This is the Grade 9 Mathematics self-learning module for Quarter 1, Module 8: Equations in Quadratic Form, set for Week 3 under learning code M9AL-Ib-7. It builds on the earlier modules on solving quadratic equations and shows learners how the same methods can solve some rational and higher-degree equations. After working through it, learners should be able to turn these equations into quadratic form and solve them.
On this page 7 sections
What learners will learn
The module states one learning competency. Learners will be able to:
- Solve equations transformable to quadratic equations, including rational algebraic equations (M9AL-Ib-7).
Parts of this module
The module follows the usual Alternative Delivery Mode (ADM) order:
| Part | What it contains |
|---|---|
| What I Need to Know | The learning competency |
| What I Know | A 10-item multiple-choice pretest |
| What’s In (Let’s Recall) | Five quadratic equations to solve by any method |
| What’s New (Higher Degree!) | A comic where students meet x⁴ − 12x² + 11 = 0 and ask how to solve an equation of degree 4 |
| What Is It | The lesson, with four worked examples |
| What’s More | Three equations to transform and solve |
| What I Have Learned | A summary of the key ideas |
| What I Can Do | Five equations to solve |
| Assessment | A 10-item multiple-choice post-test |
| Additional Activities | Finding the Average Speed, a boat-and-current word problem |
| Problem-Based Worksheet | The Water Station, a two-pipe tank problem |
| Answer Key | Answers for the activities, pretest, and assessment |
Key ideas: turning equations into quadratic form
Some equations do not look quadratic, but after a substitution or some simplifying, they can be solved like quadratic equations. The module shows three main ways.
1. Use the laws of exponents (higher-degree equations)
An equation like x⁴ − 7x² + 12 = 0 can be rewritten as (x²)² − 7(x²) + 12 = 0. If you let m = x², it becomes m² − 7m + 12 = 0, which is quadratic.
The module solves x⁴ − 12x² + 11 = 0 this way. Writing it as (x²)² − 12(x²) + 11 = 0 and using the quadratic formula gives x² = 11 or x² = 1. So x = ±√11 and x = ±1. That makes four solutions, which matches the degree of the equation.
2. Replace a repeated expression with a new variable
In (x² − 6x)² − 2(x² − 6x) − 35 = 0, the expression x² − 6x appears twice. Let r = x² − 6x, and the equation becomes r² − 2r − 35 = 0. You solve for r first. Then you put each value of r back into r = x² − 6x and solve for x.
3. Clear the denominators (rational algebraic equations)
For a rational equation, multiply both sides by the least common denominator (LCD). For 2/(y + 1) + 1/(y − 1) = 1, multiplying by (y + 1)(y − 1) gives 2(y − 1) + (y + 1) = y² − 1. This simplifies to y² − 3y = 0.
Watch out for extraneous solutions
In Example 2, the module solves x + 8/(x − 2) = 1 + 4x/(x − 2). Multiplying by the LCD (x − 2) gives x² − 7x + 10 = 0, so (x − 5)(x − 2) = 0. This gives x = 5 or x = 2.
But x = 2 is an extraneous solution. It makes the denominator x − 2 equal to zero, so it is excluded from the domain of the original equation. The module defines an extraneous root as a root of the transformed equation that is not a root of the original equation. The only real solution here is x = 5. Always check your answers against the original equation.
Summary from What I Have Learned
To decide whether an equation can be solved like a quadratic equation:
- Check if it can be written as am² + bm + c = 0, where m is an algebraic expression.
- For complicated equations, replace an expression with a new variable.
- For rational algebraic equations, simplify both sides first, including removing the denominators, so the quadratic form is easy to see.
Word problems in the module
- Finding the Average Speed! Mang Kevin, a vendor, lives 7 km upstream from town. With a current of 4 km/hr, he paddles downstream to town and back in 2 hours. Learners write a mathematical model and find his average paddling speed in still water.
- The Water Station. One pipe takes 10 minutes more than another to fill a tank. Together they fill it in 12 minutes. Using x for the first pipe’s time, the worked answer sets up 1/x + 1/(x + 10) = 1/12, which leads to x² − 14x − 120 = 0. That factors to x = 20 or x = −6. Since a time cannot be negative, the pipes take 20 minutes and 30 minutes.
How to use this module
- Learners: Take the What I Know pretest first and check it. Note the items you missed, and look for the right answers as you go through the lesson. Then work through the parts in order, and write your answers on a separate answer sheet. In the Assessment, if your answer is not among the choices, write E together with your final answer.
- Parents: Help your child follow the parts in order, and make sure the pretest is done before the lesson. The answer key on the last page is for checking work after the activities are finished.
- Teachers: The Let’s Recall items help you review solving quadratic equations before the lesson. The two word problems work well for small groups. The E-Search section points learners to extra practice on equations reducible to quadratic form.
Frequently asked questions
What grade and quarter is this module for?
It is for Grade 9 Mathematics, Quarter 1, Module 8, set for Week 3. Its learning code is M9AL-Ib-7.
What does equations in quadratic form mean?
These are equations that are not quadratic at first, such as x⁴ − 7x² + 12 = 0 or rational equations. After a substitution or clearing the denominators, they can be written as am² + bm + c = 0 and solved with quadratic methods.
What is an extraneous solution?
It is a root of the transformed equation that is not a root of the original equation, because it is excluded from the original equation’s domain. For example, a value that makes a denominator zero must be rejected.
Does the module have an answer key?
Yes. The last page has answers for What’s In, What I Know, What’s More, What I Can Do, the Assessment, the Additional Activities, and the problem-based worksheet.
How many solutions does a fourth-degree equation like x⁴ − 12x² + 11 = 0 have?
It has four solutions. In the module’s worked example, they are x = ±√11 and x = ±1.
Full text of the module
The text below is a transcription of the PDF above, as printed.
Mathematics Quarter 1-Module 8 Equations in Quadratic Form
Week 3
Learning Code – M9AL-Ib-7
[illustration: teacher and students learning online, with books, laptops, and phones]
GRADE 9
Learning Module for Junior High School Mathematics
Quarter 1 – Module 8 – New Normal Math for G9 First Edition 2020 Copyright © 2020
Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties.
Borrowed materials (i.e. songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them.
Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio
Development Team of the Module
Writers: Rowena F. Reyes – TI, Analynn M. Argel – MTII
Editor: Sally C. Caleja – Head Teacher VI, Ma. Theresa G. Mallari- Head Teacher VI, Judy Ann G. Gallo – Teacher I
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
MODULE 8
EQUATIONS IN QUADRATIC FORM
In the previous modules, you have learned about the different methods of solving quadratic equations. In this module, you will find that these ways are also necessary to solve some rational and higher degree equations.
WHAT I NEED TO KNOW
LEARNING COMPETENCY
The learners will be able to:
- solve equations transformable to quadratic equations (including rational algebraic equations). M9AL-Ib-7
WHAT I KNOW
Find out how much you already know about equations transformable to quadratic form. Write the letter that you think is the best answer to each question on your answer sheet. 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.
Which of the following mathematical equations could be transformed to quadratic form?
a. 𝑟 = 0
b. 𝑥 + 8/(𝑥−2) = 1 + 4𝑥/(𝑥−2)
c. 7𝑣 − 8 = 3𝑣 − 3
d. 4𝑥² − 4𝑥 + 11 = 3𝑥² − 6𝑥 + 𝑥²Which of the following is the quadratic transformation of the equation: 1 + (𝑥−1)/(𝑥+1) = (𝑥+1)/(𝑥−1)?
a. 𝑥² + 4𝑥 + 1 = 0
b. 𝑥² − 4𝑥 − 1 = 0
c. 𝑥² − 4𝑥 + 1 = 0
d. 𝑥² + 4𝑥 − 1 = 0Referring to the equation in item #2, which of the following are the solutions?
a. −2 ± √5
b. 2 ± √5
c. −2 ± √3
d. 2 ± √3Which of the following are the solutions of 𝑥(𝑥 + 3) − 9 = 0?
a. (3 ± 3√5)/2
b. (−3 ± 3√5)/2
c. (3 ± 3√3)/2
d. (3 ± 3√3)/2What are the solutions of the equation: 3(𝑥 + 3)² + 5(𝑥 + 3) − 2 = 0?
a. −8/3, 5
b. 8/3, −5
c. 8/3, 5
d. −8/3 − 5Which of the following are the solutions of the 4th degree equation 𝑥⁴ + 13𝑥² + 36 = 0?
a. 2𝑖, −2𝑖, 3𝑖, −3𝑖
b. 2𝑖, 3𝑖, 2, 3
c. −2𝑖, 3𝑖, −2, 3
d. 2𝑖, −3𝑖, 2, −3Which of the following rational algebraic equations is transformable to a quadratic equation?
a. (𝑣+1)/2 − (𝑣+2)/4 = 7
b. 2/𝑟 + 3/(𝑟+1) = 5
c. (2𝑏−1)/3 + 1/2 = 3𝑏/4
d. 3/(𝑧−2) + 4/(𝑧+2) = 7/𝑧What are the solutions of the equation (𝑦² − 5𝑦 + 2)² + 6(𝑦² − 5𝑦 + 2) + 8 = 0?
a. 1, 2, 3, 4
b. −1, −2, −3, −4
c. 3, 4, 5, 6
d. −3, −4, −5, −6What are the solutions of 𝑥⁶ − 9𝑥³ + 8 = 0?
a. 1, 1, 1, 2, 2, 2
b. 1, (−1±𝑖√3)/2, 2, (−1 ± 𝑖√3)
c. 1, (1±𝑖√3)/2, 2, (1 ± 𝑖√3)
d. −1, −1, −1, 2, 2, 2The algebraic rational equation 1 + (𝑥²−5𝑥−24)/(3𝑥) = (𝑥−6)/(3𝑥) is reducible into quadratic form. Which of the following are the solutions of this equation?
a. 3, 6
b. −3, −6
c. −3, 6
d. 3, −6
WHAT’S IN
LET’S RECALL
Using any method of finding the zeroes of quadratic equation, solve for the following:
- 𝑥² − 9 = 0
- 𝑥² − 5𝑥 + 6 = 0
- 6𝑥² + 7𝑥 − 15 = 0
- 2𝑥² + 3𝑥 − 5 = 0
- 3𝑥² + 7𝑥 + 4 = 0
WHAT’S NEW
Communication and collaboration
HIGHER DEGREE!
[illustration: four-panel comic of four students at a table discussing equations]
Panel 1: “−2 and +2” / “−2 and +2” / “WHAT ARE THE SOLUTIONS OF X^2 -4=0”
Panel 2: “7 and 1” / “7 and 1” / “WHAT ABOUT X^2 -8X+7=0?”
Panel 3: “…………” / “………..” / “HMMM. X^4 -12X^2+11=0?”
Panel 4: “IS IT QUADRATIC EQUATION?” / “NO!!!IT HAS A DEGREE OF 4. HOW WILL GET THE THE SOLUTION?WE WERE JUST DONE WITH QUADRATIC EQUATION!!”
1Comic generated using storyboard
The last given equation is 𝑥⁴ − 12𝑥 + 11 = 0. Obviously, it is not a second-degree equation and that is their problem. Since it is not a quadratic equation, how will they find the solution? What are the values of 𝑥?
WHAT IS IT
Communication, Critical Thinking, and Collaboration
Some equations which are obviously not quadratic in form may be solved using the same methods as in a quadratic equation after performing a few substitutions or simplifying the whole equation.
Example 1: Transform the following equations to quadratic. a. x⁴ – 7x² + 12 = 0 b. (x² – 6x)² – 2(x²-6x) – 35 = 0 c. 2/(𝑦+1) + 1/(𝑦−1) = 1
Solution
a. The equation x⁴ – 7x² + 12 = 0 can be transformed to quadratic using the law of exponent. We can see that, x⁴ – 7x² + 12 = 0 ⇒ (x²)² – 7(x²) + 12 = 0
Then, we make a substitution as follows: Let m = x² (x²)² – 7(x²) + 12 = 0 ⇒ m² – 7m + 12 = 0
b. In the given equation (x² – 6x)² – 2(x²-6x) – 35 = 0, we can clearly see that (x² – 6x) is the expression that we need to set into another variable. Thus, let r = x² – 6x, then: (x² – 6x)² – 2(x²-6x) – 35 = 0 ⇒ r² – 2r – 35 = 0
c. To transform 2/(𝑦+1) + 1/(𝑦−1) = 1 to quadratic equation, multiply both sides of the equation by the least common denominator. (𝒚 + 𝟏)(𝒚 − 𝟏)(2/(𝑦+1) + 1/(𝑦−1)) = (1)(𝒚 + 𝟏)(𝒚 − 𝟏) 2(y – 1) + (y + 1) = (y + 1)(y – 1) 2y – 2 + y + 1 = y² – 1 0 = y² – 3y Thus, 2/(𝑦+1) + 1/(𝑦−1) = 1 ⇒ y² – 3y = 0
Example 2: Given a rational algebraic equation, 𝑥 + 8/(𝑥−2) = 1 + 4𝑥/(𝑥−2), what are the values of 𝑥?
Solution: (𝑥 − 2)(𝑥 + 8/(𝑥−2)) = (1 + 4𝑥/(𝑥−2))(𝑥 − 2) Multiply both side by the LCD 𝑥(𝑥 − 2) + 8 = (𝑥 − 2) + 4𝑥 𝑥² − 2𝑥 + 8 = 𝑥 − 2 + 4𝑥 𝑥² − 2𝑥 − 𝑥 − 4𝑥 + 8 + 2 = 0 𝑥² − 7𝑥 + 10 = 0 This is now in quadratic form
You may use factoring to find the solution. 𝑥² − 7𝑥 + 10 = 0 (𝑥 − 5)(𝑥 − 2) = 0 Hence, 𝑥 = 5 or 𝑥 = 2 But 𝑥 = 2 is an extraneous solution. Why?
What is an extraneous root or solution? It is a root of a transformed equation which is not a root of the original equation since it was excluded from the domain of the original equation. https://www.varsitytutors.com/
Example 3: Solve: 𝑥⁴ − 12𝑥² + 11 = 0
Solution: The given equation can be written in the form of (𝑥²)² − 12(𝑥²) + 11 = 0. Using the quadratic formula, 𝑥² = (12 ± √((−12)² − 4(1)(11)))/(2(1)) = (12 ± √(144 − 44))/2 = (12 ± √100)/2 = (12 ± 10)/2 𝑥² = 11, 1
Therefore, 𝑥 = ±√11, ±1
Example 4: Solve: (𝑥² + 2𝑥)² − 2(𝑥² + 2𝑥) + 3 = 0
Solution: This example will show the method of solving a complicated equation by using a new variable and substituting it for an expression in the given equation.
So, in this problem, let 𝑚 = 𝑥² + 2𝑥 Substituting m to the original equation, we have: 𝑚² − 2𝑚 + 3 = 0 (𝑚 + 1)(𝑚 − 3) = 0 𝑚 + 1 = 0 or 𝑚 − 3 = 0 𝑚 = −1, 𝑚 = 3
Substituting the values of m to 𝑚 = 𝑥² + 2𝑥, we have: 𝑥² + 2𝑥 = 3 or 𝑥² + 2𝑥 = −1 𝑥² + 2𝑥 − 3 = 0 | 𝑥² + 2𝑥 + 1 = 0 (𝑥 − 1)(𝑥 + 3) = 0 | (𝑥 + 1)(𝑥 + 1) = 0 (𝑥 − 1) = 0 or (𝑥 + 3) = 0 | (𝑥 + 1) = 0 or (𝑥 + 1) = 0 𝑥 = 1 or 𝑥 = −3 | These will be both 𝑥 = −1
Since we are looking for 4 solutions, the solutions are: 𝒙 = 𝟏, −𝟑, −𝟏, −𝟏
WHAT’S MORE
Transform the following equations into quadratic form and solve for the unknown variable.
𝑦⁴ − 15𝑦² = 16 Quadratic form:___________ Solutions:_________________
(𝑥 − 4)/(𝑥 − 5) − (2𝑥 − 1)/(𝑥 + 4) = (1 − 2𝑥)/(𝑥² − 𝑥 − 20) Quadratic form:___________ Solutions:_________________
(𝒙² + 𝟏)² + 𝟑(𝒙² + 𝟏) − 𝟏𝟎 = 𝟎 Quadratic form:___________ Solutions:_________________
WHAT I HAVE LEARNED
In order to determine whether a certain equation can be solved using the same method as in quadratic equations, you may do or consider any of the following:
- The equation can be written in the form am² + bm + c = where m is an algebraic expression.
- For complicated equations, you can make it simple by replacing a new variable for an expression in the equation,
- For the solutions of a rational algebraic equation, you must simplify first the expression on both sides (includes the removal of the denominator). This will make you see clearly that the resulting equation is in quadratic form.
WHAT I CAN DO
Solve the following equations:
- 4/𝑥 − 5/(𝑥+3) = 1
- (𝑥²+1)/(𝑥²−3) + (2𝑥²)/(𝑥²−3) = 4
- (𝑟² + 2)² + (𝑟² + 2) − 6 = 0
- 𝑥⁴ − 5𝑥² + 4 = 0
- 𝑥⁴ − 3𝑥² + 2 = 0
Assessment
Write the letter of the correct answer on your answer sheet. If your answer is not among the choices, write E together with your final answer.
Which of the following rational algebraic equations could be transformed to quadratic form?
A. 1/r = 6/5r + 1
B. (h−2)/(h+3) − 1 = 3/(h+2)
C. 1/6d² = 1/2d + 7/6d²
D. (q+5)/(q²+q) = 1/(q²+q) − (q−6)/(q+1)Which of the following is the quadratic transformation of the equation: (2x−1)/(x+1) = (x+1)/(x−1)?
A. x² − 5x = 0
B. x² + 5x = 0
C. x² − 5x + 1 = 0
D. x² + 5x + 1 = 0Referring to the equation in item #2, which of the following are the solutions?
A. 0, −5
B. 0, 5
C. (3±√6)/3
D. (−3±√6)/3The rational equation 1 + (x²−5x−24)/3x = (x−6)/3x is reducible into quadratic form. Which of the following are the solutions of this equation?
A. 3, 6
B. −3, −6
C. −3, 6
D. 3, −6Which of the following are the solutions of b(b − 5) − 9 = 0?
A. (5±√61)/2
B. (−5±√61)/2
C. 5, 9
D. −5, −9What are the solutions of the equation: (x + 3)² + 2(x + 3) − 8 = 0
A. −7, −1
B. 7, 1
C. 3, 8
D. −3, −8Which of the following are the solutions of the 4th degree equation x⁴ − 8x² + 12 = 0?
A. ±6, ±2
B. ±√6, ±2
C. ±6, ±√2
D. ±√6, ±√2What are the solutions of x⁴ − 10x² + 9 = 0?
A. 1, 3, 1, 3
B. −1, −3, 1, 3
C. 1, −3, 1, −3
D. 1, 3, −1, −3Which of the following rational equations is transformable to a quadratic equation?
A. (v+1)/2 − (v+2)/4 = 7
B. 2/r + 3/(r+1) = 5
C. (2b−1)/3 + 1/2 = 3b/4
D. 3/(z−2) + 4/(z+2) = 7/zWhat are the solutions of the equation (y² − 8y + 8)² − 6(y² − 8y + 8) + 8 = 0?
A. 4 ± 2√3, 4 ± √10
B. −4 ± 2√3, 4 ± √10
C. −4 ± 2√3, −4 ± √10
D. −4 ± 2√3, 4 ± √10
In this lesson, we have found that some equations can be made look like quadratic equations especially those with higher degree and rational algebraic equations. Consequently, this can be solved using the same techniques as the quadratic equations. Think and reflect! What are the other concepts related to these techniques that can be used particularly in real life situations?
E-Search
You may also check the following links for your reference and further learnings on solving equations in quadratic form:
- https://www.khanacademy.org/math/in-in-grade-10ncert/x573d8ce20721c073:quadratic-equations-advanced/x573d8ce20721c073:solving-quadratic-equations-advanced/e/equations-reducible-to-quadratic-equations
- https://www.youtube.com/results?search_query=equations+reducible+to+quadratic+form+worksheet
Additional Activities
Communication, Critical Thinking, Creativity
Finding the Average Speed!
Read and study the given scenario. Then answer the questions that follow.
Mang Kevin is a vendor who lives 7km upstream from the town. When the current is 4 km/hr, he can paddle his boat downstream to reach the town and come back in 2 hours.
- Make a mathematical model representing the scenario.
- What is the average paddling speed in still water?
[illustration: a man paddling a boat on water, with a house and trees in the background]
References
Dugopolski, Mark. 2006. Elementary and Intermediate Algebra 2nd edition. McGraw-Hill. New York City
Morgan, F. M., & Paige, B. L. (n.d.). Algebra 2. pp 98-102. America: Henry Holt and Company.
Oronce, O. A., Santos, G. C., & Ona, M. I. (n.d.). Interactive Mathematics III (Concepts, Structures, adn Metods for High School. pp 250-256 Manila: Rex Book Store.
http://tutorial.math.lamar.edu/Classes/Alg/ReducibleToQuadratic.aspx
Problem – Based Worksheet
The Water Station
In a water refilling station, the time that a pipe takes to fill a tank is 10 minutes more than the time that another pipe takes to fill the same tank. If the two pipes are opened at the same time, they can fill the tank in 12 minutes.
[illustration: a water storage tank on a metal stand]
- If x represents the time for the first pipe to fill the tank, how will you represent the time for the second pipe to fill the same tank?
- What equation describes the situation?
- How many minutes does each pipe take to fill the tank?
Answer Key
[transcriber’s note: this page is printed upside down; it has been transcribed in reading order after rotating it. Some symbols are hard to read and are marked [?] where uncertain.]
What’s In
- ±3
- 3, 2
- (−7 ± √409)/12
- 1, −5/2
- −1, −4/3
What I Can Do
- −6, 2
- ±13
- −2, 1, (−1 ± √−11)/2 [?]
- ±√5, ±√−1
- ±2, ±1
What’s More
- Quadratic Form: (y²)² − 5y² − 16 = 0 [?]
Solutions: ±4, ±i [?] - Quadratic Form: x² − 13x − 22 = 0 [?]
Solutions: 11, 2 - Quadratic Form: Letting y = x² + 1, we have y² + 3y − 10 = 0
Solutions: ±√6i, ±1 [?]
What I Know
- B
- B
- B
- B
- D
- A
- B
- A
- B
- C
Assessment
- D
- A
- B
- C
- A
- A
- D
- B
- B
- A
Additional Activities
- 7/(x−4) + 7/(x+4) = 3 [?]
- (7 + √113)/2 kph ≈ 8.815 kph
PISA-Based Worksheet
Solutions:
- Let x = time for the pipe to fill the one tank, then x + 10 = time for the pipe to fill the second tank.
- 1/x + 1/(x+10) = 1/12
- To solve for the value/s of x:
12x(x + 10)(1/x + 1/(x+10)) = 12x(x + 10)(1/12) [?]
12(x + 10) + 12x = x(x + 10)
12x + 120 + 12x = x² + 10x
0 = x² − 14x − 120
By Factoring:
(x − 20)(x + 6) = 0
x − 20 = 0 or x + 6 = 0
x = 20 or x = −6
Since −6 is not a valid solution, 20 minutes is the answer. Thus, it takes 20 minutes for the pipe to fill one tank and 30 minutes to fill the other tank.
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