This is the Mathematics 9 self-learning module for Quarter 1, Module 6 (Week 2): Nature of Roots of a Quadratic Equation. It is for Grade 9 learners who already know how to solve quadratic equations, including with the quadratic formula. After this module, learners can describe the roots of a quadratic equation as real or imaginary, rational or irrational, and equal or unequal, without solving for the roots. The module notes that this skill helps in real-life decision making.
On this page 6 sections
What learners will learn
The module has one learning competency. Learners will be able to:
- characterize the roots of a quadratic equation using the discriminant (M9AL-Ib-3)
The discriminant and the nature of roots
For a quadratic equation in the form ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0, the quadratic formula gives two roots: (−b + √(b² − 4ac)) / 2a and (−b − √(b² − 4ac)) / 2a.
The quantity under the radical sign, b² − 4ac, is called the discriminant. Its value tells you what kind of roots the equation has:
| Value of b² − 4ac | Nature of the roots |
|---|---|
| Equal to 0 | Real, rational, and equal |
| Greater than 0 and a perfect square | Real, rational, and unequal |
| Greater than 0 but not a perfect square | Real, irrational, and unequal |
| Less than 0 | Imaginary and unequal |
The module explains the reasoning this way. Whether the roots are real or imaginary depends on whether the quantity under the radical sign is positive, zero, or negative. If that quantity is a perfect square, the roots are rational; if not, they are irrational. For the two roots to be equal, the quantity must be zero.
Worked examples from the module
Example 1 finds the nature of the roots by inspection:
- x² − 7x − 8 = 0: (−7)² − 4(1)(−8) = 49 + 32 = 81. Since 81 is positive and a perfect square, the roots are real, rational, and unequal.
- 2x² − 4x + 1 = 0: (−4)² − 4(2)(1) = 16 − 8 = 8. Since 8 is positive but not a perfect square, the roots are real, irrational, and unequal.
- 3x² + 9x + 11 = 0: (9)² − 4(3)(11) = 81 − 132 = −44. Since −44 is less than 0, the roots are imaginary and unequal.
- x² − 12x + 36 = 0: (−12)² − 4(1)(36) = 144 − 144 = 0. The roots are real, rational, and equal.
Example 2 works backward, finding an unknown value that gives a certain kind of root:
- For (3k + 1)x² + 2(k + 1)x + k = 0 to have equal roots, the discriminant is set to zero. This leads to 2k² − k − 1 = 0, so k = 1 or k = −1/2.
- For 6x² − 5x − k = 0, imaginary roots need 25 + 24k < 0, so k < −25/24. Real and unequal roots need the opposite, k > −25/24.
- For (r − 3)x² − (r − 6)x − 4 = 0 to have equal roots, the discriminant gives r² + 4r − 12 = 0, so r = 2 or r = −6.
A common mistake: rewrite the equation first
In the Spot the Error worksheet, Maria and Jose find the discriminant of 5x² − 3x = 2. Maria takes c = 2 straight from the equation and gets −31. Jose first rewrites it as 5x² − 3x − 2 = 0, so c = −2, and gets 49. Jose is correct: always write the equation in the form ax² + bx + c = 0 before reading off a, b, and c.
Parts of the module
- What I Know: a 10-item multiple-choice pretest.
- What’s In: a vocabulary review of real, imaginary, rational, and irrational numbers. It also has a hidden-number game where learners sort numbers into four “trash bins”; a number can go in two bins.
- What’s New: learners solve four equations with the quadratic formula. They then check whether each pair of roots is real or imaginary, rational or irrational, and equal or unequal. The discriminant values (9, −12, 0, and 24) are given.
- What is It: the discussion of the discriminant and the worked examples above.
- What’s More: practice in characterizing roots, finding values of k for (3k + 1)x² + (11 + k)x + 9 = 0, and writing equations with given types of roots.
- What I Have Learned: a summary of the four discriminant rules.
- What I Can Do: 10 equations to characterize and 5 problems that find unknown values.
- Assessment: a 10-item multiple-choice post-test.
- Additional Activities: three real-life scenarios (a rectangular park, two girls’ ages, a rice field). Learners decide whether each is possible and justify the answer.
- Problem-Based Worksheet: the Spot the Error task.
How to use this module
- Learners: Take the pretest first and write your answers on a separate answer sheet. Note the items you missed and look for the right answers as you work through the module. Go through the parts in order, then take the assessment. If your answer is not among the choices, write E with your final answer, as the assessment instructs.
- Parents: Help your child check work using the answer key at the end of the module after each activity is finished, not before.
- Teachers: The answer key covers the pretest, What’s New, What’s More, What I Can Do, the assessment, the additional activities, and the Spot the Error worksheet. The activities are tagged with skills such as communication, collaboration, critical thinking, and creativity, which can guide group or individual work.
Frequently asked questions
What grade and quarter is this module for?
It is for Grade 9 Mathematics, Quarter 1, Module 6, covering Week 2. Its learning competency code is M9AL-Ib-3.
What is the discriminant of a quadratic equation?
For ax² + bx + c = 0, the discriminant is b² − 4ac, the quantity under the radical sign in the quadratic formula. Its value tells you whether the roots are real or imaginary, rational or irrational, and equal or unequal.
What does it mean if the discriminant is negative?
If b² − 4ac is less than 0, the roots are imaginary and unequal, so the equation has no real roots.
How do you know if the roots are rational or irrational?
If the discriminant is positive and a perfect square, the roots are real, rational, and unequal. If it is positive but not a perfect square, the roots are real, irrational, and unequal.
Does the module have an answer key?
Yes. The last pages give answers for the pretest, What’s New, What’s More, What I Can Do, the assessment, the additional activities, and the Spot the Error worksheet.
Full text of the module
The text below is a transcription of the PDF above, as printed.
Mathematics Quarter 1-Module 6: Nature of Roots of a Quadratic Equation
Mathematics Quarter 1-Module 6 Nature of Roots of a Quadratic Equation Week 2 Learning Code – M9AL-Ib-3
[illustration: teacher on a computer screen and students learning with books, laptops, and mobile devices]
GRADE 9 Learning Module for Junior High School Mathematics
Development Team of the Module
Writers: Rowena F. Reyes – T1, Analynn M. Argel – MTII
Editors: Sally C. Caleja – Head Teacher VI Maita G. Camilon – Haed Teacher VI Melody P. Rosales – 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
Quarter 1 – Module 6 – 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
What I Need to Know
Learning Competency
The learners will be able to:
- characterize the roots of a quadratic equation using the discriminant. M9AL-Ib-3
[transcriber’s note: decorative header graphics (“SQUARE ROOTS”, “PREVIEW”) omitted]
From your previous modules, you learned how to get the roots of a quadratic equation. At this point, you will explore on describing the characteristics of the roots of a quadratic equation without solving for the roots. This knowledge would come in handy on some real-life scenarios especially in decision making.
What I Know
Find out how much you already know about the nature of roots of 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.
Which of the following is the discriminant of the quadratic equation: ax² − bx + c = 0, where a, b, c are real number and a ≠ 0?
A. b²
B. b² − 4ac
C. √b² − 4ac
D. −b ± √b² − 4acHow many real roots does the quadratic equation x² + 8x + 12 = 0 have?
A. 0 B. 1 C. 2 D. 3Which of the following quadratic equations has no real roots?
A. 2x² + 4x = 3
B. z² − 8z − 4 = 0
C. 3m² − 2m + 5 = 0
D. −2f² + f = −4Which of the following is the nature of the roots of the quadratic equation if the value of its discriminant is zero?
A. The roots are not real.
B. The roots are irrational and not equal.
C. The roots are rational and not equal.
D. The roots are rational and equal.Which of the following is the nature of the roots of the quadratic equation if the value of its discriminant is negative?
A. The roots are not real.
B. The roots are rational and equal.
C. The roots are rational and not equal.
D. The roots are irrational and not equal.Which of the following is the nature of the roots of the quadratic equation if the value of its discriminant is positive and a perfect square?
A. The roots are not real.
B. The roots are rational and equal.
C. The roots are irrational and not equal.
D. The roots are rational and not equal.Which of the following is the nature of the roots of the quadratic equation if the value of its discriminant is positive and not a perfect square?
A. The roots are not real.
B. The roots are rational and equal.
C. The roots are rational and not equal.
D. The roots are irrational and not equal.What is the value of r in the quadratic equation x² − 10x − r = 0 whose roots are equal rational and real?
A. −25
B. −10
C. 10
D. 25Which of the following quadratic equations has equal roots?
A. x² + 6x + 9 = 0
B. x² + 5x + 10 = 0
C. 2x² − 10x + 8 = 0
D. 3x² − 2x − 5 = 0What are the values of k in the quadratic equation 6x² − 5x − k = 0 whose roots are imaginary?
A. k < −25/24
B. k < 25/24
C. k > −25/24
D. k > 25/24
What’s In
Communication, Collaboration
Before we begin, let us unlock the following vocabularies. Get your dictionary or any math book and find the meaning of the following words. Give at least two examples of each.
- Real numbers
- Imaginary numbers
- Rational Numbers
- Irrational Numbers
Did you find the meaning of the words above? Now apply these words in the next activity.
Let us FIND and INVESTIGATE!
Real or Imaginary? Rational or Irrational?
This is a classroom hidden number game. In this game you must find out the hidden numbers and place it in the appropriate trash bin.
Reminder: a number can be written in two trash bins.
[illustration: classroom scene with students cleaning and decorating; hidden numbers are written on the board, walls, desks, and objects]
Hidden numbers: −2/3; 7; 2 + √6; √35; −√7/√16; (√−2)/3; (4 ± √−12)/2; 1 ± √−4; √−4; 9/√49; 5/2; √49; √5; 5; √−3; −√2
[illustration: four trash bins labeled “Real numbers”, “Imaginary numbers”, “Rational numbers”, and “Irrational numbers”]
1Source: http://clipart-library.com/clean-classroom-cliparts.html
What’s New
The quadratic equations listed can be solved by using quadratic formula as discussed from the previous topic. Using this method, find and examine the roots obtained to complete the table below. In this case, let r₁ and r₂ be the roots. Put a check in the cell that corresponds to the characteristics of the roots attained in each equation.
| Equations | Roots r₁ | Roots r₂ | Real | Imaginary | Rational | Irrational | Equal | Unequal | The value of b²−4ac |
|---|---|---|---|---|---|---|---|---|---|
| (a) x² − 7x + 10 = 0 | 9 | ||||||||
| (b) x² − 4x + 7 = 0 | -12 | ||||||||
| (c) x² − 10x + 25 = 0 | 0 | ||||||||
| (d) x² − 4x − 2 = 0 | 24 |
If the quadratic formula is used to get the roots, the values of b² − 4ac (quantity under the radical sign) for those equations are 9, −12, 0 and 24 respectively.
What is It
Communication, Critical Thinking, and Collaboration
When describing the natures or characters of the roots of a quadratic equation, it can be one of each of the following:
(a) Real or Imaginary (b) Rational or irrational (c) Equal or unequal
Given the form of the equation, ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0, then using the quadratic formula, the roots are (−b + √(b² − 4ac))/(2a) and (−b − √(b² − 4ac))/(2a).
Looking at the two roots, we can say that they are real or imaginary by using the quantity under the radical sign. It depends whether it is positive, zero or negative.
Similarly, if the value inside the radical sign is a perfect square, then the roots are rational; if not, they are irrational.
In order to have equal roots, the quantity under the radical sign must be zero.
Therefore, the nature of the roots can be decided by using the quantity b² − 4ac, which is called the discriminant of the equation ax² + bx + c = 0. Thus:
(a) If b² − 4ac = 0; the roots are real, equal and rational. (b) If b² − 4ac < 0; the roots are unequal and imaginary. (c) If b² − 4ac > 0 and a perfect square; the roots are real, unequal and rational. (d) If b² − 4ac > 0 but not a perfect square; the roots are real, unequal, and irrational.
Applying this conclusion, you may check your answer on the last activity in the “WHAT’S NEW” part.
Example 1: By inspection, determine the nature of the roots of the following equations: (a) x² − 7x − 8 = 0 (b) 2x² − 4x + 1 = 0 (c) 3x² + 9x + 11 = 0 (d) x² − 12x + 36 = 0
Solutions
(a) The equation x² − 7x − 8 = 0 gives a = 1, b = −7, and c = −8. Substituting the given values of a, b, and c in the expression we have b² − 4ac = (−7)² − 4(1)(−8) = 49 + 32 = 81
Since 81 is greater than 0 and a perfect square, then the equation x² − 7x − 8 = 0 has 2 real, rational, and unequal roots.
(b) The equation 2x² − 4x + 1 = 0 gives a = 2, b = −4, and c = 1. Substituting the given values of a, b, and c in the expression we have b² − 4ac = (−4)² − 4(2)(1) = 16 − 8 = 8
Since 8 is greater than 0 and not a perfect square, then the equation 2x² − 4x + 1 = 0 has real, irrational, and unequal roots.
(c) The equation 3x² + 9x + 11 = 0 gives a = 3, b = 9, and c = 11. Substituting the given values of a, b, and c in the expression we have
b² − 4ac = (9)² − 4(3)(11) = 81 − 132 = −44
Since −44 is less than 0, then the equation 3x² + 9x + 11 = 0 has imaginary and unequal roots.
(d) The equation x² − 12x + 36 = 0 gives a = 1, b = −12, and c = 36. Substituting the given values of a, b, and c in the expression we have b² − 4ac = (−12)² − 4(1)(36) = 144 − 144 = 0
Since the discriminant is equal to zero, then the equation x² − 12x + 36 = 0 has real, rational and equal roots.
Example 2: Answer the following (a) Given the equation (3k + 1)x² + 2(k + 1)x + k = 0, find the values of k for which the roots are equal. (b) For what values of k are the roots of 6x² − 5x − k = 0 is: i.) imaginary? ii.) real and unequal? (c) Find r so that the roots of the equation (r − 3)x² − (r − 6)x − 4 = 0 are equal?
Solutions
(a) Based on the given equation, (3k + 1)x² + 2(k + 1)x + k = 0, the expressions for a, b, c are a = 3k + 1, b = 2(k + 1), and c = k.
For equal roots, the discriminant must be equal to zero. Therefore, b² − 4ac = 0 [2(k + 1)]² − 4(3k + 1)(k) = 0 (2k + 2)² − (12k + 4)(k) = 0 4k² + 8k + 4 − 12k² − 4k = 0 −8k² + 4k + 4 = 0 8k² − 4k − 4 = 0
Dividing both sides by 4, we have: 2k² − k − 1 = 0
To solve for k, we have: (k − 1)(2k + 1) = 0
Equating both factors to zero and solving for k, k − 1 = 0 or 2k + 1 = 0 k = 1 2k = −1 k = −1/2
Thus, the values of k for which the roots are equal are 1 and −1/2.
(b) Given the equation of 6𝑥² − 5𝑥 − 𝑘 = 0, we have 𝑎 = 6, 𝑏 = −5, and 𝑐 = −𝑘
(i) In order to have imaginary roots, 𝑏² − 4𝑎𝑐 must be less than zero which is denoted by:
𝑏² − 4𝑎𝑐 < 0
Hence,
(−5)² − 4(6)(−𝑘) < 0 25 + 24𝑘 < 0 24𝑘 < −25
Therefore, 𝒌 < −𝟐𝟓/𝟐𝟒.
(ii) In order to have real and unequal roots, 𝑏² − 4𝑎𝑐 must greater than 0 which is denoted by:
𝑏² − 4𝑎𝑐 > 0
Therefore, based on (i), 𝒌 > −𝟐𝟓/𝟐𝟒. [complement of the answer in (i)]
(c) Given the equation (𝑟 − 3)𝑥² − (𝑟 − 6)𝑥 − 4 = 0, we have a = r – 3, b = -(r – 6), and c = -4.
For equal roots, the discriminant must be equal to zero. Therefore,
𝑏² − 4𝑎𝑐 = 0 [-(r – 6)]² – 4(r – 3)(-4) = 0 r² – 12r + 36 + 16r -48 = 0 r² + 4r -12 = 0
Solving for r, we have
(r – 2) (r + 6) = 0 r – 2 = 0 or r + 6 = 0 r = 2 r = -6
Therefore, the values of r for which the roots are equal are 2 and -6.
Critical Thinking, Collaboration
WHAT’S MORE
I. Using the discriminant, characterize the roots of the following quadratic equations:
- 𝑥² − 3𝑥 − 5 = 0
- 𝑥² = 5𝑥 − 11
- 5𝑥² − 6𝑥 + 13 = 0
II. Given (3𝑘 + 1)𝑥² + (11 + 𝑘)𝑥 + 9 = 0. Find the values of 𝑘 for which the roots are:
(a) equal; (b) imaginary; (c) real and unequal
III. Write a quadratic equation of each of the following types:
(a) two rational roots (b) two imaginary roots (c) two irrational roots (d) one rational root
WHAT I HAVE LEARNED
The nature of the roots of a quadratic equation using the discriminant without solving the equation. To characterize the roots of the quadratic equation 𝑎𝑥² + 𝑏𝑥 + 𝑐 = 0, where 𝑎, 𝑏, 𝑐 are real numbers and 𝑎 ≠ 0, we have the following:
- If 𝑏² − 4𝑎𝑐 = 0, the roots are real, rational and equal.
- If 𝑏² − 4𝑎𝑐 > 0 and a perfect square, the roots are real, rational, and unequal.
- If 𝑏² − 4𝑎𝑐 > 0 and not a perfect square, the roots are real, irrational and unequal.
- If 𝑏² − 4𝑎𝑐 < 0, the roots are imaginary and unequal.
WHAT I CAN DO
Critical Thinking
I. Characterize the roots of the following quadratic equations using the discriminant.
- 6𝑥² − 𝑥 + 2 = 0
- 𝑥² − 3𝑥 + 7 = 0
- 𝑚² + 8𝑚 + 16 = 0
- 3𝑥² + 8𝑥 − 8 = 0
- 3𝑥² − 4𝑥 + 2 = 0
- 7𝑥² − 10𝑥 + 1 = 0
- 𝑡² − 3𝑡 + 2 = 0
- 8𝑥² − 16𝑥 + 9 = 0
- 𝑟² − 𝑟 = 0
- 3𝑡² + 4𝑡 = 8
II. Solve the following:
- Find the values of 𝑘 that will make the roots of 4𝑥² − 𝑘𝑥 + 4 = 0 equal.
- For what values of 𝑘 are the roots of 𝑘𝑥² − 4𝑥 + 5 = 0 are imaginary?
- Determine the value of 𝑐 if 4𝑥² − 5𝑥 + 𝑐 = 0 has real and unequal roots.
- The equation 7𝑥² − 3𝑥 + 𝑚 = 0 has imaginary roots. Determine the value of 𝑚.
- Find 𝑘 so that the roots of the equation 𝑤² + (𝑘 + 3)𝑤 + 3𝑘 = 0 are equal.
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 is the discriminant of the quadratic equation: 𝑥² + 2𝑥 + 5 = 0?
- A. -16
- B. -4
- C. 4
- D. 16
How many real roots does the quadratic equation 5𝑥² − 8𝑥 + 6 = 0 have?
- A. 0
- B. 1
- C. 2
- D. 3
Which of the following quadratic equations has real roots?
- A. 𝑥² + 𝑥 + 1 = 0
- B. 2𝑥² + 5𝑥 + 3 = 0
- C. 2𝑥² − 6𝑥 + 6 = 0
- D. 4𝑥² + 3𝑥 + 3 = 0
The roots of a quadratic equation are imaginary. Which of the following statement is true about the discriminant of equation?
- A. The discriminant is negative.
- B. The discriminant is equal to zero.
- C. The discriminant is positive and a perfect square.
- D. The discriminant is positive and not perfect square.
The roots of a quadratic equation are 2 and 5.Which of the following statement is true about the discriminant of equation?
- A. The discriminant is negative.
- B. The discriminant is equal to zero.
- C. The discriminant is positive and not perfect square.
- D. The discriminant is positive and a perfect square.
Given the equation, 4𝑥² + 9𝑥 − 13 = 0, which of the following is the nature of its roots?
- A. The roots are real, rational and equal.
- B. The roots are imaginary and unequal.
- C. The roots are real, irrational and unequal.
- D. The roots are real, rational, and unequal.
The discriminant of a quadratic equation is 34. Which of the following is true about its roots?
- A. The roots are real, rational and equal.
- B. The roots are imaginary and unequal.
- C. The roots are real, irrational and unequal.
- D. The roots are real, rational, and unequal.
What is the value of 𝑡 that will make the roots of the equation 8𝑥² − 4𝑡𝑥 + 3 = 0 equal?
- A. 𝑡 = ±6
- B. 𝑡 = ±96
- C. 𝑡 = ±√6
- D. 𝑡 = ±4√6
Which of the following quadratic equations has rational, equal and real roots?
- A. 9 − 6𝑥 + 𝑥² = 0
- B. 𝑥² + 2𝑥 + 2 = 0
- C. 𝑥² + 4𝑥 − 21=0
- D. 4𝑥² + 12𝑥 + 10
What is/are the value/s of 𝑘 in the quadratic equation 2𝑥² − 5𝑥 − 𝑘 = 0 whose roots are imaginary?
- A. 𝑘 < 25/8
- B. 𝑘 < −25/8
- C. 𝑘 > −25/8
- D. 𝑘 > 25/8
ADDITIONAL ACTIVITIES
Critical Thinking, Collaboration, Creativity
The shortest way of describing the nature of the roots of a quadratic equation is not by solving for the roots but by just using the discriminant of the given quadratic equation.
Below is a specific scenario on how this lesson may help you in some decision making.
A. Apply the idea of the nature of roots of a quadratic equation to visualize, answer if it is possible or not, and then justify your answer.
SCENARIO 1 Is it possible to make a rectangular park of perimeter 80 m and area 400 m² ?
SCENARIO 2 The sum of the ages of two girls is 20 years. The product of their ages four years ago was 48.Is this possible?
SCENARIO 3 Is it possible to design a rectangular rice field whose length is twice its width and the area is 800 m²?
PROBLEM – BASED WORKSHEET
SPOT THE ERROR
Maria and Jose wanted to determine the number of real solutions of the quadratic equation 5x² – 3x = 2.
Maria 5x² – 3x = 2 b²- 4ac = (-3)² – 4(5)(2) = -31 Since the discriminant is -31, there are no real roots.
Jose 5x² – 3x = 2 5x² – 3x -2 = 0 b²- 4ac = (-3)² – 4(5)(-2) = 49 Since the discriminant is 49, positive and perfect square, there are two real roots.
Who is correct? Justify your answer.
E-Search
You may also check the following link for your reference and further learnings on nature of the roots of quadratic equations:
- https://www.khanacademy.org/math/in-in-grade-10-ncert/x573d8ce20721c073:in-in-chapter-4-quadraticequation/x573d8ce20721c073:in-in-10-quadratic-discriminant-and-number-ofsolutions/v/discriminant-for-types-of-solutions-for-a-quadratic
- https://www.toppr.com/guides/maths/quadratic-equations/nature-of-roots/
REFERENCES
Bautista, E. P., & O’dell, I. C. (n.d.).Numlock III. pp 368- 372 Quezon City: Trinitas Publishing, INC.
Dilao, S. J., & Bernabe, J. G. (2009). *Intermediate Algebra.*pp 52-54. Quezon CIty: SD Publicaton.
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 250-256Manila: Rex Book Store.
https://ya-webdesign.com/explore/trashcan-drawing-basura/ http://clipart-library.com/clean-classroom-cliparts.html https://www.freepik.com/free-vector/woman-with-long-hair-teachingonline_7707557.htm https://www.freepik.com/free-vector/kids-having-online-lessons_7560046.htm
WHAT I KNOW
- B
- C
- C
- D
- A
- D
- D
- A
- A
- A
ASSESSMENT
- A
- A
- B
- A
- D
- D
- C
- C
- A
- B
PROBLEM – BASED WORKSHEET
Solutions:
Maria made a mistake by getting the values of a, b and c without re-writing the equation in the form ax² + bx + c = 0.
Jose did the correct solution. He first re-write the equation in the form ax² + bx + c = 0 and came up with a = 5, b = -3 and c = 2.
Thus, the discriminant is 49. It is a positive number and a perfect square, therefore there are two real roots.
WHAT’S MORE
I.
- The roots are real, irrational and unequal.
- The roots are imaginary and unequal.
- The roots are imaginary and unequal.
II. a. 𝑘 = 1 or 𝑘 = 85 b. 1 < 𝑘 < 85 c. 𝑘 > 85 or 𝑘 < 1
ADDITIONAL ACTIVITIES
Yes, the dimension that will lead to an area of 400m² from a perimeter of 80m is a 20m by 20m square. Since a square is a rectangle, thus the scenario is possible.
No, if the product of the girl’s ages 4 years ago is 48, then the sum of their ages is 20 less 8.
Let x and y be the girls’ ages. The equation will be, x + y = 12 xy = 48
Solving for x and y, If x + y = 12, then y = 12 – x.
Substitute the value to the second equation, we have x(12 – x) = 48 -x² + 12x – 48 = 0
Solving for x using quadratic formula, x = (−b ± √(b²−4ac)) / 2a x = (−12 ± √((12)²−4(−1)(−48))) / (2(−1)) x = (−12 ± √(−48)) / (−2)
Since the radicand is negative, the roots are imaginary. Thus, the product of the ages four years ago is impossible.
- Yes, if we let x = width, then 2x = length. The area is:
x (2x) = 800
2x² = 800
x² = 400
x = 20
Hence x = 20. This gives a width of 20 m, and a length of 40 m.
[transcriber’s note: this page is printed upside down; the content is transcribed in its logical reading order]
Answer Key
What’s In
| Real Numbers | Imaginary Numbers | Rational Numbers | Irrational Numbers |
|---|---|---|---|
| 0, −2/3, 7, −√7/√16 [?], 2+√6, √35/√5 [?], 5/2, √49, −√2 [?] | (4±√−12)/2, 3/√−2 [?], 1±√−4, √−3 | 0, −2/3, 7, 9/49, 5/2, √49, √16 [?] | 2+√6, √35, √5, √2, −√7/√16 [?] |
[transcriber’s note: several fractions and radicals in this table are hard to read in the rotated image; those marked [?] are best readings]
What’s New
| EQUATIONS | ROOTS: r1 | ROOTS: r2 | Real or Imaginary?: Real | Real or Imaginary?: Imaginary | Rational or Irrational?: Rational | Rational or Irrational?: Irrational | Equal or Unequal?: Equal | Equal or Unequal?: Unequal |
|---|---|---|---|---|---|---|---|---|
| (a) x² − 7x + 10 = 0 | 5 | 2 | √ | √ | √ | |||
| (b) x² − 4x + 7 = 0 | (4 + √−12)/2 | (4 − √−12)/2 | √ | √ | ||||
| (c) x² − 10x + 25 = 0 | 5 | 5 | √ | √ | √ | |||
| (d) x² − 4x − 2 = 0 | 2 + √6 | 2 − √6 | √ | √ | √ |
What I Can Do
I.
- imaginary and unequal
- imaginary and unequal
- real, rational and equal
- real, irrational and unequal
- imaginary and unequal
- real, irrational and unequal
- real, rational and unequal
- imaginary and unequal
- real, rational and unequal
- real, irrational and unequal
II.
- k = ±8
- k > 4/5 [?]
- c < 25/16
- m > 9/28 [?]
- k = 3
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