Philippine Normal University Visayas

The National Center for Teacher Education

The Environment and Green Technology Education Hub
Cadiz City, Negros Occidental

A Detailed Lesson Plan

In
MATHEMATICS: GRADE 9
(RADICAL EQUATION)
Pre-demonstration
I. Objective:
At the end of a 60-minute discussion, the students will be able to solve equation involving radical
expressions.

II. Subject Matter:

A. Topic: Radical Equation
B. Mathematical Concept:
 Isolate radicals
 Combine similar terms.
 Remove the radical sign
 Solve the resulting equation.
 Check the solutions.

C. Value Focus: Patience and cooperation
D. Materials: Visual Aids, Pictures, Manila Paper, Pen & Marker, Activity Sheets, Envelope
E. References: Grade 8 Math textbook, pp. 214 – 221;

Teacher’s Activity Students Activity

A. Review
Good Morning class. Today we will be
having our new lesson. But before that. Let’s
have a recap on the topic we have discussed Good morning maam.
last meeting.

Anybody from the class who would like to

Student raise their hand.
share what he/she had learned about the topic
last meeting? Our topic last meeting was all about

Very good! Let’s give ourselves him/her

claps.

B. Motivation
Now, let’s have an activity I will divide the
class into two groups. Each group will have a
leader and a secretary. I have here the
materials and instructions that you need. You
only have 5 minutes to finish the activity, once
you’re done post your work on the board
immediately. The group who will finish it first
will get an incentive afterwards.
 Am I clear? Yes ma’am.

You may now start.

Group 1
Instruction: Inside the cross-word puzzle
find 3 words which you think related to our
new topic.

C B A L L S I S M T
A C T S A P I O N S
F R H H J C I O N S
I N V O L V I N G D
K G O A L T T D A R
E Q U I A P P T A S
L I K U E G S T A R
U P Q U I C K S R E
E E L S T R I H T S

Group 2
Instruction: Inside the cross-word puzzle
find 3 words which you think related to our
new topic.

E C B A L L S I S M T
X A C T A A P I O N S D
P F R H C J C I O N O O
O I N V I L V I N G L N
S K G O D L T T D A V E
I E Q U A A P P T A I S
E E X P R E S S I O N S
T U P Q U I C K S R G G
I E E L S T R I H T D S

What words did you find in the crossword The words are RADICAL, EQUATION and
puzzle? INVOLVING, RADICAL, EXPRESSION and
SOLVING

Students clapping.
Okay, very well! Let’s give ourselves 10 claps.
Yes ma’am.
Are you now ready for our new lesson?
 C. Presentation

Now class, what do you think is our topic for Our topic for today is all about Radical Equation.
today?

Very good class. Our topic for today is Radical

Equation.

Let’s read our objective for today, all together. . Objective:

At the end of a 60-minute discussion, the
students will be able to solve equation involving
radical expressions.

Today, you will learn more about solving radical

equation.

What is radical equation? Radical equation is an equation where the variable is

found underneath a square root, cube root or higher
root. Or in other words, an equation that contains a
variable in a radicand.

What is radicand? Radicand is the number or expression under a radical

symbol.

Here are some examples and non-examples of a

radical equations.
Examples of Radical Equations Non‐Examples of Radical
Equations
√x 5 11 √5 x2 11
3 √x 4 7 x −4 4 √16
4 √x −7 12 28 x √10−7 12 28
5 √x 225 x4 3 √27

What do we know about solving equations?

We will isolate the variables.

Right, we do need to isolate the variables. Let’s

look at an equation,
√x=6
For what value of x would you substitute to make
the equation a true statement? 36 ma’am.

The √36=6.
 1. Activity
Now, we will have a group activity. I will Students counting.
group you into three groups. Count off!

Now go to your groups and settle in. Work

silently and everyone should cooperate. You
are given 15 minutes only to finish the activity.
Am I clear? Yes ma’am.

Group 1 Group 1
Directions: Solve the following radical equation
by following the steps. Solutions:
Checking
Steps in solving radical equation.
1. Isolate the radical on the left side by applying 1. √x+5 =11 √x+5 =11
the appropriate properties of equality. (√x+5)2 = (11)2 √116+5=11
2. Combine similar terms when possible. x+5 = 121 √121 = 11
3. Remove the radical sign by raising both sides x= 116 √11=11
of the equation to the index of the radical. (e.g. 2
for √, 3 for 3√)
4. Solve the resulting equation. 2. √3x+3 - √5x-1 = 0 Checking
5. Check the solutions for a possible √3x+3=√5x-1 √3x+3 - √5x-1 =0
presence of extraneous roots. (√3x+3)2=(√5x-1)2 √3(2)+3 - √5(2)-1 =0
3x+3=5x-1 √6+3- √10-1=0
3x-5x=-1-3 √9 -√9 =0
1. √x+5 =11 -2x= -4 3-3=0
-2 -2 0=0
2. √3x+3 - √5x-1 = 0 x=2

3. √x= -3
Group 2
Group 2 Solutions:
Directions: Solve the following radical equation
Checking
by following the steps.
1. √x-4=2 √x-4=2
Steps in solving radical equation. (√x-4)2= (2)2 √8-4=2
1. Isolate the radical on the left side by applying x-4=4 √4=2
the appropriate properties of equality. x=4+4 2=2
2. Combine similar terms when possible. x=8
3. Remove the radical sign by raising both sides
of the equation to the index of the radical. (e.g. 2 2. 3√3x-1 = 2 Checking
for √, 3 for 3√) (3√3x-1)2 =(2)3 3√3x-1 = 2

3x-1 = 8 3√3(3)-1 = 2
4. Solve the resulting equation.
3x=8+1 3√9-1 = 2
5. Check the solutions for a possible
3x=9 3√8 = 2
presence of extraneous roots.
3 2=2
x=3
1. √x-4 = 2 3. 3+ 5√x =2x checking if x=1/4
5√x =2x-3 3+ 5√x =2x
2. 3√3x-1 = 2 (5√x)2 =(2x-3)2 3+ 5√1/4 =2(1/4)
25x=4x2-12x+9 3+ 5(1/2)=2/4
3. 3+ 5√x =2x 4x2-12x+9 =0 3+5/2=1/2
(4x-1) (x-9)=0 11 ≠ 1
4x-1=0 or x-9=0 2 2
x=1/4 or x=9

Group 3 Group 3
Directions: Solve the following radical equation
by following the steps. 1. √x+8 = 3 checking
(√x+8)2 =(3)2 √x+8 = 3
Steps in solving radical equation. x+8=9 √1+8 = 3
1. Isolate the radical on the left side by applying x=9-8 √9=3
the appropriate properties of equality. x=1 3=3
2. Combine similar terms when possible.
3. Remove the radical sign by raising both sides 2. 1+ √2x+3 = 6 checking
of the equation to the index of the radical. (e.g. 2 √2x+3 = 6-1 1+ √2x+3 = 6
for √, 3 for 3√) √2x+3 = 5 √2(11)+3 = 6-1
4. Solve the resulting equation. (√2x+3)2 =(5)2 √22+3 = 6-1
5. Check the solutions for a possible 2x+3=25 √25=5
presence of extraneous roots. 2x=25-3 5=5
2x=22
1. √x+8 =√ 3 2
x=11
2. 1+ √2x+3 = 6

3. √a-5 = -2

2. Analysis

Let’s take a look at the work of group 1.

Okay thank you group 1. Let’s give them excellent Students execute an excellent clap.
clap. Everybody.

How about group 2?

Students clap their hands 10 times.
Okay thank you group 2. Let’s give them 10 claps.

How about group 3? Students execute an excellent clap.

Good job group 3, you also deserve to get an

excellent clap. Students clap their hands 10 times.

All of you had a correct answer.

 3. Abstraction

 How did you find the activity? Answers may vary.

 Did you have any difficulty in Answers may vary

answering the activity?

 What have you realized from the Answers may vary

activity?

 How did you solve radical By following the steps in solving radical equation.
equation?

 What are the steps in solving radical Steps in solving radical equation.
equation?
1. Isolate the radical on the left side by applying the
appropriate properties of equality.

2. Combine similar terms when possible.

3. Remove the radical sign by raising both sides of

the equation to the index of the radical. (e.g. 2 for
√, 3 for 3√).

4. Solve the resulting equation.

5. Check the solutions for a possible presence

of extraneous roots.

4. Application
To check if you really understood our
discussion, I have here something for you to
work on. I need 2 volunteers each column
(seating arrangement).

Direction: Solve the given equation using the

steps and fill in its corresponding reasons.
 STEPS REASON STEPS REASON
√2x-1 + 5 = 8 Given Equation √2x-1 + 5 = 8 Given Equation
√2x-1 + 5-5 = 8-5 Subtraction property
Isolate square root of equality
√2x-1 = 3 Isolate square root
2x-1=9 (√2x-1)2 = (3)2 Square both sides to
eliminate the square
Division property of root
equality 2x-1=9 simplify
2x-1+1=9+1 Addition property of
equality
2x=10 Division property of
check 2 2 equality
X=5 simplify
√2x-1 + 5 = 8
√2(5)-1 + 5 = 8
√10-1 + 5 = 8 check
√9 + 5 = 8
3+5=8

STEPS REASON STEPS REASON

x-1=√5x-9 x-1=√5x-9 Given equation
(x-1)2 = (√5x-9)2 (x-1)2 = (√5x-9)2 Square both sides to
Binomial expression eliminate the square
Multiplied binomial root.
(x-1)(x-1)=5x-9 Binomial expression
(x-5)(x-3)=0 factored x2-2x+1=5x+9 Multiplied binomial
x2-7x+10=0 Set equal to zero
(x-5)(x-3)=0 factored
Check for extraneous x=5 or x=3 Used zero product
solutions property
x-1=√5x-9
5-1=√5(5)-9
4=√16
2≠√6
Check for extraneous
3-1=√5(3)-9 solutions
2=√4
2=2
 Very Good class! Is there any clarification
or question regarding our lesson today?
Okay, if none please answer the test No ma’am.
silently.

5. Evaluation

A. Multiple Choice. Choose the correct answer

in each of the following.
1.
1.

2.

3.

FALSE

TRUE
2.

3.
 FALSE

TRUE

V. Assignment/Agreement:
Directions: Write at least three application problems
using the skills learned in special product.

I. Developmental Activities: