DAILY LESSON LOG OF M9AL-Ia-b-1 (Day Four)
DAILY LESSON LOG OF M9AL-Ia-b-1 (Day Four)
DAILY LESSON LOG OF M9AL-Ia-b-1 (Day Four)
The teacher explains to the students that writing perfect square trinomials as a square of
B. Establishing a purpose for
binomial is one of the important procedure that will lead them to solve quadratic
the lesson
equations that cannot be solved by either factoring or the square root method.
The teacher lets the students stay in their respective groups and do Activity 3: (Make It
Perfect!!!), which is found on page 37 of the Learner’s module.
Answer Key:
1. 1 2. 100
3. 64 4. 144
C. Presenting examples/
121
5. 225 6.
instances of the new lesson 4
225 441
7. 8.
4 4
1 9
9. 10.
9 64
The teacher discusses with the students the process of arriving at the answer of Activity 3.
Furthermore, he/she asks the students about the mathematical skills or principles that
they used to get the correct answers. He/she tells them that to solve quadratic equations
D. Discussing new concepts using Completing the Square method, you have to simply make the expression x 2 + bx into
and practicing new skills #1 a perfect square by simply adding the square of one-half the coefficient of the second
term or
()
b 2
2
.
E. Discussing new concepts and The teacher discusses and illustrates thoroughly the steps on how to solve quadratic
practicing new skills #2 equations by Completing the Square as presented on page 38-41 of the Learner’s Module.
F. Developing mastery (leads to
formative assessment 3)
The teacher asks the students to answer the problem below with their group mates.
An open box is to be formed out of a rectangular piece of cardboard whose length is 8 cm
longer than its width. To form the box, a square of side 4 cm will be removed from each
corner of the cardboard. Then the edges of the remaining cardboard will be turned up.
1. Draw a diagram to illustrate the given situation.
G. Finding practical applications 2. How would you represent the dimensions of the cardboard?
of concepts and skills in daily w = width(in cm), w + 8 = length
living 3. What expressions represent the length, width, and height of the box?
w, w - 8, and 4
4. If the box is to hold 448 cm 3, what mathematical sentence would represent the given
situation?
4w(w - 8) = 448 w 2 – 32 w = 448 w2 – 8w = 112
5. What are the dimensions of the rectangular piece of cardboard?
width = 4 + 8√ 2 cm; length = 12 + 8√ 2 cm
The teacher summarizes the lesson by asking the students to answer the following
questions:
1. When are you going to use Completing the Square Method in solving a quadratic
equation? When the given quadratic equation cannot be solved using factoring or
extracting the square root method.
2. How do you solve quadratic equations by completing the square? Enumerate the steps
on how to solve quadratic equations by completing the square.
Divide both sides of the equation by a then simplify.
H. Making generalizations and Write the equation such that the terms with variables are on the left side of the
abstractions about the lesson equation and the constant term is on the right side.
Add the square of one-half of the coefficient of x on both sides of the resulting
equation. The left side of the equation becomes a perfect square trinomial.
Express the perfect square trinomial on the left side of the equation as a square
of a binomial.
Solve the resulting equation by extracting square the square root.
Solve the resulting linear equations.
Check the solutions obtained against the original equation.
The teacher lets the students answer individually the formative assessment.
A. Find the solutions of each of the following quadratic equations by completing the
square.
1. x2 – 2x = 3
I. Evaluating Learning 2. r2 – 10r = -17
Answer Key
1. x = -1 or x = 3
2. r = 5 + 2√ 2 or r = 5 - 2√ 2
J. Additional activities or
remediation
V. REMARKS
VI. REFLECTION
Determine a number that must be added to make each of the following a perfect square trinomial.
Summative Assessment
A. Find the solutions of each of the following quadratic equations by completing the square.
1. x2 – 2x = 3
2. r2 – 10r = -17