This is the Grade 9 Mathematics self-learning module for Quarter 1, Module 4 (Week 1), “Solving Quadratic Equation by Completing the Square.” It is for Junior High School learners who already know how to solve quadratic equations by factoring. Factoring does not work when the trinomial cannot be factored. This module teaches a method that works for any quadratic equation: completing the square.
On this page 7 sections
What learners will learn
The module has one learning competency. Learners will be able to:
- solve quadratic equations by completing the square (M9AL-Ia-b-1).
Key idea 1: Perfect square trinomials
Completing the square is built on the perfect square trinomial. The module says a trinomial is a perfect square if:
- The first and third terms are perfect squares with positive coefficients or positive numbers.
- The second term is twice the product of the square roots of those two terms.
For example, in 4x² + 12x + 9, the first term is (2x)², the last term is (3)², and the middle term is 2(2x)(3) = 12x.
To factor a perfect square trinomial:
- Find the positive square roots of the first and last terms.
- Join them with the sign of the middle term.
- Write the binomial as a square, since it is used twice as a factor.
Example: x² − 6x + 9 factors as (x − 3)², because √x² = x, √9 = 3, and the middle term is negative.
Key idea 2: Completing the square
An expression like x² + 8x is an “incomplete” square. Adding 16 gives x² + 8x + 16, which equals (x + 4)². The number you add is the square of half the coefficient of x: half of 8 is 4, and 4² = 16.
| Incomplete square | Completed square | Factored form |
|---|---|---|
| x² + 4x | x² + 4x + 4 | (x + 2)² |
| y² + 6y | y² + 6y + 9 | (y + 3)² |
| b² − 8b | b² − 8b + 16 | (b − 4)² |
| m² − 10m | m² − 10m + 25 | (m − 5)² |
The module also shows fractional cases. For d² − 7d, you add (−7/2)², or 49/4, to get (d − 7/2)².
Steps for solving a quadratic equation by completing the square
The module’s summary lists these steps:
- Move the constant term to the right side. Keep all terms with the unknown on the left.
- If the coefficient of x² is not 1, divide every term by it.
- Take the coefficient of x, divide it by 2, and square it. Add the result to both sides.
- Factor the perfect square trinomial.
- Take the square root of both sides. The right side gives two values (plus and minus).
- Set the linear expression equal to each value.
- Solve each linear equation.
- Check your answers in the original equation.
Worked example from the module
Solve x² − 8x − 9 = 0:
- x² − 8x = 9
- Add (−8/2)² = 16 to both sides: x² − 8x + 16 = 25
- Factor: (x − 4)² = 25
- Take square roots: x − 4 = ±5
- So x = 9 or x = −1. The solution set is {−1, 9}.
The module also solves 3x² − 4x + 1 = 0, which needs dividing by 3 first (solution set {1/3, 1}). It also solves x² − 4x − 8 = 0, whose answers involve a radical: {2 − 2√3, 2 + 2√3}.
Parts of the module
- What I Know: a 10-item multiple-choice pretest.
- What’s In: a review of perfect square trinomials, with a table to complete.
- What’s New: the pattern behind completing the square, with three examples.
- What Is It: three fully worked equations, with checking.
- What’s More: five equations to solve and check.
- What I Have Learned: the eight-step summary.
- What I Can Do: ten more equations to solve, showing solutions.
- Assessment: a 10-item multiple-choice post-test.
- Additional Activities: a word problem about a 60-inch TV screen that is 12 inches wider than it is tall. There is also an exploration task that previews deriving the quadratic formula, which is covered in Module 5.
- Problem-Based Worksheet: “An Open Box,” where learners model a box made from a rectangular board and find its dimensions by completing the square.
- Answer key for all parts.
How to use this module
Learners: Take the pretest first and check it. Note the items you missed, then look for the right answers as you work through the module. Study the examples in order. Do the What’s More and What I Can Do exercises, and write your full solution for each. Always check your roots by substituting them into the original equation. Write your answers on a separate answer sheet.
Parents: You do not need to be good at algebra to help. Ask your child to explain each of the eight steps on one problem, then compare the final answer with the answer key at the end of the module.
Teachers: The module fits Week 1 of Quarter 1. Its open-box worksheet and TV-screen problem work well as performance tasks. Please double-check the answer key before you use it for grading. On pretest item 6, the value that makes x² − bx + 49/4 a perfect square is b = 7 (choice B), not D. For What’s More item 5 (7x² − 2x = 1), the roots work out to (1 ± 2√2)/7. In the What’s In table, 9z² + 24z − 16 has a negative third term, so by the module’s own rule it is not a perfect square trinomial.
Frequently asked questions
What grade and quarter is this module for?
It is for Grade 9 Mathematics, Quarter 1, Module 4. It is meant for Week 1 of the quarter.
What competency does the module cover?
It covers solving quadratic equations by completing the square, with learning code M9AL-Ia-b-1.
What does completing the square mean?
It means adding a constant to an expression like x² + bx so that it becomes a perfect square trinomial. The constant is the square of half the coefficient of x. For example, adding 16 to x² + 8x gives (x + 4)².
Does the module have an answer key?
Yes. It includes answers for the pretest, the What’s In table, What’s More, What I Can Do, the assessment, the additional activity, and the open-box worksheet. A few items are worth double-checking.
Why learn completing the square if I already know factoring?
Factoring only works when the trinomial can be factored. Completing the square can solve any quadratic equation. It is also used to derive the quadratic formula, which is covered in the next module.
Full text of the module
The text below is a transcription of the PDF above, as printed.
Mathematics Quarter 1-Module 4 Solving Quadratic Equation By Completing the Square
Week 1
Learning Code – M9AL – Ia-b- 1
[illustration: grade level badge showing “9”]
[illustration: teacher on a video screen with students learning at a laptop, tablet, and desk]
GRADE 9 is omitted here as a running header.
Learning Module for Junior High School Mathematics
Development Team of the Module
Writers: Jhon Edward M. Valera – T1 Rowena F. Reyes – T1 Analyn M. Argel – MTII Roderick F. Borja- TIII
Editors: Sally C. Caleja – Head Teacher VI Marissa B. Aniban – Head 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 4 – 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
MODULE 4 SOLVING QUADRATIC EQUATION BY COMPLETING THE SQUARE
You already learned how to solve quadratic equation by factoring. However, there are limitations in using this method especially if the trinomial is not factorable. In this module, you will learn another way of solving quadratic equation – by completing the square.
WHAT I NEED TO KNOW
LEARNING COMPETENCY
The learners will be able to:
- solve quadratic equation by completing the square. M9AL – Ia-b- 1
WHAT I KNOW
Find out how much you already know about solving quadratic equation by completing the square. 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 answers as you go through this module.
What should be the middle term to make this a perfect square trinomial m² + _____ + 100?
- A. 5m
- B. 10m
- C. 20m
- D. 50m
What must be added to x² – 4x + _____ to make it a perfect square trinomial?
- A. 2
- B. 4
- C. 8
- D. 16
In solving the equation x² – 8x = −1 by completing the square, what must be added to both sides to make the left side a perfect square trinomial.
- A. 1
- B. 4
- C. 16
- D. -2
Which of the following is a perfect square trinomial?
- A. x² + 10x + 25
- B. x² + 10x – 25
- C. x² − 5x + 25
- D. x² + 10x + 5
What is the factored form of x² – 6x + 9?
- A. (x + 3) (x + 3)
- B. (x − 3) (x − 3)
- C. (x − 3) (x + 3)
- D. (x + 3) (x + 1)
In the expression x² – bx + 49/4, what is b to make the expression a perfect square.
- A. 49
- B. 7
- C. 4
- D. 1
What is the square of binomial in question number 6?
- A. (x − 7/2)²
- B. (x − 7/4)²
- C. (x + 7/2)²
- D. (x + 7/4)²
Find the solution set of quadratic equation b² – 2b − 3 = 0 by completing the square.
- A. 1, -3
- B. -2, -4
- C. 2, -4
- D. -1, 3
What are the roots of the quadratic equation x² + 5x = − 6?
- A. 2, 3
- B. 3, -4
- C. -1, 6
- D. -2, -3
What must be added to x² + 22x + _____ to make it a perfect square trinomial?
- A. 100
- B. 110
- C. 112
- D. 121
WHAT’S IN
Collaboration and Critical Thinking
The concept of “perfect square trinomial” is the basis in the process of “completing of the square”. In this case, let us have a recall to illustrate this.
A trinomial is a perfect square if:
- The first and third terms are perfect squares containing positive numerical coefficients or positive numbers.
- The second term is twice the product of the square roots of the two perfect square terms.
[chart: diagram; factoring a perfect square trinomial 4x² + 12x + 9]
4x² + 12x + 9 (the terms 4x² and 9 are circled, and 12x is boxed) (2x)² under 4x²; (3)² under 9 2(2x)(3) below both
Try this!
Complete the table below.
| Trinomial | Is it a perfect square trinomial? | Factors or Reason |
|---|---|---|
| x² + 8x + 16 | Yes | (x + 4)² |
| y² + 10y − 25 | No | The sign of the third term (25) is negative. |
| p² – 24p + 144 | ||
| r² + 6r + 36 | ||
| 9z² + 24z – 16 | ||
| a² – _____ + 121 | Yes | (a − 11)² |
| _____ + 40b + 100 | Yes | (2b + 10)² |
To factor a perfect square trinomial:
- Find the positive square roots of each of the two perfect square terms.
- Connect the two squares roots with the sign of the numerical coefficient of the remaining term (the second term) of the given trinomial.
- Indicate that this binomial is used twice as a factor.
Example:
x² − 6x + 9 √x² = x √9 = 3 (x − 3)²
WHAT’S NEW
Communication, Critical Thinking, and Collaboration
An expression such as x² + 8x is an “incomplete” square. But if you add 16 to x² + 8x, we will get x² + 8x + 16 which is a perfect square trinomial and equivalent to (x + 4)² in factored form. When we add 16 to x² + 8x, we are completing the square.
Look for a pattern in the following table.
| Incomplete Square | Completed Square | Factored Form |
|---|---|---|
| x² + 4x | x² + 4x + 4 | (x + 2)² |
| y² + 6y | y² + 6y + 9 | (y + 3)² |
| b² – 8b | b² – 8b + 16 | (b – 4)² |
| m² – 10m | m² – 10m + 25 | (m – 5)² |
Notice that the constant term in the completed square is the square of half the coefficient of x. For example:
[chart: diagram; x² + 6x + 9 with annotations]
x² + 6x + 9 Half of 6 is 3 3 when squared (3²) is equal to 9
This suggests the procedure of “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.
In x² + 2bx, if we take ½(2b) or b, and square it, we will get the third term b².
Examples
Complete the square. Then write each completed square in factored form.
x² + 18x
Solution: To complete the square, add (18/2)² or 81.
Completed square: x² + 18x + 81
Factored form: (x + 9)²d² – 7d
Solution: To complete the square, add (−7/2)² or 49/4.
Completed square: d² – 7d + 49/4
Factored form: (d − 7/2)²w² − 2/3 w
Solution: To complete the square, add (−2/3 ÷ 2)² or 1/9.
Completed square: w² − 2/3 w + 1/9
Factored form: (w − 1/3)²
WHAT IS IT
Solving quadratic equations by factoring is simple and easy to do. However, there are some quadratic equation that are very difficult to factor. We can solve any quadratic equation using the method called completing the square.
Example 1: Solve x² – 8x – 9 = 0 by completing the square.
Solution:
x² – 8x – 9 = 0 (Original equation)
x² – 8x = 9 (Add 9 to both sides)
x² – 8x + 16 = 9 + 16 (Add 16 to both sides. The right side is a perfect square trinomial.)
Note: To get the constant that will complete the square take the coefficient of x, divide it by 2, then square it. We have -8 as the numerical coefficient of -8x. Thus, (−8/2)² = 16
x² – 8x + 16 = 25
(x – 4)² = 25 (Factor the trinomial square.)
x – 4 = √25 (Extract the square root of both sides.)
x – 4 = ± 5
x – 4 = 5 or x – 4 = −5 (Equate the linear expressions to each of the two values. Solve each of the resulting linear equations.)
x = 9 or x = −1
The solution set is {−1, 9}.
Checking:
For x = −1
x² – 8x – 9 = 0
(−1)² – 8(−1) – 9 = 0
1 + 8 – 9 = 0
0 = 0
For x = 9
x² – 8x – 9 = 0
(9)² – 8(9) – 9 = 0
81 – 72 – 9 = 0
0 = 0
Example 2: Solve 3x² – 4x + 1 = 0 by completing the square.
Solution:
3x² – 4x + 1 = 0 (Original equation)
3x² – 4x = −1 (Add -1 to both side of the equation.)
3x²/3 − 4x/3 = −1/3 (Divide by 3.)
x² − 4x/3 = −1/3 (The coefficient of x is −4/3.)
(−4/3 ÷ 2)² = (−2/3)² = 4/9
x² − 4/3 x + 4/9 = −1/3 + 4/9 (Add 4/9 to both sides.)
x² − 4/3 x + 4/9 = 1/9
(x − 2/3)² = 1/9 (Factor the trinomial square.)
√(x − 2/3)² = √1/9 (Extract the root of both sides.)
x − 2/3 = ±1/3
x − 2/3 = −1/3 or x − 2/3 = 1/3
x = 1/3 or x = 1
The solution set is {1/3, 1}
*Checking the solution set will serve as your exercise.
Example 3: Solve x² – 4x – 8 = 0 by completing the square.
Solution:
x² – 4x – 8 = 0 (Original equation)
x² – 4x = 8 (Adding 8 on both side of the equation.)
The coefficient of x is -4.
(−4 ÷ 2)² = (−2)² = 4
x² – 4x + 4 = 8 + 4 (Add 4 on both sides.)
x² – 4x + 4 = 12
(x – 2)² = 12 (Factor the trinomial square.)
x – 2 = ± √12 (Extract the square root of both sides.)
x – 2 = ± √4 ∙ 3
x – 2 = ±2√3
x = 2 ± 2√3
The solution set is {2 − 2√3, 2 + 2√3}
Checking:
For x = 2 – 2√3
x² – 4x – 8 = 0
(2 – 2√3)² – 4 (2 – 2√3) – 8 = 0
4 – 8√3 + 12 – 8 + 8√3 – 8 = 0
0 = 0
For x = 2 + 2√3
x² – 4x – 8 = 0
(2 + 2√3)² – 4 (2 + 2√3) – 8 = 0
4 + 8√3 + 12 – 8 – 8√3 – 8 = 0
0 = 0
WHAT’S MORE
To master your skills in solving quadratic equation by completing the square, solve and check your solutions for the following:
- x² + 6x + 5 = 0
- 2x² + 3x − 2 = 0
- x² − 4x + 12 = 0
- 5x² + 26x + 5 = 0
- 7x² − 2x = 1
WHAT I HAVE LEARNED
To have a better understanding, below are the summarized steps in “completing the square”.
- Place the constant term on the right side of the equation. All the terms with unknowns are on the left side.
- If the coefficient of the quadratic term, a ≠ 1, divide each term of the equation by the numerical coefficient.
- To get the constant term needed to complete the square, get the numerical coefficient of linear term, divide it by 2 and square it. Add the result to both sides of the equation.
- Factor the perfect square trinomial.
- Extract the square root from both sides. Two values will be obtained for the right side of the equation.
- Equate the linear expressions to each of the two values.
- Solve each of the resulting linear equations.
- Check your answer by substituting to the original equation.
WHAT I CAN DO
Find the solution set of each of the following quadratic equations using the process of “completing the square”. Show your solution.
- $x^2 + 10x + 2 = 0$
- $b^2 – 2b – 15 = 0$
- $m^2 – 7m + 12 = 0$
- $2w^2 + 2w – 4 = 0$
- $3n^2 + 24n = -48$
- $2y^2 – 3 = 5y$
- $p^2 + 2p – 15 = 0$
- $3y^2 + 12y + 12 = 0$
- $2x^2 + 4x + 1 = 0$
- $a = 2a^2 – 1$
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 asnwer.
Which of the following will make the expression $x^2 + 6x + __$ a perfect square trinomial?
a. 3
b. 6
c. 9
d. 12What is the missing term to make the expression $p^2 + _____ + 49$ a perfect square trinomial?
a. 14p
b. 4p
c. 16p
d. 9pWhich of the following will have a perfect square trinomial on one side of equation if 36 is added to both sides of equation?
a. $x^2 – 12x = -1$
b. $x^2 – 6x = -1$
c. $x^2 – 9x = -1$
d. $x^2 + 24x = -1$Which of the following is a perfect square trinomial?
a. $x^2 – 2x + 1$
b. $x^2 – 4x + 6$
c. $x^2 – 4$
d. $x^2 + 9$In the expression x2 – bx + 100, what is the value of b to make the expression a perfect square?
a. 10
b. 5
c. 15
d. 20To find the value of b in question number 5, get the ______ of the product of the square roots of the $x^2$ and 100.
a. thrice
b. once
c. twice
d. tripleWhat is the squared binomial in question number 5?
a. $(x – 5)^2$
b. $(x – 10)^2$
c. $(x + 20)^2$
d. $(x + 10)^2$Find the solution set of quadratic equation $6x^2 = 10 + 11x$ by completing the square.
a. $\frac{5}{2}, -\frac{2}{3}$
b. $-\frac{5}{2}, -\frac{2}{3}$
c. $\frac{5}{2}, \frac{2}{3}$
d. $\frac{2}{5}, -\frac{2}{3}$What are the roots of quadratic equation $x^2 + \frac{5}{3}x + 1 = 0$?
a. $\sqrt{11}$
b. $\frac{5 \pm \sqrt{11}}{8}$
c. $\frac{-5 \pm \sqrt{11}}{4}$
d. no real rootsWhat constant should be added to both sides of $x^2 + 6x – 1 = 0$ a make the left-hand side a perfect square trinomial?
a. 4
b. 6
c. 9
d. 12
E. 7m [transcriber’s note: the stray mark “E. 7m” is printed in the left margin, just above the ADDITIONAL ACTIVITIES box]
ADDITIONAL ACTIVITIES
Critical Thinking and Creativity
A. The scenario below can be represented by a quadratic equation. Write a mathematical model showing the situation then solve for what is asked on the problem. If the process of “completing the square” is appropriate to solve for the unknown, then use it. Otherwise, use other methods that you know.
Led television screens are usually measured by the length of its diagonal. The television set has 60-inch diagonal. The screen is 12 inches wider than its height. Find the dimensions of the screen.
[photo: flat-screen television with a landscape image on the screen]
Exploration
In the next module, we will solve the value of $x$ of a quadratic equation $ax^2 + bx + c = 0$ ,where $a, b, c$ and $a \neq 0$ by using a specific formula. One way of deriving this formula is by completing the square. Try it now without looking ahead!
PROBLEM – BASED WORKSHEET
An Open Box
A rectangular board is 6 in longer than its width. An open box was made by cutting 3 in. squares at each corner and folding the sides.
[photo: dark gray rectangle, a placeholder for the picture of the board]
Draw a diagram to illustrate the situation.
[transcriber’s note: an empty rectangle is printed here for the drawing]
Write expressions that represent the width, length and height of the box.
_________________________________________________________________
Suppose the volume of the box is 336 cubic in., write the mathematical sentence that will represent the situation.
__________________________________________________________________
Use completing the squares to find the dimensions of the rectangular board.
What are the dimensions of the box? __________________________________
E-Search
You may also check the following link for your reference and further learnings on solving quadratic equation using completing square.
- https://www.youtube.com/watch?v=0IFPixtQpYM
- https://www.youtube.com/watch?v=9UgvReIRsp8
- https://www.youtube.com/watch?v=prx_Bf2hakw
REFERENCES
Dugopolski, Mark.2006.*Elementary and Intermediate Algebra 2nd edition.*MCGraw-Hill.New York City
E-Math Worktext in Mathematics 9, Revised Edition by O. Oronce & M. Mendoza
Wizard Mathematics, Intermediate Algebra Worktext II by A. DIgnadice
ADDITIONAL ACTIVITIES
A. Height=36 inches Width= 48 inches B. Please see module 5
WHAT I KNOW
- C
- B
- C
- A
- B
- D
- A
- D
- D
- D
ASSESSMENT
- C
- A
- A
- A
- D
- C
- B
- A
- D
- C
PROBLEM – BASED WORKSHEET
Let x = width of the board
then x + 6 = length of the boardIf squares with side 3 in. will be cut from the corners, the box will have,
width = x − 6
length = x
height = 3[chart: diagrams of the open box and the board; unnumbered]
Labels on the board: x + 6 (length), x (width), 3 (corner squares).
Labels on the cut-out net: x − 6 (width), x (length), 3 (side folded up).
Labels on the box: x − 6, x, 3.With V = lwh, the volume of the box will be
3x(x − 6) = 336To find the dimensions on the board using completing the square:
3x² − 18x = 336
3(x² − 6x + 9) = 336 + 27
3(x − 3)² = 363
(x − 3)² = 121
x − 3 = ±11
x = 11 + 3 and x = −11 + 3
x = 14 and x = −8But a negative value of x does not fit the conditions of the problem, thus the value of x is 14. Thus, the board is 14 in. by 20 in.
[transcriber’s note: the printed line reads “co B. in in. by 20”; the letter “B” and the repeated “in” look like stray marks, so the sentence is transcribed as it reads in context]
The dimensions of the box are:
width = 8 in
length = 14 in
height = 3 in
ANSWER KEY
WHAT’S IN
| Trinomial | Is it a perfect square trinomial? | Factors or Reason |
|---|---|---|
| $x^2 + 8x + 16$ | Yes | $(x + 4)^2$ |
| $y^2 + 10y – 25$ | No | The sign of the third term (25) is negative. |
| $p^2 – 24p + 144$ | YES | $(p – 12)^2$ |
| $p^2 + 6p + 36$ | NO | The square root of 36 is ±6 and if you multiply this by 2, the coefficient of the middle term must be 12 or -12 |
| $9z^2 + 24z – 16$ | YES | $(3z + 4)^2$ |
| $a^2 – _____ + 121$ | Yes | $(a – 11)^2$ |
| $4b^2 + 40b + 100$ [?] | Yes | $(2b + 10)^2$ |
[transcriber’s note: the first trinomial in the last row is partly cut off at the edge of the cell; the reading “4b² + 40b + 100” is inferred from the printed factors]
WHAT’S MORE
- ${-5, -1}$
- ${-2, \frac{1}{2}}$
- no real solution
- ${-5, -\frac{1}{5}}$
- $\frac{1}{7} \pm \frac{2\sqrt{7}}{7}$
WHAT I CAN DO
- ${-5 \pm \sqrt{23}}$
- ${-3, 5}$
- ${4, 3}$
- ${-2, 1}$
- ${-4}$
- ${-\frac{1}{2}, 3}$
- ${-5, 3}$
- ${-2}$
- $\left{\frac{-2 \pm \sqrt{2}}{2}\right}$
- ${-\frac{1}{2}, 1}$
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