.
6. Equate the linear expressions to each of the two values.
7. Solve each of the resulting linear equations.
8. Check your answer by substituting to the original equation.

Here is how to derive the Quadratic Formula, simply follow the steps in completing the
squares.
𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 step 1
2
𝑎𝑥 + 𝑏𝑥 = −𝑐
𝑏 𝑐
𝑥2 + 𝑥 = − step 2
𝑎 𝑎
𝑏 𝑏2 𝑐 𝑏2
𝑥2 + 𝑥 + 2 = + 2 step 3
𝑎 4𝑎 𝑎 4𝑎
𝑏 2 𝑏2 −4𝑎𝑐
(𝑥 + ) = step 4
2𝑎 4𝑎2
𝑏 𝑏2 −4𝑎𝑐
𝑥 + = ±√ step 5
2𝑎 4𝑎2
𝑏 2
√𝑏 −4𝑎𝑐
𝑥 + = ±
2𝑎 √4𝑎2
𝑏 √𝑏2 −4𝑎𝑐
𝑥 + = ± step 6
2𝑎 2𝑎
𝑏 √𝑏2 −4𝑎𝑐
𝑥 = − ±
2𝑎 2𝑎
−𝑏±√𝑏2 −4𝑎𝑐
𝑥 = , 𝑓𝑜𝑟 𝑎 ≠ 0 step 7
2𝑎

−𝒃±√𝒃𝟐 −𝟒𝒂𝒄
QUADRATIC FORMULA: 𝒙 = , 𝒇𝒐𝒓 𝒂 ≠ 𝟎
𝟐𝒂

Example 1: Solve for the real roots of each equation using quadratic formula.
a. 𝑥 2 – 2𝑥 – 19 = 0
b. 2𝑥 2 + 6𝑥 = −3
c. 2𝑥 2 – 3𝑥 + 4 = 0

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Solutions
a. In 𝑥 2 – 2𝑥 – 19 = 0, 𝑎 = 1, 𝑏 = −2, and 𝑐 = −19,
−𝑏±√𝑏2 −4𝑎𝑐
𝑥 = Quadratic formula
2𝑎
−(−2)±√(−2)2 −4(1)(−19)
𝑥 = Substitute the values of a, b and c
2(1)
2 ±√2+76
𝑥 = Simplify
2
2 ±√78
𝑥 = Evaluate the square root, if possible.
2
2 ±√78
Since 78 does not have any perfect square factor, therefore, the roots are { }
2

b. Change 2𝑥 2 + 6𝑥 = −3 to standard form, we have 2𝑥 2 + 6𝑥 + 3 = 0. The value

of a = 2, b = 6 and c = 3.
−𝑏±√𝑏2 −4𝑎𝑐
𝑥 = Quadratic formula
2𝑎
6 ±√(6)2 −4(2)(3)
𝑥 = Substitute the values of a, b and c
2(2)
6 ±√36−24
𝑥 = Simplify
4
6 ±√12
𝑥 = Evaluate the square root, if possible.
4
6 ± √(4)(3)
𝑥 =
4
6 ± 2√3
𝑥 = Simplify
4
3 ± √3
𝑥 =
2
3 ± √3
The roots are { }
2

c. In 2𝑥 2 – 3𝑥 + 4 = 0, 𝑎 = 2, 𝑏 = −3, and 𝑐 = 4
−𝑏±√𝑏2 −4𝑎𝑐
x= Quadratic formula
2𝑎
−(−3)±√(−3)2 −4(2)(4)
x= Substitute the values of a, b and c
2(2)
3±√−23
x= Simplify
4

Notice that the radicand is negative. It indicates that the roots are non-real.

Example 2: Erick throws the ball from the roof onto the ground 9.14m below. He
tosses the ball with an initial downward velocity of 3.05 m per second. With H as the
height of an object after t seconds, v as the initial velocity, and h as the initial height,
the condition above is given by the equation 𝐻 = −16𝑡 2 + 𝑣𝑡 + ℎ, how long does it
take for the ball to hit the ground?

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Learning Module for Junior High School Mathematics

Solution:
This is the story model we had earlier. Given the equation 𝐻 = −16𝑡 2 + 𝑣𝑡 + ℎ,
if the ball hits the ground then the height, H, is zero. Thus, the quadratic equation is
−16𝑡 2 + 3.05𝑡 + 9.14 = 0. Solving for the time it takes the ball to hit the ground,
−𝑏±√𝑏2 −4𝑎𝑐
𝑥 = Quadratic formula
2𝑎
−(3.05)±√(3.05)2 −4(−16)(9.14)
𝑥 = Substitute the values of a, b and c
2(−16)
−3.05 ±√9.3025+584.96
𝑥 = Simplify
2
−3.05 ±√594.2625
𝑥 = Evaluate the square root, if possible.
−32
−3.05 ±24.38
𝑥 =
−32
𝑥 = −0.67 or 𝑥 = 0.86

Since -0.67 second is not possible, thus the ball will heat the ground in 0.86 seconds

WHAT’S MORE Critical Thinking

TEST YOURSELF!
I. Use the quadratic formula to solve each of the following quadratic equations:
a. 𝑥 2 − 9𝑥 = −10
b. 2𝑚2 + 13𝑚 + 20 = 0
c. 2𝑦 2 + 8𝑦 = 9
d. 9𝑡 − 5𝑡 2 = 12
e. 3𝑏 2 + 13𝑏 + 20 = 0
II. Solve for the unknown in the problem using the quadratic formula.
The area of a table cloth is 7 square ft.If the wifth is 3 feet shorter than the
length,then what are the dimensions of the table cloth?

WHAT I HAVE LEARNED

Quadratic Formula

For any quadratic equation of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0, where a, b, and c are real
number s and 𝑎 ≠ 0, the solutions are:
−𝑏 ± √𝑏 2 − 4𝑎𝑐
𝑥 =
WHAT I CAN DO 2𝑎

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Learning Module for Junior High School Mathematics

WHAT I CAN DO Critical Thinking

A. Solve for the real roots of each equation using quadratic formula.
1. 𝑥 2 – 2𝑥 – 4 = 0 6. 𝑥 2 + 3𝑥 – 8 = 0
2. 3𝑥 + 2𝑥 – 21 = 0
2 7. 4𝑥 2 – 4𝑥 + 11 = 0
3. 𝑥 – 6𝑥 + 21 = 0
2 8. −2𝑥 2 + 4𝑥 + 1 = 0
4. 𝑥 = 8𝑥 + 4
2 9. 4𝑥 2 = −7𝑥 + 15
5. 𝑥 2 + 4 = − 12𝑥 10. 3𝑥 2 + 1 = 5𝑥

B. Solve.
A rocket is launched straight up from the side of a cliff 43.89 m high. If the
initial velocity (v) is 34.14 m/sec and the height (H) is given by the formula
𝐻 = −16𝑡 2 + 𝑣𝑡 + ℎ, find the time at which the rocket will hit the ground.

ASSESSMENT

Write the letter of the correct answer on your answer sheet. If your answer is
not found among the choices, write the correct answer.

1. If the quadratic equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 is solve using quadratic formula, the

solutions are:
−𝑏±√𝑏2 −4𝑎𝑐 −𝑏±√𝑏2 +4𝑎𝑐
A. 𝑥 = C. 𝑥 =
2𝑎 2𝑎
𝑏±√𝑏2 −4𝑎𝑐 −𝑏±√𝑏2 −4𝑎𝑐
B. 𝑥 = D. 𝑥 =
𝑎 𝑎
2. Which is the correct substitution of the values of a, b and c in the quadratic
formula of the equation 𝑥 2 = −7𝑥 − 5?
−(−7)±√(−7)2 −4(1)(−5) −7±√72 −4(1)(5)
A. 𝑥 = C. 𝑥 =
2(1) 2(1)
−(−7)±√(−7)2 −4(−1)(−5) −7±√72 −4(1)(−5)
B. 𝑥 = D. 𝑥 =
2(1) 2(1)

3. Solve for the roots of 2𝑥 2 + 6𝑥 = −3 using quadratic formula.

−3 ± √3 3 ± √3 2 ± √3 −2 ± √3
A. B. C. D.
2 2 2 2

4. What is/are the positive real root/s of the quadratic equation 𝑥 2 – 5𝑥 + 3 = 0?

5+ √13 −5− √13 5± √13 −5± √13
A. B. C. D.
2 2 2 2

5. How will you describe the roots of the quadratic equation 𝑥 2 + 3 = 5𝑥 when
solved using quadratic formula?
A. Equal B. Irrational C. Rational D. Non-real
6. Use quadratic formula in solving for the roots of 10𝑥 2 = −7𝑥 + 6.
1 6 1 6 1 6 1 6
A. , − B. − , − C. , D. − ,
2 5 2 5 2 5 2 5

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Learning Module for Junior High School Mathematics

7. Find the roots of 𝑥 2 = 8𝑥 + 4 using quadratic formula.

A. −4 ± √5 B. 4±2√5 C. 4±√5 D. -4±2√5

8. Solve for the positive real roots of 5𝑥 2 + 21𝑥 = −18 using quadratic formula.
A. 3 C. No positive roots
6
B. D. No real roots
5

9. What are the roots of 𝑥 2 + 6𝑥 = 10?

A. −3 ± √19 B. 3 ± √19 C. −3 ± √−4 D.3 ± √−4

10. A rocket is launched straight up from the side of a cliff 40m high. If the initial
velocity (v) is 30 m/sec and the height (H) is given by the formula 𝐻 = −16𝑡 2 +
𝑣𝑡 + ℎ, find the time at which the rocket will hit the ground.
A. 2.78sec B. -0.90sec C. 3.1sec D.-0.6sec

ADDITIONAL ACTIVITIES
Critical Thinking and
Think it up! Creativity
A. There are four ways of solving quadratic equations presented in this module. List
down the advantages and disadvantages of using each method.
Method of solving quadratic Advantages Disadvantages
equation
Extracting the square root
Factoring
Completing the Squares
Quadratic Formula

B. Solve the following quadratic equation using the most appropriate method. Give
your reasoning on choosing that method.
1. 𝑥 2 + 8𝑥 − 4 = 0
2. 𝑥 2 – 54 = 0
3. 6𝑥 2 + 11𝑥 – 255 = 0
4. 𝑥 2 + 7𝑥 + 12 = 0

C. Reflect! There are several options to choose in solving for

the roots of quadratic equation. For a given equation it
may be more effective to use one method instead of
another. In this situation we need be careful and
decisive in choosing the most efficient method.
In real life, have you ever encountered situation
where your decision-making is needed? What options
did you consider? Did you choose the best option?

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Learning Module for Junior High School Mathematics

PISA – BASED WORKSHEET

The Cliff Diver

A cliff diver jumps from about 17m above the

water. His height in meters from the water t seconds
after he jumps is given by the formula h = -4.9t2 + 1.5t
+ 17.

How long will it take for the diver to reach the water?

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Learning Module for Junior High School Mathematics

E-Search

You may also check the following link for your reference and further learnings
on solving quadratic equation using quadratic formula.

• https://www.khanacademy.org/math/algebra/x2f8bb11595b61c86:quadratic-
functions-equations/x2f8bb11595b61c86:quadratic-formula-a1/v/using-the-
quadratic-formula
• https://www.math-only-math.com/word-problems-using-quadratic-
formula.html
https://www.math-only-math.com/methods-of-solving-quadratic-
equations.html

REFERENCES

Dugopolski, Mark.2006.Elementary and Intermediate Algebra 2nd edition.MCGraw-

Hill.New York City
Kaufmann, Jerome, Intermediate Algebra for College Students, PWS-KENT Pubishing
Company, Boston USA
MacKeague, Charles, Intermediate Algebra, Concepts and Graphs. Saunders College
Publishing, USA
Mathematics 9 Learner’s Material, Department of Education
Ogena, Ester, et. al. Our Math Grade 9. Mc Graw Hill, Vibal Group. Inc.
https://www.freepik.com/free-vector/woman-with-long-hair-teaching-
online_7707557.htm
https://www.freepik.com/free-vector/kids-having-online-lessons_7560046.htm
https://www.freepik.com/free-vector/illustration-with-kids-taking-lessons-online-
design_7574030.htm

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