This is Mathematics 9 Quarter 1 Module 9, Solving Problems Involving Quadratic Equations, a Week 4 self-learning module for Grade 9 learners under DepEd’s Alternative Delivery Mode. It builds on the earlier lessons on solving quadratic equations, the nature of the roots, and the relationship between roots and coefficients. After working through it, learners should be able to turn real-life word problems into quadratic equations and solve them.
On this page 7 sections
What learners will learn
The module states one learning competency. Learners will be able to:
- solve problems involving quadratic equations (M9AL-1e-1)
Parts of the module
| Part | What it contains |
|---|---|
| What I Know | A 10-item multiple-choice pretest |
| What’s In | “Math Translator!”: turning word phrases into equations |
| What’s New | “Hope to Heal”: a problem about the size of a room in Quezon City’s HOPE 1 quarantine facility |
| What Is It | Four worked examples |
| What’s More | Two practice problems |
| What I Have Learned | Six steps for solving quadratic word problems |
| What I Can Do | Two application problems on a billboard and a table |
| Assessment | A 10-item multiple-choice post-test |
| Additional Activities | A ladder problem and a task to write your own problems |
| Problem-Based Worksheet | Two work problems: “Painting the House” and “Car Wash” |
Answers for the pretest, activities, worksheet, and assessment are at the end of the module, along with a rubric for checking learner-made word problems.
Steps for solving quadratic word problems
The module gives these steps:
- Read and analyze the problem carefully.
- Identify what is given and what is asked.
- Represent the unknown using algebraic expressions.
- Write the mathematical sentence. This is your “working quadratic equation.”
- Find the roots using any method from the earlier lessons (factoring, the quadratic formula, and so on).
- Check by substituting the roots into the working equation.
The module adds an important note: the work does not end with finding the roots. Because these are real-life problems, you must decide whether both, one, or none of the solutions make sense. For example, a length or a time cannot be negative.
Types of problems in the worked examples
Area problems
A HOPE 1 room has an area of 120 square meters, and its length is 1 less than twice its width. With w as the width, the length is 2w − 1, so (2w − 1)(w) = 120. This becomes 2w² − w − 120 = 0, which factors to (2w + 15)(w − 8) = 0. Since a measurement cannot be negative, w = 8 m and the length is 2(8) − 1 = 15 m.
Perimeter and area together
Mang Juan’s rectangular lot has a perimeter of 90 m and an area of 450 m². From the perimeter, L + W = 45. From the area, LW = 450. Writing L = 45 − W and substituting gives (45 − W)(W) = 450, or W² − 45W + 450 = 0. This factors to (W − 30)(W − 15) = 0, so the lot measures 30 m by 15 m. The example shows how the sum and product of the roots help find both sides.
Cost problems
A pipe costs ₱30.00. If it were 1 meter shorter and each meter cost ₱1 more, the total cost would stay the same. With x as the length, the price per meter is 30/x, so 30/(x − 1) = 1 + 30/x. Clearing fractions leads to x² − x − 30 = 0, or (x − 6)(x + 5) = 0. The negative value is rejected, so the pipe is 6 m long.
Work problems
Jack and Jill together finish a job in 6 days, and Jack takes 5 days less than Jill. With x as Jill’s time, the equation is 1/x + 1/(x − 5) = 1/6, which leads to x² − 17x + 30 = 0, or (x − 2)(x − 15) = 0. The value 2 is rejected because it would give Jack a negative number of days. So Jill alone needs 15 days and Jack needs 10 days.
The worksheet uses the same idea. In “Painting the House,” Philo takes x hours and Chino takes x + 5 hours, and together they take 6 hours. Solving 1/x + 1/(x + 5) = 1/6 gives Philo 10 hours and Chino 15 hours.
How to use this module
- Learners: Start with the pretest in What I Know and write your answers on a separate answer sheet. Note the items you missed and look for the right answers as you go through the module. Study the four examples in What Is It before you try What’s More and What I Can Do, then take the Assessment. For the Assessment, if your answer is not among the choices, write E with your final answer.
- Parents: Ask your child to explain the six steps out loud and to say why a negative answer was rejected in a problem. Check answers only after the work is done.
- Teachers: Use the answer key at the end for the pretest, activities, and post-test. Use the rubric (requirements, content, solution, organization, and mechanics, scored 1 to 4 points each) to grade Additional Activity B, where learners write and solve their own quadratic word problems.
Frequently asked questions
What grade and quarter is this module for?
It is for Grade 9 Mathematics, Quarter 1, Module 9, and it is meant for Week 4. Its learning code is M9AL-1e-1.
What does the module teach?
It teaches learners how to solve real-life word problems using quadratic equations. These include problems about area, perimeter, cost, and work done together.
Does the module have an answer key?
Yes. The last pages give answers for the pretest, What’s In, What’s More, What I Can Do, the Problem-Based Worksheet, the Assessment, and the Additional Activities, plus a rubric for learner-made word problems.
Why are negative roots rejected in these problems?
The module explains that real-life measurements such as length, width, and time cannot be negative. After solving, you must decide whether both, one, or none of the roots make sense for the situation.
What are the steps for solving a quadratic word problem?
Read and analyze the problem, identify what is given and what is asked, represent the unknown with expressions, and write the working quadratic equation. Then solve it with any method you have learned and check the roots by substituting them back.
Full text of the module
The text below is a transcription of the PDF above, as printed.
[illustration: grade level badge showing 9]
Mathematics Quarter 1-Module 9 Solving Problems Involving Quadratic Equations
Week 4 Learning Code – M9AL-1e-1
[illustration: people learning online with a computer, books and a tablet]
Quarter 1 – Module 9 – 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: Junalisa A. Bartolome- TI Rowena F. Reyes- TI Judy Ann G. Gallo – TI Marvin G. Sollera – MTII
Editor: Sally C. Caleja – Head Teacher VI Rene V. Salgado – Head Teacher VI Corazon T. Misa – 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
WHAT I NEED TO KNOW
MODULE 9 SOLVING PROBLEMS INVOLVING QUADRATIC EQUATIONS
In the previous lessons, you learned about the different ways in solving quadratic equations, the concepts of the “nature of its roots” and the relationship of its roots and coefficients. Your mastery of the lessons is an important tool to solve many real-world problems related to quadratic equations. In this module, you will learn how to solve word problems that involves quadratic equations.
LEARNING COMPETENCY
The learners will be able to:
- solve problems involving quadratic equations. M9AL-1e-1
WHAT I KNOW
Find out how much you already know about quadratic equation. 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.
The length of the garden is 2 m more than twice its width and its area is 24 m². Which of the following equations represents the given situation?
a. x² + x = 12
b. x² + 2x = 12
c. x² + x = 24
d. x² + 2x = 24A piece of iron rod costs ₱ 60.00. If the rod was 2 meter shorter and each meter costs ₱ 1 more, the cost would remain unchanged. What is the length of the rod?
a. 12
b. 10
c. -12
d. -10The hypotenuse of a right-angled triangle is 20 cm. The difference between its other two sides is 4 cm. Find the length of the sides.
a. 5 cm and 10 cm
b. 6 cm and 8 cm
c. 12 cm and 16cm
d. 8 cm and 12 cmThe length of a tarpaulin is 3 feet more than thrice its width and its area is 126 ft². What is the length of the tarpaulin?
a. 18 ft
b. 16 ft
c. 19 ft
d. 21 ftThe sides of an equilateral triangle are shortened by 12 units, 13 units and 14 units respectively and a right-angle triangle is formed. Find the length of each side of the equilateral triangle.
a. 13 units
b. 17 unite
c. 5 units
d. 2 unitsDifference between a number and its positive square root is 12. Find the number.
a. 8
b. 9
c. 22
d. 16
For Question 7 & 8: Jonna throw an object upward from the top of a 1200 ft. tall building. The height of the object, (measured in feet) t seconds after she threw it is h = -16t² + 160t + 1200.
Where is the object 3 seconds after Jonna threw it?
a. 1356 ft.
b. 1536 ft.
c. 1563 ft.
d. 1653 ft.How long does it take for the object to hit the ground?
a. 5 seconds
b. 12 seconds
c. 10 seconds
d. 15 secondsSmith and Johnson together can do a piece of work in 4 days. If they had to work separately, the time taken by Johnson to do the work would be more than that of Smith by 6 days. In how many days can Smith alone do the work?
a. 6 days
b. 10 days
c. 8 days
d. 7 daysAn office has two envelope stuffing machines. Working together they can stuff a batch of envelopes in 4 hours. Working Separately, it will take the second machine 6 hour longer than the first machine. How long would it take for the second machine to stuff a batch of envelopes?
a. 11
b. 20
c. 12
d. 10
WHAT’S IN
Creativity, collaboration and critical thinking
ACTIVITY: MATH TRANSLATOR!
Let us translate word phrases into mathematical equations using the given variable. The first item is done for you.
- The length of a rectangular tarpaulin is 3ft. more than thrice its width and the area is 126 ft². Use w to represent the width.
Solution: The product of the length and width of a rectangular tarpaulin represents its area. Hence, length (l) times width (w) = 126, or lw = 126 and since the length is 3ft more than its thrice its width, then l = 3w + 3. Therefore, the equation is (3w + 3)(w) = 126.
Try this!
The area of a garden is 160 m². The length of the garden is 3 m more than twice its width. Use w to represent the width.
The base of a triangle is 4 dm longer than the altitude and its area is 16 dm². Use h to represent the altitude.
The area of a concrete rectangular pathway is 350 m² and its perimeter is 90 m. Use w to represent the width.
Were you able to represent each situation by an equation?
WHAT’S NEW
Communication, creativity and critical thinking
H O P E T O H E A L
In response to the rising number of COVID-19 positive cases in the country, the local government of Quezon City has set-up HOPE 1, an alternative medical facility. It has been identified to complement local government hospitals as they prepare for the possible overloading of their facilities in the coming days.
HOPE 1 will accept referrals from the Quezon City General Hospital, Rosario Maclang Bautista General Hospital and Novaliches District Hospital. The facility will be led by a team of city health professionals. Each room will house several patients and has dividers to isolate one patient from the others.
If the area of a quarantine facility for COVID-19 patients is 120 square meters and the length is 1 less than twice the width. Find the length and the width of the quarantine facility.
[photo: hallway of a quarantine facility with white curtain dividers between beds]
Guide Questions:
- What expression represents the width of the room? How about the expression that represents its length?
- What equation can be formed that would relate the width, length, and the area of the room?
- How will you describe the equation you formulated?
WHAT IS IT
Communication and critical thinking
WHAT ARE MY DIMENSIONS?
Example 1: The area of a room of HOPE1 for a COVID-19 patient is 120 square meters and the length is 1 less than twice the width. Find the length and the width of the room.
[illustration: rectangle with the vertical side labeled Width and the horizontal side labeled Length]
Solution: The product of the length and width represents the area of the room or (l)(w) = 120.
Let w = width l = 2w – 1
Equation: lw = A
(l)(w) = 120 Equation (2w − 1)(w) = 120 Substitution 2w² − w = 120 Using distributive property and simplifying 2w² − w − 120 = 0 Adding the inverse of 120 on both sides (2w + 15)(w − 8) = 0 Factoring and principle of zero product property w = 8, w = −15/2 Solve for w. (Use the positive value of w since there is no negative value for measurements.)
Thus, w = 8
Solving for l, l = (2w − 1) = 2(8) − 1 = 15 meters
Example 2: Mang Juan owns a rectangular lot. The perimeter of the lot is 90 m and its area is 450 m².
QUESTIONS:
a. What equation represents the perimeter of the lot? How about the equation that represents its area? b. Using the idea of the sum and product of the roots of a quadratic equation, how would you determine the length and the width of the rectangular lot?
Solutions:
Perimeter a. 2L + 2W = 90 m 2(L+W)/2 = 90 m/2 L + W = 45 m
Area (L)(W) = 450 m²
b. L + W = 45 EQUATION 1 LW = 450 EQUATION 2
Get L in terms of W: L + W = 450 → L = 450 − W Substitute to EQUATION 1: LW = 450 (45 − W)(W) = 450 → W² − 45W + 450 = 0 Factor: (W − 30)(W − 15) = 0 Thus, W = 30; W = 15
Example 3: A piece of pipe costs ₱ 30.00. If the pipe was 1 meter shorter and each meter costs ₱ 1 more, the cost would remain unchanged. What is the length of the pipe?
Equation: Let ‘x’ be the length of the piece of pipe in meters
Since the cost of the piece of pipe is ₱ 30.00, then the price per meter of the pipe is 30/x.
Based on the problem we will have the following equation,
30/(x − 1) = 1 + 30/x
Solution: 30/(x−1) = 1 + 30/x Equation 30/(x−1) − 30/x = 1 Using Addition Property of Equality (30x − 30(x−1))/(x(x−1)) = 1 Find the LCD (30x − 30x + 30)/(x² − 1) = 1 Using distributive property 30/(x² − 1) = 1 Simplifying 30 = x² − 1 Using Multiplicative Property of Equality x² − 1 − 30 = 0 (x − 6)(x + 5) = 0 Factoring and principle of zero product property x = 6 or x = −5
Because length cannot be a negative number, we can ignore “- 5”. So, the length of the given pipe is 6 m.
Example 4: Jack and Jill working together can do a work in 6 days. If Jack takes 5 days less than Jill to finish the work, in how many days can Jill alone do it?
Equation: Let x be the time of Jill take to finish the work x – 5 will be the time Jack take to finish the work.
1/x + 1/(x − 5) = 1/6
Solution: (x+x−5)/(x(x−5)) = 1/6 Find the LCD (2x−5)/(x²−5x) = 1/6 Using distributive property and simplify 12x − 30 = x² − 5x Using Multiplicative Property of Equality x² − 17x + 30 = 0 x² − 15x − 2x + 30 = 0 x(x − 15) − 2(x − 15) = 0 Factoring and principle of zero product property (x − 2)(x − 15) = 0 x = 2 or x = 15
We can ignore “2” because it will make the number of days negative, Since x = 15, Jill can work in 15 days, while Jack can work in 10 days.
WHAT’S MORE
Critical Thinking
Solve the following problems:
- The square of a number plus the twice number is equal to 1. Find the number.
- A school has a quadrangle that is 20 ft by 30 ft. The alumni association wants to plant a strip set of sunflowers around the edge of the quadrangle. They have enough seed to cover 336 feet. How wide will the strip be?
[illustration: rectangle with a strip around its edge; “x” marked at the top-left and bottom-right corners of the strip]
WHAT I HAVE LEARNED
In solving problems involving the quadratic equation, you may use the following procedures:
- Read and analyze the problem carefully.
- Identify the given and what is asked in the problem
- Represent the unknown using algebraic expressions.
- Formulate the mathematical sentence. This will serve as the “working quadratic equation”.
- Find the solution set or roots of the quadratic equation using any method discussed on the previous lessons.
- To check, substitute the obtained roots to the working quadratic equation.
NOTE: Solving problems involving quadratic equations does not end in finding and checking the roots. Since, it is a real-life problem you must decide whether both or one or none of the solutions is reasonable.
WHAT I CAN DO
Critical Thinking
Answer the following.
- An amusement park wants to place a new rectangular billboard to inform visitors of their new attractions. Suppose the length of the billboard to be placed is 4 m longer than its width and the area is 96𝑚². What will be the length and the width of the billboard?
- A rectangular table has an area of 27𝑓𝑡² and a perimeter of 24𝑓𝑡. What are the dimensions of the table?
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.
The length of a rectangle is 2 m less than twice the width. If the area of the rectangle is 112 𝑚², what is the width?
a. 5 m
b. 8 m
c. 9 m
d. 14 mA piece of pipe costs ₱ 20.00. If the pipe was 1 meter shorter and each meter costs ₱ 1 more, the cost would remain unchanged. What is the length of the pipe?
a. -4
b. -5
c. 5
d. 4The hypotenuse of right-angled triangle is 10 cm. The difference between its other two sides is 2 cm. Find the length of the sides.
a. 5 cm and 7 cm
b. 6 cm and 8 cm
c. 6 cm and 12 cm
d. 8 cm and 10 cmThe perimeter and area of a rectangle are 48 cm and 140 𝑐𝑚² respectively. Find the length and width of the rectangle.
a. 12 cm by 10 cm
b. 10 cm by 8 cm
c. 12 cm by 8 cm
d. 14 cm by 10 cmThe sides of an equilateral triangle are shortened by 2 units, 3 units and 4 units respectively and a right-angle triangle is formed. Find the length of each side of the equilateral triangle.
a. 3 units
b. 7 unite
c. 5 units
d. 12 unitsDifference between a number and its positive square root is 6. Find the number.
a. 8
b. 4
c. 9
d. 16
For Question 7 & 8: Jhondy throw an object upward from the top of a 300 ft. tall building. The height of the object, (measured in feet) t seconds after he threw it is h = -4t² + 40t + 300.
- Where is the object 5 seconds after Jhondy threw it?
a. 500 ft.
b. 160 ft.
c. 1000 ft.
d. 530 ft. - How long does it take for the object to hit the ground?
a. 5 seconds
b. 12 seconds
c. 10 seconds
d. 15 seconds - John, Luke and Mark can finish painting the fence in 2 hours. If John does the job alone, he can finish it in 5 hours. If Luke does the job alone, he can finish it in 6 hours. How long will it take for Mark to finish the job alone?
a. 5 hrs.
b. 7.5 hrs.
c. 3 hrs.
d. 1 hr. - Two pipes working together can fill a tank in 35 minutes. If the large pipe alone can fill the tank in 24 minutes less than the time taken by the smaller pipe then find the time taken by large pipe working alone to fill the tank.
a. 60 mins.
b. 20 mins.
c. 50 mins.
d. 84 mins.
ADDITIONAL ACTIVITIES
Critical Thinking and Creativity
A. A ladder 5 m long is leaning against a wall. The distance from the bottom of the ladder to the wall is 1 m less than the distance from the top of the ladder to the ground. How far is the bottom of the ladder from the wall?
[illustration: a ladder leaning against a brick wall]
B. Formulate problems that involve quadratic equations. Provide an illustrations and guide questions, then solve in as many ways as possible.
PROBLEM – BASED WORKSHEET
A.
Painting the House
After the ECQ, Chino and Philo are planning to paint a house together. Chino thinks if he works alone, it would take him 5 hours more than the time Philo takes to paint the entire house. Working together, they can complete the job in 6 hours.
[illustration: a worker painting a wall with a roller]
- If Philo can finish the job in x hours, how long will it take Chino to finish the job?
- What should be the equation to solve the given situation?
- How would you represent the amount of work that Chino can finish in 1 hour?
- What amount of work can Philo finish in 5 hours?
- How long does it take Chino to paint the house working alone?
B.
Car Wash
Mario and Kenneth work in a car wash station. The time that Mario takes in washing a car alone is 16 minutes less than the time that Kenneth takes in washing the same car. If both of them work together in washing the car, it will take them 15 minutes.”
[illustration: a person washing a car with a bucket and sponge]
- If Mario can finish the job in y minutes, what expression will represent Kenneth’s rate?
- What should be the equation to solve the problem?
- How long will it take Mario to finish washing the car working alone?
- How long will it take Kenneth to finish washing the car working alone?
- Working together, what amount of the job can they finish in 10 minutes?
E-Search
You may also check the following link for your reference and further learnings on solving problems involving quadratic equations:
- http://sites.millersville.edu/bikenaga/basic-algebra/quadratic-word-problems/quadratic-word-problems.html
- https://www.vivaxsolutions.com/maths/gcseqdrtcwrdproblems.aspx
- https://www.youtube.com/watch?v=IGGnn9oa4QY
- https://www.youtube.com/watch?v=LMApqDGjOr4
REFERENCES
Fadrigalan, Catalina M, et.al.College Algebra Mathematics 9 Learner’s Material, Department of Education
Ali Lafcioǧlu, Ismail Ersȍzoǧlu, Quadratics, Modular System of Equations, Inequalities and Functions, Zambak Publishing
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
https://www.polk.k12.ga.us/userfiles/313/Two%20step%20word%20problem%20rubric.pdf
PROBLEM – BASED WORKSHEET
A.
If x represents Philo’s completion of the job, then Chino can finish the job in (x + 5) hours.
Chino’s rate of work is 1/(x+5), thus, the amount of work that he can finish in 1 hour is represented by the expression = (rate)(time) = (1/(x+5))(1 hour) while Philo’s rate is 1/x, the working equation is,
1/x + 1/(x+5) = 1/6
Multiplying both sides by the least common denominator. [LCD = 6x(x+5)]
6(x+5) + 6x = x(x+5) 12x + 30 = x² + 5x x² − 7x − 30 = 0 (x − 10)(x + 3) = 0 x = 10 or x = −3
The equation has two solutions. However, we only consider the positive value of x since the situation involves measures of time. Thus, it takes 15 hours for Chino to paint the house working alone, while it takes Philo 10 hours. Also, for 5 hours, what part of the house has already painted by Philo.
B.
If Mario can finish the job in y minutes, thus his rate is 1/y, it follows that Kenneth’s rate is 1/(y+16). The working equation is:
1/y + 1/(y+16) = 1/15
Multiplying both sides by the least common denominator. [15y(y+16)]
15(y+16) + 15y = y(y+16) 30y + 240 = y² + 16y y² − 14y − 240 = 0 (y − 24)(y + 10) = 0 y = 24 or y = −10
We only consider the positive value of y since the situation involves measures of time. Thus, it takes 24 minutes for Mario to finish the job alone, while it takes Kenneth 40 minutes working alone. Working together, the amount of work they can finish in 10 minutes is
1/24 (10) + 1/40 (10) = 5/12 + 1/4 = (5+3)/12 = 8/12 or 2/3
ASSESSMENT
- B
- C
- B
- D
- B
- C
- A
- D
- B
- A
ADDITIONAL ACTIVITIES
A. 3 meters
B. Use the given Rubric to check your work.
https://www.polk.k12.ga.us/userfiles/313/Two%20step%20word%20problem%20rubric.pdf
Rubric: Writing Math Word Problems
| 4 pts | 3 pts | 2 pts | 1 pts | |
|---|---|---|---|---|
| Requirements | Student has all required word problems for two steps problem. | Student has created 1 step problems. | Student created the problem but did not solve it. | Student could not create a word problem. |
| Content | Appropriate content is used for each word problem. Student clearly understands the mathematical concepts. | Appropriate content is used for each word problem. Student shows some understanding of the mathematical concepts. | Appropriate content may be used. Student shows little understanding of the mathematical concepts. | Appropriate content is not observed. Student does not demonstrate an understanding of the mathematical concepts. |
| Solution | Solution is correct and all work is shown. | Solution is correct, however there is no work shown for the problem. | Solution is incorrect, however the student showed their work and made an error in calculations. | Solution is incorrect and student either didn’t show the work or worked the problem incorrectly. |
| Organization | The word problem is written in clear and coherent language. It is easy to follow and read. | The word problem is written in clear and coherent language. It is easy to read most of the time. | The word problem is not written in clear and coherent language. It is easy to follow and read some of the time. | The word problem is not written in clear and coherent language, or may not be observed. It is difficult to read and follow. |
| Mechanics | Proper language, capitalization and punctuation are present with no mistakes. | Proper language, capitalization and punctuation are present with no more than two mistakes. | Proper language, capitalization and punctuation may be used, but more than two mistakes. | Proper language, capitalization and punctuation are not observed. |
ANSWER KEY
WHAT I KNOW
- A
- A
- C
- A
- B
- D
- B
- D
- A
- C
WHAT’S IN
- (2w + 3)(w) = 160
- 16 = (h+4)(h)/2
350 = (45 − w)(w) - Perimeter: l + w = 90/2 = 45
Area: lw = 350 ; l = 45 − w
WHAT I CAN DO
- Length = 12m, width = 8m
- The dimensions of the table are 3ft and 9ft, respectively.
WHAT’S MORE
- Let x be the unknown number.
x² + 2x = 1 x² + 2x − 1 = 0; a=1, b=2, c=−1
x = (−b ± √(b²−4ac)) / (2a) = (−2 ± √((2)²−4(1)(−1))) / (2(1)) = (−2 ± √8) / 2 = (−2 ± 2√2) / 2 = −1 ± √2
- Let x be the width of the strip set of sunflowers.
20 + 2x = width of the quadrangle with strip set of sunflowers 30 + 2x = length of the quadrangle with strip set of sunflowers
[chart: diagram; area of the quadrangle with strip set of sunflowers] Area of the large rectangle = l × w = (30 + 2x)(20 + 2x) Area of Quadrangle = l × w = 30 × 20 = 600 Area of strip = (30 + 2x)(20 + 2x) − 600 = 336
SOLUTION:
(30 + 2x)(20 + 2x) − 600 = 336 4x² + 100x + 600 − 600 = 336 4x² + 100x − 336 = 0 x² + 25x − 84 = 0 (x + 28)(x − 3) = 0 x = −28 or x = 3
since there is no negative value for measurements, the strip set of sunflowers around the edge of the quadrangle is 3 feet wide.
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