SS TWO FIRST TERM MATHS NOTE pdf.pdf

1
WKS SCHEME OF WORK FOR SS 2
FIRST TERM (MATHEMATICS)
1. Logarithms: (i) revision of logarithm of numbers greater than one (ii) characteristics of logarithm
of numbers greater than one and less than one and standard form of numbers (iii) logarithm of
numbers less than one; multiplication, division power and roots. (iv) Solution of simple
logarithmic equation (v) Accuracy to use of logarithm table in calculation.
2. APPROXIMATION: - (i) rounding up and down of numbers to significant figures, decimal places
and nearest whole number. (ii) Application of approximation to everyday life (iii) percentage
error (iv) application of percentage error of everyday life
3. ALGEBRAIC FRACTION 1: (i) simplification of fraction (ii) operation in algebraic fraction
4. ALGEBRAIC FRACTION 2: (i) equation involving fraction (ii) substitution in fraction (iii)
simultaneous equation involving fraction and undersigned value of faction.
5. SEQUENCE AND SERIES 1 (AP):- (i) first term (a) common difference (d) and the nth term of an
AP (ii) arithmetic sum and mean of an AP (iii) practical problem solving involving real life
situation on arithmetic mean of an. (a) practical problem solving involving real life situation on
sum of AP.
6. SEQUENCE AND SERIES 2 (GP): (i) first term (a), common ratio (r) and nth term of a GP (iii) sum
of geometric mean and sum of term of a GP (iii) sum of infinity of GP (iv) practical problem
involving real life situation on GP.
7. QUADRATIC EQUATION1: (i) revision of factorization and perfect squares. (ii) quadratic equation
using completing the squares method (iii) deducing the quadratic formula from completing the
squares and its application to solving problems.
8. Quadratic equation 2: (i) graphical method of solving equation (ii) word problems leading to
quadratic equation (iii) application of quadratic equation to real life equation.
9. SIMULTANEOUS LINEAR AND QUADRATIC EQUATION 1: (i) revision of simultaneous linear and
quadratic equation (ii) simultaneous linear and quadratic equation by elimination (iii)
simultaneous linear and quadratic equation by substitution (iv) word problem leading to
simultaneous linear and quadratic equation
10. Simultaneous linear and quadratic equation 2: (i) Revision of linear and quadratic graph (ii)
graphical method
11. Gradient of a curve: (i) revision of straight line graph (ii) gradient of a straight line (iii) drawing
target to curve (iv) determination of gradient of a curve.
12. Revision
13. Examination

2
THEME:
DATE:
CLASS:
TIME:
PERIOD:
DURATION:
SUBJECT: Mathematics
TOPIC: LOGARITHM
SPECIFIC OBJECTIVES: At the end of the lesson the student should be able to:-
i. Express numbers on standard form
ii. Obtain characteristics of the logarithm from the standard form obtained on (i)
iii. Use mathematics tables to obtain the decimal fraction
iv. Write down the logarithm of numbers greater than one
v. Write down the logarithm of numbers less than one
INSTRUCTIONAL RESOURCES: (i) a chart showing the standard form A x 10n
(ii) mathematics tables
STEP 1:- IDENTIFICATION OF PRIOR IDEAS.
MODE: Individual
Teachers Role:- Leads the students to express numbers in standard form (numbers greater than one) (ii)
the teacher leads the students to express a number less than one in standard form (iii) the teacher
explain that in practice, the characteristics of a number less than one in written in ‘bar’ notation
Examples
465 = 4.65 x 102
(standard form)
= 100.6675
x 102
(from table)
= 100.6675 + 2
(indices)
= 102.6675
Log 465 = 2.6675
Also
0.00541 = 5.41 x 0.001

3
= 5.41 x 10-3
(standard form)
= 100.7332
x 10-3
(from tables)
= 10-3 + 0.7332
Log 0.00541 = - 3 + 0.7332
Note that in practice the logarithm of 0.00541 and read ‘bar 3 point 7332 the logarithm 3.7332 has a
negative integer 3 and a positive fraction (0.7332). as before the integer of a logarithm can be found by
expressing the given number in standard form
STUDENTS ACTIVITIES
I. The teacher gets the students to participate in expressing the number in standard form
II. Read the decimal fraction from the mathematics tables
STEP II: EXPLORATION
MODE: INDIVIDUAL
Teachers Role/Activities:
i. The teacher leads the students to express numbers in standard form
ii. Find the characteristics of numbers greater than one and less than one
Students’ activities
i. The teacher gets the students to participate in the activities
STEP III: DISCUSSION
MODE: whole class
Teacher’s role:
i. The teacher leads the students to obtain the logarithm of number
ii. He also lead the students to check the decimal fractions from the mathematical tables
Examples:
i. Find the logarithm of (a) 0.03862 (b) 0.00000032
Solution:
a. Log 0.03862 = 2.5868
b. Log 0.00000032 = 7.5051
Students Activities

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i. The students will be made to state the characteristics in 0.03862 and log 0.00000032
ii. The students will be made to find from the mathematical table the decimal fractions .5868
and .5051
STEP IV: APPLICATION
Mode: Group
Teacher's role: The teacher gives the students a related problem to solve in group, while he goes round
to help those that specific difficulties
1. Express the following in standard form and state the characteristics of the logarithm of the number
a. 4325
b. 0.00002548
c. 0.3424
Students Role: the teacher ensures that the groups are doing or solving the problems above
STEP V EVALUATION
Mode: Individual
Teachers Role: the teacher gives the students a related problem to solve individually while he monitors
their progress helping those with specific difficulty.
i. Use tables to find the logarithm of the following numbers
a. 0.006642 b. 0.00075 c. 34.5
Assignment: use antilogarithm tables to find the number whose logarithm are a. 0.3645 b. 2.4997 c.
5.8226
References: (1) New general Mathematics for SSS book 2 page 22-23
(2) STAN Mathematics for SSS book 2 page 5-7

5
THEME:
DATE:
CLASS:
TIME:
PERIOD:
DURATION:
SUBJECT: mathematics
TOPIC: logarithm
SUB-TOPIC: multiplication & division with numbers less than 1
SPECIFIC OBJECTIVES: at the end of the lesson, the students should be able to
i. Compare the characteristics of logarithm and standard form of numbers as expressed on
power of 10.
ii. Determine the characteristic of any given number without first expressing the number in
standard form
iii. Solve problems involving multiplication and division with numbers less than one
iv. Apply the laws of malice when multiplying logarithms of numbers.
INSTRUCTIONAL RESOURCES: logarithm table and collectors
LESSON PRESENTATION:
STEP 1: identification of prior ideas
MODE: individual
TEACHERS ROLE: teacher leads the students to identify numbers less than one.
EXAMPLE: write the following numbers in standard form
O.5678=5.678x10-1
O.05678=5.678x10-2
Characteristics of 0.5678 =1 and0.05678=2
STUDENTS ACTIVITIES: the teacher directs the students to write the characteristics of these numbers.
(i) 0.349 (ii) 34.9 (iii) 0.0000349

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STEP 2: Exploration
Mode: whole class
TEACHERS ROLE:- the teacher leads the students to explain further determine the characteristics of
each of the following numbers
(i) 0.0502 = 2
(ii) 0.5261 = 1
(iii) 0.0011 = 3
(iv) 0.00000843 = 6
STUDENTS ROLE: the teacher tells them to answer these questions. Find the logarithm of each of the
following numbers (i) 0.2885 (ii) 0.002886
STEP 3: DISCUSSION
Mode : whole
TEACHER’S ROLE: Teacher leads the student to solve this question. Evaluate the following
(i) 0.4523 x 0.02472
(ii) 0.04523 x 0.006392
Solution
1. No Log
0.4523 1.6554
0.02472 2.3931
2. 0485
Antilog 0.0118
STUDENTS ROLE: Teacher tells the students to solve the second question above
Evaluate 0.04523 x 0.006392
Mode: Individual
TEACHER’S ROLE: Teacher continues on his explanation
Evaluate 0.0453 ÷ 0.006392
NO. Log
0.0453 2.6554
0.006392 3.805

7
7.-076 0.8498
STUDENTS’ ACTIVITY: teacher ask the students to solve the question
i. evaluate 463x00049
STEP 5: Evaluation
MODE: individual
TEACHERS R OLE: the teacher evaluates the lesson by given them some question
i. Express 0,00567 in standard form and obtain its characteristic
ii. What is the characteristic in log 0.0567
iii. Evaluate 463 X 0.000493
iv. Evaluate 0.0043 ÷ 4.006
v. Use the laws of indices in (3) and (4) above
STUDENT ROLE: teacher asks the student to solve the problem above.
Use logarithm table to evaluate each of the figure
i. 0.626x2.383
ii. 6.735x0.8945
iii. 1.428x84.37
REFERENCES :-(i) new general mathematics for sss book 2 page 22-23
(ii)STAN mathematics for sss book 2 page 5-7

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THEME:
DATE:
CLASS:
TIME:
PERIOD:
DURATION:
TOPIC: LOGARITHM
SPECIFIC OBJECTIVES: At the end of the lesson the student should be able to:-
vi. Express numbers on standard form
vii. Obtain characteristics of the logarithm from the standard form obtained on (i)
viii. Use mathematics tables to obtain the decimal fraction
ix. Write down the logarithm of numbers greater than one
x. Write down the logarithm of numbers less than one
INSTRUCTIONAL RESOURCES: (i) a chart showing the standard form A x 10n
(ii) mathematics tables
STEP 1:- IDENTIFICATION OF PRIOR IDEAS.
MODE: Individual
Teachers Role:- Leads the students to express numbers in standard form (numbers greater than one) (ii)
the teacher leads the students to express a number less than one in standard form (iii) the teacher
explain that in practice, the characteristics of a number less than one in written in ‘bar’ notation
Examples
465 = 4.65 x 102
(standard form)
= 100.6675
x 102
(from table)
= 100.6675 + 2
(indices)
= 102.6675
Log 465 = 2.6675
Also
0.00541 = 5.41 x 0.001

9
= 5.41 x 10-3
(standard form)
= 100.7332
x 10-3
(from tables)
= 10-3 + 0.7332
Log 0.00541 = - 3 + 0.7332
Note that in practice the logarithm of 0.00541 and read ‘bar 3 point 7332 the logarithm 3.7332 has a
negative integer 3 and a positive fraction (0.7332). as before the integer of a logarithm can be found by
expressing the given number in standard form
STUDENTS ACTIVITIES
III. The teacher gets the students to participate in expressing the number in standard form
IV. Read the decimal fraction from the mathematics tables
STEP II: EXPLORATION
MODE: INDIVIDUAL
Teachers Role/Activities:
iii. The teacher leads the students to express numbers in standard form
iv. Find the characteristics of numbers greater than one and less than one
Students’ activities
ii. The teacher gets the students to participate in the activities
MODE: whole class
Teacher’s role:
iii. The teacher leads the students to obtain the logarithm of number
iv. He also lead the students to check the decimal fractions from the mathematical tables
Examples:
ii. Find the logarithm of (a) 0.03862 (b) 0.00000032
Solution:
c. Log 0.03862 = 2.5868
d. Log 0.00000032 = 7.5051
Students Activities

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iii. The students will be made to state the characteristics in 0.03862 and log 0.00000032
iv. The students will be made to find from the mathematical table the decimal fractions .5868
and .5051
Mode: Group
Teacher's role: The teacher gives the students a related problem to solve in group, while he goes round
to help those that specific difficulties
1. Express the following in standard form and state the characteristics of the logarithm of the number
d. 4325
e. 0.00002548
f. 0.3424
Students Role: the teacher ensures that the groups are doing or solving the problems above
STEP V EVALUATION
Mode: Individual
Teachers Role: the teacher gives the students a related problem to solve individually while he monitors
their progress helping those with specific difficulty.
ii. Use tables to find the logarithm of the following numbers
b. 0.006642 b. 0.00075 c. 34.5
Assignment: use antilogarithm tables to find the number whose logarithm are a. 0.3645 b. 2.4997 c.
5.8226
References: (1) New general Mathematics for SSS book 2 page 22-23
(2) STAN Mathematics for SSS book 2 page 5-7

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THEME:
DATE:
CLASS:
TIME:
PERIOD:
DURATION:
TOPIC: logarithm
SUB-TOPIC: multiplication & division with numbers less than 1
SPECIFIC OBJECTIVES: at the end of the lesson, the students should be able to
v. Compare the characteristics of logarithm and standard form of numbers as expressed on
power of 10.
vi. Determine the characteristic of any given number without first expressing the number in
standard form
vii. Solve problems involving multiplication and division with numbers less than one
viii. Apply the laws of malice when multiplying logarithms of numbers.
INSTRUCTIONAL RESOURCES: logarithm table and collectors
STEP 1: identification of prior ideas
MODE: individual
TEACHERS ROLE: teacher leads the students to identify numbers less than one.
EXAMPLE: write the following numbers in standard form
O.5678=5.678x10-1
O.05678=5.678x10-2
Characteristics of 0.5678 =1 and0.05678=2
STUDENTS ACTIVITIES: the teacher directs the students to write the characteristics of these numbers.
(ii) 0.349 (ii) 34.9 (iii) 0.0000349

12
STEP 2: Exploration
Mode: whole class
TEACHERS ROLE:- the teacher leads the students to explain further determine the characteristics of
each of the following numbers
(v) 0.0502 = 2
(vi) 0.5261 = 1
(vii) 0.0011 = 3
(viii) 0.00000843 = 6
STUDENTS ROLE: the teacher tells them to answer these questions. Find the logarithm of each of the
following numbers (i) 0.2885 (ii) 0.002886
STEP 3: DISCUSSION
Mode : whole
TEACHER’S ROLE: Teacher leads the student to solve this question. Evaluate the following
(iii) 0.4523 x 0.02472
(iv) 0.04523 x 0.006392
Solution
2. No Log
0.4523 1.6554
0.02472 2.3931
2. 0485
Antilog 0.0118
STUDENTS ROLE: Teacher tells the students to solve the second question above
Evaluate 0.04523 x 0.006392
Mode: Individual
TEACHER’S ROLE: Teacher continues on his explanation
Evaluate 0.0453 ÷ 0.006392
NO. Log
0.0453 2.6554
0.006392 3.805

13
7.-076 0.8498
STUDENTS’ ACTIVITY: teacher ask the students to solve the question
ii. evaluate 463x00049
STEP 5: Evaluation
MODE: individual
TEACHERS ROLE: the teacher evaluates the lesson by given them some question
vi. Express 0,00567 in standard form and obtain its characteristic
vii. What is the characteristic in log 0.0567
viii. Evaluate 463 X 0.000493
ix. Evaluate 0.0043 ÷ 4.006
x. Use the laws of indices in (3) and (4) above
STUDENT ROLE: teacher asks the student to solve the problem above.
Use logarithm table to evaluate each of the figure
iv. 0.626x2.383
v. 6.735x0.8945
vi. 1.428x84.37
REFERENCES :-(i) new general mathematics for sss book 2 page 22-23
(ii)STAN mathematics for sss book 2 page 5-7

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THEME: Numerical processes
DATE:
CLASS:
TIME:
PERIOD:
DURATION:
TOPIC: Logarithm
SUB-TOPIC: Logarithm of Numbers Less Than One, Power and Roots
SPECIFIC OBJECTIVES: at the end of the lesson, the student should be able to
i. Find the characteristic of a logarithm of number.
ii. find the mantissa
iii. Use anti logarithm table.
iv. Multiply and divide the logarithm of number lees than 1.
v. Evaluate problem on power at roots.
INSTRUCTIONAL RESOURCES: mathematical table and calculator.
STEP 1: identification of prior ideas:
MODE: individual
TEACHER’S ROLE: the teacher ask the student to say the characteristic of the following logarithms
(a) 0.075 (b) 0.0075 (c) 0.0075 (d) 0.00000075
STUDENTS’ ROLE: the student supplies the answer to the question above; i.e.
(a)2(b)3(c)4(d)7
STEP 2: Exploration
MODE: whole class.
TEACHER ROLE: the teacher leads the student to solve a problem on power.
Example: evaluate 0.67253
SOLUTION
NO Log
0.725 1. 8277
0.67253
1.8277x3=1. 4831
0.3042 1.4831
:- 0.67253
= 0.3042
STUDENTS’ ROLE: the student supplies all the values in solving the problems in the example above i.e.
getting the characteristics and the mantissa of 0.6725.
STEP 3: Discussion
MODE: whole class
TEACHER’S ROLE: the teacher lead the students to solve a problem involving root while the participate
in supplying the Values OF THE logarithm.
Example 2: evaluate √0.06357

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SOLUTION
NO log
0.06357 2.8033
√0.006357 2.8033÷2=1.4017
0.2522 1.4017
0.06357=0.2522
STUDENTS’ ROLE: The students participate in the supply of the characteristics and mantissa and also
in the division of bar notation. The teacher gives the students a problem on multiplication and
division of bar notation.
Simplify the following:-
(1)4.3x2 (2)5.8÷3
SOLUTION
(i) 4. 3 x 2 (2) 5.8 ÷ 3
(1) 4.3 x 2 = 2 (4+ 0.3)
= 8 + 0.6
= 8.6
(2) 5.8 ÷ 3 = 3) 5.8
3) 5 + 0.8
3) 6 + 1.8
2+ 0.6
2.6
MODE: whole class
TEACHER’S ROLE: the teacher leads the students in a applying the concept of power and root in a
more challenging problem
Examples 3: Evaluate by using tables
√
0.0763
309 𝑋 0.008465
4
SOLUTION:
No Log
0.0763 2.8825
309 2.4900
0.008465 3.9277
309 x 0.008465 0.4177
NUM 2.8825 – 0.4177 = 2.4648
+

16
den
2.4648 ÷4 = 1.6162
√
𝑁𝑢𝑚
𝐷𝑒𝑛
4
0.4132 1.6162
√
0.0763
309 𝑋 0.008465
4
= 0.4132
STUDENTS’ ROLE:- as before the students responds to question arising from the solution above
STEP IV :- EVALUATION
MODE: Individual
TEACHER’S ROLE: the teacher gives the students a similar work to do individual
1. Find the characteristics in the logarithm of 0.6784
2. Use the log table to find the mantissa of the lo g 0.6784
3. Use the antilog table to find the number whose logarithm is 2 .1563
4. Use table to evaluate 0.67842
5. Use table to evaluate √0.00489
3
STUDENTS’ ROLE: - Teacher instructs the students to solve the problem above.
ASSIGNMENT: - The Teacher gives the students a problem to solve at home i.e.
EVALUATE
18.62 x 0.00735
0.03842
REFFERENCES: -
1. New general Mathematics for Senior Secondary 2 by M.F macraeetal pages 21-22.
2. Man Mathematics for SS2 Page 5-7.

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DATE:
CLASS:
TIME:
PERIOD:
DURATION:
UNIT TOPIC: Logarithm of numbers less than one
LESSON TOPIC: Multiplication, division, power and roots of numbers less than one
SPECIFIC OBJECTIVES: At the end of the lesson the students should be able to:
1. Find the characteristics of logarithm
2. Find mantissa of logarithm
3. Use antilog table
4. Evaluate problems on multiplication, division, power and roots
INSTRUSTIONAL RESOURCES: Mathematical table and calculator
STEP1: Identification of prior ideas
MODE: Individual
TEACHER’S ROLE: The teacher ask the students to say the characteristics of the following numbers less
than one: (a) 0.5 (b) 0.05 (c) 0.005
STUDENTS’ ROLE: The students respond to the question above: (a) 1 (b) 2 (c) 3
STEP 2: EXPLORATION
MODE: - Whole Class
TEACHER’S ROLE: - The teacher leads the students to solve 0.00574 x 0.56782
SOLUTION:
No Log
0.00574 3.7589
0.56782
1. 7542 x 2= 1. 5084
0.00574 x 0.56782
3. 2673
STUDENTS ROLE: - The teacher ask the students to supply all the values
on the log column of the example above
STEP III: - Discussion
MODE: - Whole Class
TEACHERS ROLE: - The teacher teach the students to solve the problem 0.42162
x 3.45
SOLUTION:
No Log
0.42162 1. 6249 x 2= 1.2498
3. 45 0.5778 0.5778
0.42162
x 3.45 3.4797
STUDENTS’ ROLE: - The students supply the logarithm of the numbers above
STEP IV: - APPLICATION
MODE: - Whole Class
TEACHER’S ACTIVITIES: - The teacher leads the students to above a problem involving multiplication,
Division, Power and roots following

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EXAMPLE: use table to evaluate
√
0.00574𝑥0.56782
0.42162x 3.45
5
SOLUTION:
NO Log Log
0.00574 3.7589 3.7589
0.56782 1.7542 x 2 1. 50844
Num 3.2673
0.42162
1.6249 x 2 1.2498
3.45 0.5778 0.5778
Den 1.8276
𝑁𝑢𝑚
𝐷𝑒𝑛
3.2673
1.8276
√
𝑁𝑢𝑚
𝐷𝑒𝑚
5
3.3397 ÷ 5 1.4957
0.3132 1.4959
STUDENT’S ROLE: - The students respond to all the question answering from the solution above.
STEP V: - Evaluation:
MODE: Individual
TEACHER’S ROLE: - The teacher gave the students a similar problem to solve.
1. Given √
0.75𝑋0.0453
0.00639
3
a. Find the characteristics in log 0.75, log 0.045 and log 0.00639
c. Find the mantissa in log 0.75, log 0.045 and log 0.00639
d. Add the logarithm of the numerator
e. Subtract the logarithm of the numerator by the logarithm of the denominator (law of
indice)
f. Use the antilog table to evaluate the given problem
SOLUTION:
No Log Log
-
+
+

19
STUDENT’S ROLE: - The
student solve the
problem above.
ASSIGNMENT: - The teacher gives the students additional problem
to solve at home.
Use table to evaluate
0.67842𝑋 √0.00489
3
0.0057
REFFERENCES: - (1) New general Mathematics for Senior Secondary 2 by M.F. macraeetal pages 21-22
(2) Stan mathematics for SS 2 pages 5-7.
0.75 1.8751 1.8751
0.0453
2.6532 x 3 5.9596 +
Num 5.8347-
0.00629 3.7987 3.7987
(
𝑁𝑢𝑚
𝐷𝑒𝑚
)
1
3
2.0360÷ 3 = 1.3453
0.2215 1. 3453

20
THEME: Number and Numeration
DATE: -
CLASS: -
TIME: -
PERIOD: -
DURATION: 40 minutes
SUBJECT: - Mathematics
TOPIC: -logarithms
SUB- TOPIC: Solution of simple logarithmic equation
SPECIFIC OBJECTIVES: - At the end of the lesson, the student should be able to: -
i. State the theory of logarithm
ii. Express logaN =x in the form N = ax
iii. Simplify ( 11) above
iv. Solve problems on simple logarithms equations
INSTRUCTIONAL RESOURES: - the theory of logarithm written on a cardboard sheet
LESSON PRESENTATION: -
STEP I: - Identification of prior ideas
MODE: - Individual
TEACHER’S ROLE: - The teacher ask the students to list the forms of equations they are familiar with e.g
simple linear, quadratic, simultaneous, cubic equations etc
STUDENTS’ ROLE: - The students attempt to list the equations as in above.
Step II: Exploration
Mode: whole class
TEACHER’S ROLE: the teacher reminds the students that the statement 1000 = 103
and log1000 = 3 are
two ways of writing the same thing. Also 8 = 23
, then log2 8 = 3 etc.
STUDENTS’ ROLE: the students are made to give more examples of the the statements above and
deduce that in general if N = ax
, then loga N = x.
STEPIII: discussion
MODE: entire class
TEACHER’s ROLE: the teacher leads the students to apply the theory of logarithm to simple logarithmic
equation.
Example 1: solve the following equations for x. (a) logx 64 =2 (b) x =log5 25 (c) log10 x = 5
Solution: (a) logx 64 =2
∴ 64 = x2
82
= x2
Equating the base gives
X = 8
(b) x =log5 25 (c) log10 x = 5
25 = 5x
x = 105
52
= 5𝑥
X = 2

21
STUDENTS’ ROLE: the students contribute in the application of the theory in the problems above.
STEP 1V: application
MODE : the entire class
TEACHER’S : the teacher leads the students to apply the theory of logarithm in solving more logarithmic
equation.
Example2: solve the following equations for x. (a) 2logx (3
3
8
)= 6 (b) log10x = - 2
Solution: 2logx (3
3
8
)= 6
(
27
8
)2
= x6
[ (
3
2
)3
]2
= x6
(
3
2
)6
= x6
∴ x =
3
2
(b) log10x = - 2 X = 10-2
X=
1
102 X =
1
100
STUDENTS’ ROLE: the students supplies all the requirements in the solving the problems above.
STEP V : Evaluation
MODE: Individual
TEACHER’S ACTIVITY: the teacher gives the students similar work to do.
1. state the theory of logarithm
2. Express(a) log2 x = 7 (b) log x 32 = 5 (c) log 3 9 = x in the form N = 𝑎𝑥
3. Simplify the form N = 𝑎𝑥
(a), (b) and (c)
4. Solve (a), (b) and (c) above
SOLUTION: (a) log2 x = 7
X = 27
(b) log x 32 = 5
X = 128 32 = x5
25
= x5
x = 2
(c) log 3 9 = x
9 = 3x
32
= 3x
X =2
STUDENTS’ ACTIVITIES: the students `solves the problems above under the teacher supervision
ASSIGNMENT: the teacher gives the students assignment to do at home.
Solve the following equations
(a) log10 x= -3 (b) logx 81 = 4.
REFERENCES: (1) new general mathematics for ss2 by M. F macrae et al page 24 (2) STAN mathematics
for ss2

22
DATE: -
CLASS: -
TIME: -
PERIOD: -
DURATION: - minutes
SUBJECT: - Mathematics
TOPIC: - approximation
SUB- TOPIC: rounding up and down
SPECIFIC OBJECTIVES: - At the end of the lesson, the student should be able to: -
i. Round numbers to the nearest tens
ii. Round numbers to certain significant figures
iii. Round numbers to a certain decimal places
iv. Round numbers to the nearest whole number.
v. Round numbers to certain significant figures
INSTRUCTIONAL RESOURES: - Measuring devices like tapes and chalk board ruler
PRESENTATION: -
STEP I: - Identification of prior ideas
MODE: - Individual
TEACHER’S ROLE: - The teacher ask the students why they think it is sometimes necessary to estimate or
approximate in calculation and to list all the forms of rounding up and down they know eg to the
nearest whole number, to 1 decimal places etc.
STUDENTS’ ROLE: - The students supply the answers to the questions asked.
STEP II: exploration
MODE: the entire class
TEACHER’s ROLE: the teacher explains that approximation are very important when calculating.
Numbers can be approximated in order to obtain rough estimate of calculations. For example if the tax
deducted from person’s salary is ₦93. 80k per month, the amount deducted in a year is ₦93. 80x12. A
rough estimate is ₦90x10 = ₦900. Rough estimates are not accurate. However they give an idea of size
or magnitude of the correct result of calculation. Thus they are often used to check the correctness of
answers to calculation.
STUDENTS’ ROLE: the students listen attentively and ask questions on areas of doubt
STEP III: discussion
TEACHER’s ROLE: the teacher also explains that number can be rounded off to the nearest tents, whole
numbers etc or to a given number of decimal places and significant figures. The digits 1,2,3 and 4 are
rounded down and the digits 5,6,7,8,9 are rounded up.
STUDENTS’ ROLE : the students listens attentively asking questions on area doubt.
STEP 1V: application

23
TEACHER’s ROLE: the teacher leads the students to solve problems.
Example 1: round off the numbers 341. 774 to (a) 2d.p (b) 1 dp (c) to the nearest whole number (d) the
nearest hundred
Solution: (a) 341. 774 = 341.77 to 2dp
(b) 341. 774 = 341.8 to 1dp
(c) 341. 774 = 342 to the nearest whole number
(d) 341. 774 = 300 to the nearest hundred
Example 2: round off 164. 90 to 1, 2, 3, and 4 significant figures
Solution: (1) 164.90 = 200 to 1 s.f
(2) 164.90 = 160 to 2 s.f.
(3) 164.90 = 165 to 3 s.f
(4) 164.90 = 164.9 4s.f
Example 3 : round off the number o.oo9281 to 1, 2 and 3 significant figures
Solution: (1) 0.009281 = 0.009 to 1 s .f
(2) 0.009281 = 0.0093 to 2 s. f
(3) 0.009281 = 0.00928 to 3 s .f
STUDENTS’ ACTIVITIES: the students supply all the answers in the three examples above under the
leadership of the teacher.
STEP V: Evaluation
MODE: Individual
TEACHER’s ACTIVITY: the teacher gives the students a similar work to do
Round off 82.7502 to (1) 1 d.p (2) the nearest tens (3) the nearest whole number (4) 3 d. p
(2) Round off 129. 7 and 0.0005239 to (a) 1 s.f (b) 2 s. f (3) 3 s .f (4) 4 s. f
Solution: 1 (a) 82.7502 = 82.8 to 1 d.p
(b) 82.7502 = 80 to nearest tens
(c) 82.7502 = 83 to the nearest whole number
(d) 82.7502 = 82.750 to 3 d.p
2 (a) 129.7 = 100 to 1 s.f (b) 129.7 = 130 to 2 s. f (c) 129.7 = 130 to 3 s. f (d) 129.7 = 129. 7 to 4 s. f
(e) 0.0005239 = 0.0005 s.f (f)0.0005239 = 0. 00052 s.f (g) 0.0005239 = 0.000524
(h) 0.0005239 = 0.0005239 4 s. f
STUDENTS’ ROLE: the students solve the problems above.
ASSIGNMENT: the teacher gives assignment to do at home
Round the following to 4 and 5 s.f
(a) 165.623 (b) 0.00692634
REFERENCES: (1) new general mathematics for sssbk 2 by M.F macrae et al pages 44- 49 (2)
comprehensive mathematics for sss by D.B Adu.

24
THEME: Number and numeration
DATE:
CLASS:
TIME:
DURATION:
UNIT TOPIC: Approximation
LESSON TOPIC: Percentage error
SPECIFIC OBJECTIVES: at the end of the lesson the students should be able to
(1) Find the range of actual measurement.
(2) Find the error in the measurement
(3) State the formula for calculating percentage error.
(4) Substitute in the formula to calculating the percentage error.
INSTRUCTIONAL RESOURCES: A chart explaining how to find the range of measurements.
STEP I: Identification of prior ideas
MODE: Individual
TEACHER’S ROLE: the teacher revises the last lesson on approximation/ rounding up and down by asking
the students to round off 0.007258 to 1d.p, 2 s.f. etc.
STUDENTS ROLE: the students supplies the answers to the questions above i.e. 0.007 and 0.0073 etc.
STEP II: Exploration
TEACHER’S ROLE: the teacher explains that no measurement however carefully made is exact.
Measurements are approximate. If the width of a room is measured as 4.6m to 2 s.f, the actual width of
the room may lie between 4.55m and 4.65m. The error of this measurement is 4.55 – 4.6 or 4.65 – 4.6 ie
± 0.05.
STUDENTS’ ROLE: the students listen attentively and respond to questions that the teacher could ask in
the course of his explanations.
STEP III: Discussion
MODE : the entire class

25
TEACHER’S ROLE: the teacher uses his chart to explain actual measurement and state the formula for
percentage error ie percentage error =
𝑒𝑟𝑟𝑜𝑟
𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑚𝑒𝑛𝑡
x 100%
In the example above percentage error = ±
0.05
4.6
x 100%= ± 1.08696%
= ± 1.1 % ( to 2 s.f)
Suppose the width of the same room is measured to the nearest centimeter as 4m 60cm. the range of
actual measurement is between 4m 59.5cm and 4m 60.5cm.
Error = ± 0.5cm
Percentage error = ±
0.5
460𝑐𝑚
x 100%
= ± 0.1o8696%
= ± 0.11% (to 2 s.f)
From the above it can be seen that the percentage error is smaller when the centimeter is used. Thus
when measuring, the degree of accuracy improves as the unit of measurement decreases in size.
STUDENTS’ ROLE: the students listen and supplies answers to calculations done above.
STEP IV: application
TEACHER’S ROLE: the teacher leads the students to solve some problems
Example 1: the distance from school to the market is measured as 9.8km to 1d.p. what is the range of its
distance. Calculate the percentage error.
Solution: since the measurement is given to 1 d.p. the error = ± 0.05km
% error =
𝑒𝑟𝑟𝑜𝑟
𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑚𝑒𝑛𝑡
x 100%
= ±
0.05𝑘𝑚
9.8𝑘𝑚
x 100%
= ± 0.510204%
= ± 0.5% (to 1 d.p)
Example 2: the circumference of a circle is measured and given as 500m. Find the percentage error if the
length is measured.
(a) To the nearest metre
(b) To the nearest 10m
(c) to one significant figure

26
Solution: (a) since the measurement is done to the nearest metre, the range of actual measurement lies
between 499.5m and 500.5m
Error = ± 0.5m
% error = ±
0.5𝑚
500𝑚
x 100%
= ± 0.1%
(b) The range of actual measurement is between 495m and 505m
∴ ERROR = ± 5m
% error =
± 5m
500𝑚
x 100%
= ± 1%
(c) The range of actual measurement is between 450m and 550m
∴ ERROR = ± 50m
% error =
± 5m
500𝑚
x 100%
= ± 10%
STUDENTS’ ROLE: the students contributed to the solutions above and say what they have observed
from the three answers above.
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives a similar problem to the students to solve
Question: 1. a pole is 125cm high. Find the of the actual length of the pole
2. Find error in the measurement above
3. State the formula for calculating percentage error
4. Using the formula stated above calculate the percentage error in (1)
SOLUTION: the range of the actual measure is between 122.5cm to 127.5cm
Error = ± 2.5
% error =
± 2.5cm
125𝑐𝑚
x 100%
= 2%
The range of actual measurement is between 124. 5 To 125.5
Error = ± 0.5

27
% error =
± 0.5cm
125𝑐𝑚
x 100%
= ± 0.4%
The range of actual measurement lie between 124.5 to 125.5cm
Error = ± 0.5cm
% error =
± 0.5cm
125𝑐𝑚
x 100%
= ± 0.4%
STUDENTS’ ROLE: The student solves the problem as above while the teacher supervises their work.
ASSIGNMENT: A student draws a line and says that it is 10cm long. When it is carefully measured, the
true length is 10.2cm. What is the % error in the drawing?
REFERENCES: new general mathematics for sssbk 2 by M.F macrae et al pages 44- 47.

28
DATE:
CLASS:
TIME:
PERIOD:
DURATION:
UNIT TOPIC: approximation
LESSON TOPIC: percentage error
SPECIFIC OBJECTIVES: At the end of the lesson, the students should be able to:
(1) Find errors of calculations
(2) State the formula for percentage error
(3) Substitute correctly in the formula above
(4) Simply to obtain the percentage error
INSTRUCTIONAL RESOURCES: A chart carrying the formula for calculating percentage error
STEP I: Identification of prior ideas.
MODE: Whole class
TEACHER’S ROLE: The teacher ask the students to state the formula for percentage error.
Percentage error =
𝒆𝒓𝒓𝒐𝒓
𝒎𝒆𝒂𝒔𝒖𝒓𝒆𝒎𝒆𝒏𝒕
x 100 %.
But that if the true value of a measurement is known, then
% error =
𝒕𝒓𝒖𝒆 𝒗𝒂𝒍𝒖𝒆
x 100 %.
STUDENTS’ ROLE: the students give the formula of percentage error as in above.

29
MODE: The entire class
TEACHER’S ROLE: The teacher leads the students to solve a problem in which the true value is given
Example 1: the length of a stick is 8cm. a student measured the length as 8.5cm.
Find the percentage error in the measurement.
Solution: error in the measurement = (8.5 − 8) cm
= 0.5cm
% error =
𝒕𝒓𝒖𝒆 𝒗𝒂𝒍𝒖𝒆
x 100 %.
=
𝟎.𝟓𝒄𝒎
𝟖𝒄𝒎
x 100 %.
= 6.25%
STUDENTS’ ROLE: The students listens attentively and supplies the necessary ingredients needed in the
calculation above ie the students find the error 0.5cm, states the formula for percentage error and lead
in the substitution
STEP III: Discussion
MODE: The entire class
TEACHER’S ROLE: The teacher leads the students to solve another problem.
Example 2: A square is 10cm by 10cm. A girl measure a side of the square as 9.9cm and used it to
calculate the area of the square. Find the percentage error in (a) the length of the side (b) the area of of
the square
Solution: (a) error in measurement = (9.9 − 10) cm
= - 0.1cm
% error in length =
− 𝟎.𝟏𝒄𝒎
𝟏𝟎𝒄𝒎
x 100 %
= - 1%
= 1 %
(b) Actual area of square = (10𝑥10) cm2
= 100 cm2
Calculated area = (9.9 − 9.9) cm2
= 98.01cm2
Error in the area = (9.801 − 100) cm2
= - 1.99cm2

30
% error in area =
− 𝟏.𝟗𝟗𝐜𝐦𝟐
𝟏𝟎𝟎𝐜𝐦𝟐
x 100 %.
= - 1.99%
= 1.99%
STUDENTS’ ROLE: The students supply the values of the error, actual area, calculated area and the error
in area.
STEP IV: Application
TEACHER’s ROLE: the teacher leads the students to solve a more challenging problem.
Example 3: A surveyor measures a road as being 69.3km long, however there is a – 1 % error in this
measurement. What is the true length of the road?
Solution: let the true length be x
Percentage error =
𝐞𝐫𝐫𝐨𝐫
𝐭𝐫𝐮𝐞 𝒍𝒆𝒏𝒈𝒉𝒕
x 100 %.
=
𝟔𝟗.𝟑𝐤𝐦−𝐱𝐤𝐦
𝒙𝒌𝒎
x 100 % = - 1%
=
𝟔𝟗.𝟑−𝒙
𝐱
x 100= - 1
= 100(69.3 − 𝑥) = - x
6930 – 100x = - x
6930 = - x + 100x
6930 = 99x
X =
𝟔𝟗𝟑𝟎
𝟗𝟗
X = 70km
∴The true length of the road is 70km.
STUDENTS’ ROLE: The students help in the solving the problem by supplying all the values needed.
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives the students a similar work to do
(1) A pencil is 18cm long. Someone measures its length as 20cm. find the percentage error.
(2) State the formula you will use to calculate the percentage error in (1) above
(3) Substitute correctly in the formula above
(4) Calculate the percentage error

31
STUDENTS’ ROLE: the students solve the problem above.
ASSINGMENT: the teacher gives the following problem as assignment.
Question: A stick is 20cm long. A student makes a 5% error in measuring the stick. Find two possible
values for the students’ measurement.
REFERENCES: New general mathematics for senior secondary school book2 by M.F. macrae et al pages
45- 48
THEME: Algebra
DATE:
CLASS:
TIME:
DURATION:
UNIT TOPIC: Algebraic fractions

32
LESSON TOPIC: simplification of algebraic fractions
(1) Factorize algebraic expression
(2) Identify common factors
(3) Use the concept of difference of two squares
(4) Simplify algebraic expressions
INSTRUCTIONAL RESOURCES: A chart showing difference of two squares and steps involved in factoring
quadratic expressions.
MODE: Individual
TEACHER’S ROLE: The teacher gives the students some expressions to factorizee.g.h2
– hk and m2
– 2am
– 3a2
and h2
–k2
.
STUDENTS’ ROLE: the students respond to the teachers question by solving the three problems above
i.e.
(1) h2
– hk = h(h – k )
(2) m2
– 2am – 3a2 =
m2 – 3am + am – 3a2
= m( m – 3a) + a(m + 3a)
= (m – 3a) (m + 3a).
(3) h2
– k2
= (h + k ) (h – k ) difference of two squares
MODE: Whole class
TEACHER’S ROLE: the teacher leads the students to explain the three types of factorization use above
explaining the steps involved in each.
STUDENTS’ ROLE: the students listen attentively and ask questions in areas of doubt.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to simplify some algebraic fractions.

33
Example 1: simplify
𝐡𝟐 – 𝐡𝐤
𝒉𝒌
Solution:
𝐡𝟐 – 𝐡𝐤
𝒉𝒌
=
𝐡(h – k)
𝒉𝒌
=
𝒉−𝒌
𝒌
Example 2: simplify
𝐡𝟐 – 𝐤
(h−k)2
Solution:
h2 – k2
(𝐡−𝐤)𝟐 =
(ℎ−𝑘)(ℎ+𝑘)
(ℎ−𝑘)(ℎ−𝑘)
(difference of two squares)
=
𝒉+𝒌
𝒉−𝒌
STUDENTS’ ROLE: the students supplies the answers in each step in the simplification above
MODE: Whole class
TEACHER’S ROLE: the teacher leads the students to solve more challenging problems
Example 3: simplify
9a2 – m2
m2 −2am −3a2
Solution
9a2 – m2
m2 −2am −3a2=
(3a)2 – m2
m2 −2am −3a2
=
(3a−m)(3a+m)
(m−3a)(m+a)
=
(3a−m)(3a+m)
−(m−3a)(m+a)
= −
(3a+m)
a+m
Example 4: simplify
u2 – 5uv + 6v2
u2 + uv − 12v
Solution:
u2 – 5uv+ 6v2
u2 + uv − 12v2=
u2 – 3uv −2uv+ 6v2
u2 +4uv −3uv− 12v2
=
u(u –3v)−2v(u−3v)
u(u+4v)−3v(u+4v)
=
(u−3v)(u−2v)
(u+4v)( u−3v)

34
=
u−2v
u+4v
STUDENTS’ ROLE: the students factorize the algebraic expressions in all the cases in the examples above.
STEPIV: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives the students similar work to do in the class.
1. What is the common factor in the expression 𝑥2
- 2x
2. Factorize 𝑥2
- 2x
3. Use the concept of difference of two squares to factorize 𝑑2
– 9
4. Simplify as far as possible the following
(a)
x2− 2x
x2− 4
(b)
d2 −9
d2−7d+12
(c)
8− 2a –a2
2a2 – 3a −2
STUDENTS’ ROLE: the students solve the problem above.
Solution:
(1)
x2− 2x
x2− 4
=
x(x−2)
x2−22 =
x(x−2)
(x−2)(x+2)
= =
x
x+2
ASSIGNMENT: the teacher gives the students assignment.
Simplify
a2 – am – an +mn
a2 – am+ an−mn
REFERENCE: new general mathematics for sss 2 by M.F macrae et al pages 193- 194

35
THEME: Algebraic processes
DATE:
CLASS:
TIME:
DURATION:
LESSON TOPIC: operation in algebraic fractions
(1) Factorize quadratic expressions
(2) Factorize algebraic expressions using difference of two squares
(3) Multiply algebraic fractions
(4) Divide algebraic fractions
INSTRUCTIONAL RESOURCES: A chart showing difference of two squares and steps involved in factoring
quadratic expressions.

36
TEACHER’S ROLE: the teacher ask the students to factorize the following (1) 𝑛2
− 9 (2) 𝑛2
− 3𝑛 + 2
STUDENTS’ ROLE: the students the expressions above i.e.
(1) 𝑛2
− 9 = (n + 3)( n – 3)
(2) 𝑛2
− 3𝑛 + 2 = ( n – 2)(n-1)
MODE: Whole class
TEACHER’S ROLE: the teacher informs the students that algebraic fractions can be multiplied or divided
and gives examples
Example 1:simply
𝑛2− 9
𝑛2− 𝑛
x
𝑛2− 3𝑛+2
𝑛2+ 𝑛−6
Solution:
𝑛2− 9
𝑛2− 𝑛
x
𝑛2− 3𝑛+2
𝑛2+ 𝑛−6
=
(𝑛+2)(𝑛−3)
𝑛(𝑛−1)
x
(𝑛−2)(𝑛−1)
(𝑛+3)(𝑛−2)
=
𝑛−3
𝑛
STUDENTS’ ROLE: the students gives the factorized form of 𝑛2
− 𝑛 and 𝑛2
− 𝑛 + −6 and participated in
the simplification
MODE: whole class
TEACHER’S ROLE: the teacher again informs the students that it is possible to also divide algebraic
fractions.
Example 1: simplify
𝑒2− 5𝑒+6
𝑒2+ 2𝑒−3
÷
3𝑒−9
2𝑒2+ 6𝑒
Solution: to divide an algebraic fraction, multiply by its reciprocal i.e.
𝑒2− 5𝑒+6
𝑒2+ 2𝑒−3
÷
3𝑒−9
2𝑒2+ 6𝑒
=
(𝑒−2)(𝑒−3)
(𝑒+3)(𝑒−1)
𝑥
2𝑒(𝑒+3)
3(e−3)
=
2𝑒(𝑒−2)(𝑒−3)(𝑒+3)
3(𝑒+3)(𝑒−1)(𝑒−3)
=
2𝑒(𝑒−2)
3(𝑒−1)
STUDENTS’ ROLE: the students gives the factorized forms (e -2) (e -3), (e +3) (e-1), 2e (e+3) and 3(e-3)
and participate in the simplification
TEACHER’S ROLE: the teacher leads the students to solve a problem involving both multiplication and
division.

37
Example 3: simplify
𝑎2+𝑎𝑏− 2𝑏2
𝑎2− 2𝑎𝑏− 3𝑏2 x
𝑎2− 𝑏2
𝑎𝑏+2𝑏2 ÷
𝑎2− 2𝑎𝑏+𝑏2
𝑎2− 3𝑎𝑏
Solution:
(𝑎+2𝑏)(𝑎−𝑏)
(𝑎+𝑏)(𝑎−3𝑏)
x
(𝑎+𝑏)(𝑎−𝑏)
𝑏(𝑎+2𝑏)
x
𝑎(𝑎−3𝑏)
(𝑎−𝑏)(𝑎−𝑏)
=
𝑎
𝑏
STUDENTS’ ROLE: the students factorized the algebraic expression in the problem above.
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the gives the student’s problem to solve
1. Factorize 𝑢2
+ 3u – 10
2. Use difference of two squares to factorize 𝑎2
- 𝑏2
3. Simplify
𝑎2−𝑏2
𝑎𝑏+𝑎2 x
2𝑎3
𝑎𝑏−𝑎2
4 simplify
𝑢2+3𝑢+ 10
3 𝑢2+12𝑢
÷
𝑢2− 25
𝑢2−𝑢−20
STUDENTS’ ROLE: the students solve the problems above.
Solution: (a)
𝑎2−𝑏2
𝑎𝑏+𝑎2 x
2𝑎3
𝑎𝑏−𝑎2 =
(𝑎−𝑏)(𝑎+𝑏)
𝑎(𝑏+𝑎)
x
2𝑎3
𝑎𝑏−𝑎2
(b)
𝑢2+3𝑢+ 10
3 𝑢2+12𝑢
÷
𝑢2− 25
𝑢2−𝑢−20
=
(𝑢+5)(𝑢−2)
3𝑢(𝑢+4)
x
(𝑢−5)(𝑢+4)
(𝑢+5)(𝑢−5)
=
𝑢−2
3𝑢
ASSIGNMENT: the teacher gives the students assignment.
Simplify
5
𝑑−4
+
2
𝑑+4
REFERENCES: new general mathematics for sss bk2 by M.F. macrae et al pages 194-195

38
DATE:
CLASS:
TIME:
PERIOD:
DURATION:
LESSON TOPIC: Operation in algebraic fractions
(1) Expand algebraic fractions
(2) Find L C M of the denominators of fraction
(3) Add algebraic fractions
(4) Subtract algebraic fractions
(5) Simplify to obtain the simplified forms.
INSTRUCTIONAL RESOURCES: LCM of algebraic fractions written on a cardboard sheet
MODE: Individual
TEACHER’S ROLE: the teacher test the understanding of LCM by the students by asking them to find the
LCM of
1
2
+
1
3
and
1
2𝑒
+
3
3𝑓
etc.

39
STUDENTS’ ROLE: the student’s responds to the teachers question i.e. the LCM of 2 and 3 is 6 and the
LCM of 2e and 5f is 10ef etc.
MODE: Whole class
TEACHER’S ROLE: the teacher leads the students in applying the concept of LCM in simplifying algebraic
fractions
EXAMPLE 1: simplify
4
5𝑑
+
1
3𝑒
SOLUTION: the LCM of 5d and 3e is 15de
∴
4
5𝑑
+
1
3𝑒
=
4
5𝑑
x
3𝑒
3𝑒
+
7
3𝑒
x
5𝑑
5𝑑
=
12𝑒+
15𝑑
+
35𝑑
15𝑑𝑒
=
12𝑒+35𝑑
15𝑑𝑒
STUDENTS’ ROLE: the students supplies the LCM of 5d and 3e i.e. 15de and participate in the
simplification.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve another problem
EXAMPLE 2: simplify
5
𝑑−4
+
2
𝑑+4
Solution:
5
𝑑−4
+
2
𝑑+4
=
5
𝑑−4
x
𝑑+4
𝑑+4
+
2
𝑑+4
x
𝑑−4
𝑑−4
=
5(𝑑+4)
(𝑑+4)(𝑑−4)
+
2(𝑑−4)
(𝑑+4)(𝑑−4)
=
5𝑑+20+2𝑑− 8
(𝑑+4)(𝑑−4)
=
7𝑑+12
𝑑2− 16
STUDENTS’ ROLE: the students supply the LCM of (d – 4) and (d +4) and use it to simplify the problem.
TEACHER’S ROLE: the teacher leads the students to solve more problems.
EXAMPLE 3: simplify
3𝑎−𝑏
5𝑎𝑏
-
2𝑏+3𝑐
6𝑏𝑐
+
3𝑐−2𝑎
15𝑎𝑐

40
SOLUTION: the LCM of 5ab, 6bc and 15ac is 30abc.
Express each fraction with denominator 30abc
3𝑎−𝑏
5𝑎𝑏
x
6𝑐
6𝑐
-
2𝑏+3𝑐
6𝑏𝑐
x
5𝑎
5𝑎
+
3𝑐−2𝑎
15𝑎𝑐
x
2𝑏
2𝑏
=
6𝑐(3𝑎−𝑏)
30𝑎𝑏𝑐
-
5𝑎(2𝑏+3𝑐)
30𝑎𝑏𝑐
+
2𝑏(3𝑐−2𝑎)
30𝑎𝑏𝑐
=
18𝑎𝑐−6𝑏𝑐−10𝑎𝑏−15𝑎𝑐+6𝑏𝑐−4𝑎𝑏
30𝑎𝑏𝑐
=
3𝑎𝑐−14𝑎𝑏
30𝑎𝑏𝑐
=
𝑎(3𝑐−14𝑏)
30𝑎𝑏𝑐
STUDENTS’ ROLE: the students find the LCM of 5ab, 6bc and 15ac I.e. 30abc and helps the teacher to
simplify the fractions
STEP IV: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives the students a similar problem to solve
1. Expand 4( u –v)
2. Find the LCM in
1
4(𝑢−𝑣)
-
1
5(𝑣−𝑢)
3. Add (3) above
4. Subtract
1
2𝑏−𝑎
from
𝑎+𝑏
(𝑎−2𝑏)2
5. Simplify (3) and (4) to its simplest form
STUDENTS’ ROLE: the students solves the problem above
ASSIGNMENT: Simplify
4𝑚−9𝑛
16𝑚2 -
9𝑛2+1
4𝑚−3𝑛
REFERENCES :(1) New general mathematics for sss book 2 by M.F. macrae et al pages 195 – 197
(2) New concept mathematics for sss schools 2 by HN odogwu et al

41
THEME: Algebra processes
DATE:
CLASS:
TIME:

42
PERIOD:
DURATION:
LESSON TOPIC: Equations with fractions
(1) Find least common multiples of denominators
(2) Clear fractions by multiplying by the LCM
(3) Form an equation from (3)
(4) Solve the equation to find the value of the unknown
INSTRUCTIONAL RESOURCES: examples of LCM of some algebraic expressions written on a cardboard
sheet.
MODE: Individual
TEACHER’S ROLE: the teacher highlights the concept of LCM and ask the students to say what they
understand by equation and what it means to solve an equation
STUDENTS’ ROLE: the students’ responds to the teacher’s question and explain that to solve an
equation means that the value of the unknown is found.
MODE: Whole class
TEACHER’S ROLE: the teacher leads the students to solve and explain in details an equation with fraction
Example 1: solve
3
𝑐−2
=
3
2𝑐−5
SOLUTION:
3
𝑐−2
=
3
2𝑐−5
Multiply each fraction by (c – 2)(2c -5) i.e.
3
𝑐−2
x (c -2)(2c – 5) =
3
2𝑐−5
x (c – 2)(2c – 5)
3(2c – 5) = 3(c – 2)
6c – 15 = 3c – 6

43
6c – 3c = - 6 +15
3c = 9
3𝑐
3
=
9
3
C = 3
STUDENTS’ ROLE: the students supplies the LCM (2c – 5)(c – 2) and help the teacher in the simplification
at each step.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve a more challenging problem
EXAMPLE 2:solve
11
𝑚+3
=
5
2𝑚
-
1
𝑚−4
SOLUTION:
11
𝑚+3
x2m (m + 3)(m – 4) =
5
2𝑚
x2m (m +3)(m – 4) -
1
𝑚−4
x 2m(m+ 3)(m – 4)
22m (m -4) = 5(m+3)(m-4) – 2m(m +3)
22m2
– 88m = 5m2
– 5m – 60 – 2m2
– 6m
19m2
– 77m + 60 = 0
19m2
– 57m – 20m + 60= 0
19m (m – 3) – 20 (m- 3) = 0
(19m – 20)(m-3) = 0
19m -20 = 0 or m-3 =0
M =
20
19
or m = 3
STUDENTS’ ROLE: the students again supplies the LCM 2m (m+3) (m-4), helps the teacher to expand,
factorize and simplify the problem at each step.
TEACHER’S ROLE: the teacher leads the students to solve more problems.
EXAMPLE 3:solve
4𝑎−1
𝑎+4
- 2 =
2𝑎−1
𝑎+2
SOLUTION:
4𝑎−1
𝑎+1
x (a + 4)(a + 2) - 2(a + 4)(a + 2) =
2𝑎−1
𝑎+2
x (a + 4)(a + 2)
(4a -1)(a + 2) – 2(a + 4)(a - 2) = (2a -1)(a +4)

44
4a2
- a + 8a -2 – 2a2
– 12a -16 = 2a2
– a + 8a – 4
-12a – 14 = 0
- 12a = 14
a= -
14
12
= -
7
6
STUDENTS’ ROLE: the students supplies the LCM (a+4)(a+2) and helps the teacher to simplify the
problem at each step.
STEP IV: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives the students a similar problem to do.
Given the equation
2
𝑑−2
=
3𝑑
4𝑑+12
1. Find the LCM of the denominators
2. Clear the fractions by multiplying both sides by the LCM
3. Form a simple linear equation in (2) above
4. Solve the equation in (3) to find the value of d
Solution: (1)
2
𝑑−2
x (d- 2)(4d +12) =
3𝑑
4𝑑+12
x (d -2)(4d +12)
2(4d + 12) = 3d (d – 2)
8d + 24 = 3d2
– 6d
0 = 3d2
– 6d – 8d – 24
0 = 3d2
– 18d + 4d – 24
0 = 3d (d – 6) + 4(d – 6)
0 = (3d +4)(d – 6)
Either 3d +4 = 0 or d – 6 = 0
d= -
4
3
or d= 6
ASSIGNMENT: solve the equation
2
𝑢+2
=
2
𝑢+1
-
1
𝑢+4
REFERENCES: (1) New general mathematics for sss book 2 by M.F. macrae et al pages 198 – 199

45
DATE:
CLASS:
TIME:
PERIOD:
DURATION: minutes
LESSON TOPIC: simultaneous equations involving fraction
(1) Find least common multiples (LCM)
(2) Clear fractions by multiplying by the denominator by the LCM
(3) Form linear equations
(4) Solve the two linear equations simultaneously
(5) use elimination or substitution method

46
INSTRUCTIONAL RESOURCES: least common multiples of a algebraic expressions written on a cardboard
sheet
MODE: Individual
TEACHER’S ROLE:the teacher ask the students to find the LCM of 1and2, 2, 3 and 6, 1 and 3, 2 and 4
STUDENTS’ ROLE: the students supply the answers i.e. 2, 6, 3 and 4 as LCM of the numbers above.
MODE: whole class
TEACHER’S ROLE: the teacher ask the students to say in their opinion why they think we need the LCM
in solving simultaneous equations involving fractions
STUDENTS’ ROLE: the students’ responds to the teacher’s question above. I.e. we use the LCM to clear
the fraction.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve problem.
EXAMPLE 1:solve the pair of equation
× -
𝑦
2
= 1 and
𝑥
2
+
𝑦
3
=2
5
6
Solution:first clear the fraction
× -
𝑦
2
= 1 (1)
𝑥
2
+
𝑦
3
=2
5
6
(2)
(1) x2, 2x – y = 2 (3)
(2) x6, 3x + 2y =17 (4)
Solving by elimination
(3) X 2, 4x – 2y = 4 (5)
(4) X 1, 3x +2y= 17 (6)

47
Add 7x = 21
X = 3
Subst. x = 3 to find y in (5)
4(3) – 2y = 4
- 2y = 4 – 12, 2y = - 8, Y = 4
STUDENTS’ ROLE: the students pays attention and responds to the teachers in the in the example
above.
TEACHER’S ROLE: the teacher leads the students to solve another example.
EXAMPLE 2: solve
1
3
(m – 3m) = 2 and
𝑚+𝑛
4
=
1
2
Solution:
1
3
(m – 3m) = 2 (1)
𝑚+𝑛
4
=
1
2
(2)
(1) x 3, m – 3n = 6 (3)
(2) x 4, m + n = 2 (4)
Subtract - 4n= 4
n= - 1
subst. n= -1 in (3)
m – 3(-1)= 6
m +3= 6
m= 6-3= 3
∴ m=3 and n= -1
STUDENTS’ ROLE: the students give each step in the solution above.

48
STEP IV: Evaluation
MODE: whole class
Given the equations x+
𝑦
2
=
1
2
and
𝑥
2
-
𝑦
6
=1
1
2
1. Find the LCM of the denominator in each case
2. Use LCM to clear the fractions
3. Form the linear equation arising from (2)
4. Solve the two linear equations in (2) simultaneously
5. Use either elimination or substitution method to solve (4) above
Solution: (1) x+
𝑦
2
=
1
2
(1)
𝑥
2
-
𝑦
6
=1
1
2
(2)
(1) X 2, 2x+y= 1 (3)
(2) X 6, 3x –y= 9 (4)
Adding, 5x= 1o
X= 2
Subst. x= 2 in (3)
2(2) + y= 1
4+y= 1
y= 1- 4
y=-3
∴ x= 2 and y=-3
ASSIGNMENT: solve
2
𝑒
-
3
𝑓
= 1 and
8
𝑒
+
9
𝑓
=
1
2
REFERENCES: (1) New general mathematics for sss book 2 by M.F. macrae et al pages 73– 75(2) New
concept mathematics for sss schools 2 by HN odogwu et al

49
DATE:
CLASS:
TIME:
PERIOD:
DURATION: minutes

50
LESSON TOPIC: substitution in fractions
(1) Interpret x:y as
𝑥
𝑦
(2) Write x: y= 9:4 as
𝑥
𝑦
=
9
4
(3) Substitute examples in (2) in a given fraction
(4) Simplify (3) above
INSTRUCTIONAL RESOURCES: A chart showing the ratio x:y = 9:4 is written as
𝑥
𝑦
=
9
4
MODE: whole class
TEACHER’S ROLE: the teacher ask the students to explain the concept of ratio like x:y = 2:3
STUDENTS’ ROLE: the students answers the question posed by the teacher
MODE: whole class
TEACHER’S ROLE: the teacher explain the concept of ratio that x:y = 2:3 can be written as
𝑥
𝑦
=
2
3
STUDENTS’ ROLE: the students pay attention to grasp the concept of ratio as explained above.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve examples
Example 1: Given x:y = 2:7, evaluate
7𝑥+𝑦
𝑥−
1
𝑦
𝑦
(Wace)
SOLUTION: if x:y= 2:7, then
𝑥
𝑦
=
2
7

51
Divide the numerator and denominator of
7𝑥+𝑦
𝑥−
1
𝑦
𝑦
by y i.e.
7𝑥+𝑦
𝑥−
1
𝑦
𝑦
=
7(
𝑥
𝑦
) + 1
𝑋
𝑌
−
1
7
Subst.
2
7
for
𝑥
𝑦
=
7(
2
7
) + 1
2
7
−
1
7
=
2+1
1
7
= 3x7
= 21
STUDENTS’ROLE: the students participate in the simplification in each step in the solution.
EXAMPLE 2: if x=
3𝑤−1
𝑤+2
, express
2𝑥−3
3𝑥−1
in terms w.
SOLUTION: Subst.
3𝑤−1
𝑤+2
for x in
2𝑥−3
3𝑥−1
2𝑥−3
3𝑥−1
=
2 (
3𝑤−1
𝑤+2
)−3
3(
3𝑤−1
𝑤+2
)−1
=
2(3𝑤−1)−3(𝑤+2)
3(3𝑤−1)−(𝑤+2)
=
6𝑤−2−3𝑤−6
9𝑤−3−𝑤−2
,
=
3𝑤−8
8𝑤−5
STUDENTS’ ROLE: the students pay attention and contribute their quota at each state of the solution
above.
STEP V: Evaluation
MODE: Individual

52
1. Interpret x: y
2. Write p:q = 9:5 in the form
𝑝
𝑞
=
9
5
3. Substitute
𝑝
𝑞
=
9
5
in
15𝑝−2𝑞
5𝑝+16𝑞
4. Simplify (3) above
STUDENTS’ ROLE: the students solves the problem above while the teacher goes round to inspect and
mark accordingly.
ASSIGNMENT: if x =
𝑎+3
2𝑎
- 1, express
2𝑥+1
3𝑥+1
in term of a
REFERENCES: (1) New general mathematics for sss book 2 by M.F. macrae et al pages 197 – 198.(2) New
concept mathematics for sss schools 2 by HN odogwu et al page 106
DATE:
CLASS:
TIME:
PERIOD:
DURATION: minutes
LESSON TOPIC: undefined fractions
(1) Explain when a fraction is said to be undefined
(2) Solve examples of undefined fraction
(3) Find values for which fractions are undefined
(4)Factorize algebraic expressions
(5)Simplify simple linear equations
INSTRUCTIONAL RESOURCES: The statement that fractions are said to be undefined if their
denominators are zero written on a cardboard sheet.
MODE: Individual

53
TEACHER’S ROLE: the teacher ask the students to solve simple equation like 2x + 5 = 0, 𝑥2
− 3𝑥, 𝑥2
− 9
and 𝑥2
- 3x – 10 = 0
STUDENTS’ ROLE: the students solve the problems above:
(1) 2x + 5= 0 (2) 𝑥2
− 9 = 0
2x = -5 𝑥2
-32
= 0
X = -5/2 (x – 3)(x+3) = 0
Either, x- 3 = 0 or x+3 = 0
X = ±3
(3) 𝑥2
- 3x – 10 = 0
𝑥2
- 3x – 12 = 0
𝑥2
- 5x + 2x – 10 = 0
X( x -5) +2(x -5) = 0
(x -5)(x+2)= 0
X= 5 or x= - 2
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to explain to that fraction are said to be undefined if
their denominators are equal to zero e.g.
1
0
and that to find the value for which a fraction is undefined
means equating the denominator to zero and solving to find the value of the unknown.
STUDENTS’ ROLE: the students give more examples of undefined fraction
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve some examples
EXAMPLE 1:Find the value of x for which the fraction
3𝑥+2
𝑥+7
is undefined
SOLUTION:
3𝑥+2
𝑥+7
is undefined if x+7 =0
I.e. x =7
∴The fraction is undefined if x = - 7

54
EXAMPLE 2: find the value of x for which
𝑥2+ 12𝑥+36
𝑥2− 3𝑥−10
is undefined
SOLUTION:
The fraction is undefined if 𝑥2
- 3x -10 =0
I.e. if x = 5 or x = -2
STUDENTS’ROLE: the students solves the equation x+7 = 0 and 𝑥2
-3x – 10 = 0
EXAMPLE 3: Find the value of x for which the fraction
18
𝑥
+
𝑥2+ 1
𝑥2− 9
is undefined
SOLUTION: the expression is undefined if any of the fractions has denominator of 0.
18
𝑥
Is undefined when
x = 0
𝑥2+ 1
𝑥2− 9
=
𝑥2+ 1
(𝑥+3)(𝑥−3)
is undefined when x= 0, -3 or 3.
EXAMPLE 4: find the value of x for which the fraction
𝑥2− 8𝑥−20
6𝑥−1
is
(a) Undefined (b) zero
Solution: (a) the fraction is undefined when 6x -1 = 0 i.e. when x =
1
6
(b) The fraction is zero if
𝑥2− 8𝑥−20
6𝑥−1
= 0
i.e. 𝑥2
- 8x -20 = 0
(x +2)(x -10) = 0
I.e. x= -2 or x = 10
∴The expression is zero when x = -2 or 10.
STUDENTS’ ROLE: the students listen attentively and provide the solution in 𝑥2
- 9 = 0, 6x -1 = 0 and 𝑥2
-
8x -20 = 0
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives the students a similar problem to do

55
(1) When is a fraction is said to be undefined
(2) Give examples of undefined fraction
(3) Factorize 𝑥2
+ 8x + 15 = 0
(4) Simplify x + 3= 0 and x + 5 = 0
(5) What value of x is
2𝑥−1
𝑥2+8𝑥+15
undefined ?
ASSIGNMENT: Find the value(s) of x for which the expression
𝑥2+12𝑥+36
𝑥2− 3𝑥−10
(a) undefined (b) zero
REFERENCES (1) New general mathematics for sss book 2 by M.F. macrae et al pages 199 – 200
(2) New concept mathematics for sss schools 2 by HN odogwu et al page 107

56
DATE:
CLASS:
TIME:
DURATION: minutes
UNIT TOPIC: sequence and series
LESSON TOPIC: Arithmetic progression.
(1) Identify the first term of a sequence
(2) Obtain the common difference of a sequence
(3) Find any term of a sequence
(4)Find the nth term
of a sequence
INSTRUCTIONAL RESOURCES: A chart showing the general AP and nth
term of an AP
MODE: Individual
TEACHER’S ROLE: the teacher willask the students to list the next three terms in each of the following
numbers
(a) 2, 4, 6, 8, 10 …
(b) 20, 10, 0,-10,-20 …
(C) 1,
1
2
,
1
4
,
1
8
,………
(d) 99, 9.9, 0. 99…
(e) 1
1
2
, 2
1
3
, 3
1
4
…
STUDENTS’ ROLE: the students list the next three numbers.

57
(a) 12, 14, and 16
(b) -30, -40, -50
(c)
1
16
,
1
32
, and
1
64
(d) 0. 099, 0.0099 and 0. 00099
(e) 4
1
5
, 5
1
6
, and 6
1
7
MODE: Individual
TEACHER’S ROLE: the teacher ask the students to identify the rules followed to obtain the next terms in
each numbers above.
STUDENTS’ ROLE: the students state the rules individually
(a) Any term +2 gives the next term
(b) Any term ÷ 0.1 or any term x
1
10
= next term
(c) Any term x
1
2
= next term
(d) Any term ÷ 10 or any term x 0.1 gives next term
(e) 1 added to the whole number and denominator of the fraction in turn.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to explain the concept of Ap. Thus: A sequence in
which the terms either increase or decrease in equal step is called an arithmetic progression AP. E.g. 9,
12, 15, 18, 21 … is an AP. It has 9 as its first term and 3 as its common difference (d)
STUDENTS’ROLE: the students will be asked to give more examples of AP and identify the first and
common difference (d)
TEACHER’S ROLE: the teacher leads the students to generate the nth
term of an AP and solve some
problems on AP. Thus:
The following sequences represent a general AP a, a+d, a+2d, a+3d…
If 𝑢𝑛 denotes the nth
term of a sequence
1𝑠𝑡
Term = a

58
2𝑛𝑑
Term = a+d
3𝑟𝑑
Term = a+2d
.
.
.
𝑛𝑡ℎ
Term = a + (n – 1) d
Example 1: given the AP 18, 12, 6, 0, -6…Findits (a) 8𝑡ℎ
term (b) 20𝑡ℎ
term (C) 𝑛𝑛𝑡
term.
S0lution:
a=18, d=-6, n=8
𝑢𝑛= a+(n- 1)d
𝑢8 = 18 +(8 – 1) x – 6
= 18 – 42
= -24
(b) n=20
𝑢20 = 18 – 6(20 -1)
= 18 – 114
= - 96
(C) 𝑢𝑛 = 18 – 6( n -1)
= 18 – 6n +6
= 24 – 6n
Example 2: the28th
termofan AP is – 5. Find its common difference if its first term is 31.
Solution:
a=31, n=28, 𝑢28 = 5
𝑢28 = a+ (28 – 1)d
-5 = 31 + 27d
-5- 31 = 27d
-36 = 27d

59
d=
−36
27
∴The common difference is
−4
3
STUDENTS’ ROLE: the students participate in generating nth
term and identify 1st
term and common
differences in the examples above.
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives the students the following to do.
Given the AP 0, 5, 10, 15…
(1) What is the first term of the sequence?
(2) Find the common difference
(3) Find the 9th
term
(4) Find the nth
term of the AP
STUDENTS’ ROLE: The student supplies the answers in the problem above.
Solution:
(1) a= 0
(2) d= 5
(3) n = 9
𝑢𝑛= 9 + (n -1) d
𝑢9= 0 + (9 -1) x 5
= 0 + 8 x 5
= 40
(4) 𝑢𝑛= a+ (n – 1) d
= 0 + 5 (n – 1)
= 0 + 5n -5
= 5n – 5
ASSIGNMENT: the 12th
term of an AP is 51. Find its common difference if its first term is 7.
REFERENCES (1) New general mathematics for sss book 2 by M.F. macrae et al pages 202 – 204

60
DATE:
CLASS:
TIME:

61
PERIOD:
DURATION: 40 minutes
LESSON TOPIC: nth
term of an AP
(1) State the first term of an AP
(2) State the common difference of an AP
(3) State the nth
term of an AP
(4)Use the nth
term to find the first, common difference number of terms etc.
INSTRUCTIONAL RESOURCES: A chart showing the general AP and the nth
term
MODE: Individual
TEACHER’S ROLE: the teacher ask the students to state the nth
term of an AP individually.
STUDENTS’ ROLE: the students state the nth
term of an AP as 𝑢𝑛 = a + (n-1) d
MODE: Whole
TEACHER’S ROLE: the teacher leads the students in using the nth
term of an AP through an example.
Example 1:Given the AP 18, 12, 6, 0, -6, … Find its (a) 8th
term (b) 20th
term (c) nth
term
Solution:
a=18, d=12-18=6-12=0-6=-6, n=8
𝑢𝑛 = a+ (n-1) d
𝑢8 = 18 + (8-1) x -6
= 18-42
= -24
(b) n=20
𝑢20 = 18 + (20-1) x -6

62
= 18 – 6(19)
= 18 – 114
= -96
𝑢𝑛 = 18 + (n-1) x -6
= 18 – 6(n-1)
= 18 -6n +6
= 24 – 6n
STUDENTS’ ROLE: the students state the values of a, d and n and participate in the simplification leading
to the answers.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve more examples
Example 2: the 28th
term of an AP is -5.find its common difference if its first term is 31.
Solution:
a=31, n=28, 𝑢28 = 5
𝑢28 = a+ (28 – 1)d
-5 = 31 + 27d
-5- 31 = 27d
-36 = 27d
d=
−36
27
∴ The common difference is
−4
3
Example 3: The 18th
term of an AP is 25. Find its first term if its common difference is 2.
Solution:
n=18,𝑢18 = 25 (given), d=2
𝑢𝑛 = a + (n – 1) d
𝑢18 = 25 = a+ (18 – 1) x 2
25 = a + 17 x 2
25 = a + 34

63
25 -34 = a
-9 = a
∴The first term is -9
STUDENTS’ROLE: the students ‘supplies all the ingredients needed in solving the problems
TEACHER’S ROLE: the teacher leads the students to solve another problem.
Example 4: Find the number of terms and an expression for the nth
term in the AP
1
2
,
3
4
, 1, … , 5
1
2
Solution:
A=
1
2
, d=
1
4
, 𝑢𝑛= 5
1
2
𝑢𝑛 = a + (n -1) x
1
4
5
1
2
=
1
2
+ (n -1) x
1
4
11
2
=
1
2
+
1
4
(n -1)
22 = 2 +n – 1
22 – 2 +1 = n
21 = n
∴ There are 21 terms in the AP.
To find an expression for the nth
substitute a and d in 𝑢𝑛 = a + (n – 1) d i.e.
𝑢𝑛 =
1
2
+
1
4
(n -1)
=
1
2
+
𝑛
4
-
1
4
=
𝑛
4
+
1
4
=
1
4
(n+1)
STUDENTS’ ROLE: the students listen and supply the values of a, d and the nth
term and help the teacher
to simplify the problem
STEP V: Evaluation
MODE: Individual

64
TEACHER’S ROLE: the teacher gives the students the following to do.
1. Find the first term of the AP 0, 5, 10, 15…
2. What is its common difference
3. State the 𝑛𝑡ℎ
term and find the 𝑛𝑡ℎ
termof the AP 0, 5, 10, 15…
4. The 28th
term of an AP is -5.find its common difference if its first term is 31.
STUDENTS’ ROLE: students solves the problems by given the teacher as in above
ASSIGNMENT: The first and last terms of an AP are 6.7 and 17.1 respectively. If thereare 14 terms in the
sequence. Find its common difference.
REFERENCES (1) New general mathematics for sss book 2 by M.F. macrae et al pages 178–179
DATE:
CLASS:
TIME:
PERIOD:
DURATION: minutes
LESSON TOPIC: Sum of Arithmetic progression

65
(1) Identify the first term of a series
(2) Identify the common difference
(3) State correctly the formula for the sum of AP
(4)Substitute in the formula above
(5) Simplify (iv) above to obtain the sum of an AP
INSTRUCTIONAL RESOURCES: A chart carrying the formula for sum of AP
MODE: Individual
TEACHER’S ROLE: the teacher ask the students to list the values of a and d in the sequences below:
(a) 6, 3, 0, -3, -6,…
(b) 1,
1
2
,
1
4
,
1
8
, …
(c) 100, 96, 92, 98, 84,…
STUDENTS’ ROLE: the students supply the answers to the questions above i.e.
(a) -3 = d, a=6
(b)a=1, the sequence is not an AP
(c) a=100, d= -4
MODE: Whole
TEACHER’S ROLE: the teacher explains the concept of series: and lead the students to obtain the sum of
AP when the terms of a sequence are added, the resulting expression is called series e.g. (a)
1+2+3+4+5+… is an infinite series (b) 9 +12+15+…+ 45 is a finite series.
The following expression represents a general arithmetic series where the terms are added:
a+ (a+d) +(a+2d) +…. +(l +d)
Let S = a + (a+d) + (a+2d) +…. + (l-d) +l
In reverse:
S = l + (l-d) + (l-2d) +…. + (a+d) +a
Adding:

66
2s = (a+l) + (a+l) +…+ (a+l)
2s = n (a+l)
S =
1
2
n (a+l) where n is the number of terms……………….. (1)
But l=a+ (n-1) d
∴ S =
1
2
n[2𝑎 + (𝑛 − 1)𝑑] …………….. (2)
To find the sum of AP either use formula (1) or (2)
STUDENTS’ ROLE: the students pay attention while contributing to the derivation of the formula
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve a problem.
Example 1: Find the sum of first 20 terms of the AP 3+ 6+9+…
Solution:
a=3, d=3, n=20
Using S =
1
2
n[2𝑎 + (𝑛 − 1)𝑑]
S=
1
2
x 20[2(3) + (20 − 1)𝑥3]
S= 10[6 + 19 𝑥3]
= 10(6 +57)
= 10 (63)
= 630
STUDENTS’ROLE: the students’ supplies the values of a, d, and n and participate in the substitution and
hence the simplification of the problem.
Example: the first and the last term of an AP are 3 and 48. If the sum of the series is 255, find
(a) The number of terms in the AP
(b) The common difference between them

67
Solution:
a=3, last term =48, sum =255
Using S =
1
2
n (a+l)
255=
n
2
(3 + 48)
510 = 51n
510
51
= n
10 = n
∴ AP has 10 terms
(b) using l = a +(n – 1) d
48 = 3 + (10 -1) d
48 = 3 + 9d
48 -3 = 9d
45 = 9d
5 = d
∴ the common difference is 5.
STUDENTS’ ROLE: the students will again supply the values of a, l, and the sum of the AP and partake in
the simplification
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives a similar work to the students
Given the series
Question: find the sun of the series 60 +91 + 122 + 153 + 184
1. Identify the first term of the series
2. Identify the common difference
3. State the formula for sum of an AP
4. Substitute in the formula
5. Simplify (4) above to find the sum of the AP
STUDENTS’ ROLE: students answers the problems above by solving accordingly
ASSIGNMENT: An AP has 15 terms and a common difference of -3. Find its first and last term if its sum is
-120

68
DATE:
CLASS:
TIME:
PERIOD:
DURATION: minutes
LESSON TOPIC: Practical problems on AP
(1) Identify a practical problem on AP
(2) Identify the first term (a) of AP
(3) Identify the common difference (d) and last term (l)
(4)Identify which formula to use
(5) Substitute correctly in the formula variable in the question
(6) Find the unknown variable in the question

69
INSTRUCTIONAL RESOURCES: The formula for nth
term and sum of AP
MODE: Individual
TEACHER’S ROLE: the teacher ask the students one after the other to state the nth
term and the sum of
an AP
STUDENTS’ ROLE: the students state the two formulas 𝑢𝑛 = a + (n-1) d and sum =
1
2
n (a +l) or s =
1
2
n[2𝑎 + (𝑛 − 1)𝑑]
MODE: Whole
TEACHER’S ROLE: the teacher informs the students that it is possible to use the concept or knowledge of
arithmetic progression i.e. the nth
term and sum of AP to solve real life practical problems. He explains
that when quantities increase or decrease by regular intervals, the knowledge of AP can be used
STUDENTS’ ROLE: the students listen and ask questions if any.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve some practical problems.
Example 1:A sum of money is shared among nine people so that the first gets ₦75, the next ₦150, the
next ₦225 and so on.
(a) How much money does the 9th
person gets?
(b) How much money is shared altogether?
Solution:
1st
person gets ₦75
2nd
person gets ₦150
3rd
person gets ₦225 and so no
This is an AP in which
a=₦75, d=₦150 - ₦75=₦75
(a) Here n=9

70
Using the nth
term formula
𝑢𝑛 = a +( n-1) d
U9 = ₦75 + (9-1) x ₦75
= ₦75 + 8 x₦75
= ₦75 +₦600
=₦675
(b) Using S =
1
2
n (a+l)
S=
1
2
9 x9 (₦75 +₦675)
=
9
2
x ₦750 = ₦3375 was shared altogether
STUDENTS’ROLE: the students identify the values of a, d and n from the question given and participates
in the simplification
Example2: The starting salary for a particular job is ₦1.2m per annum.The salary increase each year by
₦75,000 to a maximum of ₦1.5m.
(a) In which year is the maximum salary reached?
(b) How much money would a person earn in that year?
Solution:
a=₦1.2m, d=₦75000, l=₦1.5m
(a) using l= a +(n-1)
₦1.5m = ₦1.2m +(n-1) x ₦75000
₦1.5m -₦1.2m= ₦75000n - ₦75000
₦0.3m +₦75000 = ₦75000n
₦300000 +₦75000= ₦75000n
₦375000= ₦75000n
n=
₦375000
₦7500

71
n=5
∴ It will take 5years for the maximum salary to be reached
(b) Using S=
1
2
n(a+l)
=
1
2
x5(₦1.2m +₦1.5m)
=
1
2
x ₦13.5m
= ₦6.75m
STUDENTS’ ROLE: the students will again supply the values of a, l, and the sum of the AP and partake in
the simplification
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives a similar work to the students
Question: The salary scale for a clerical officer starts at ₦1.1m per annum. A rise of ₦72,000 is given at
the end of each.
1. Is the question above a practical/ real life problem?
2. What is the first term of the AP?
3. What is the common difference?
4. State the formula for the sum of AP
5. Substitute in the formula in (4) above
6. Find the total amount of money earned in 12 years
STUDENTS’ ROLE: students answers the problems above by solving accordingly
ASSIGNMENT: exercise 18d number 9 of new general mathematics for ss2 page209

72
DATE:
CLASS:
TIME:
PERIOD:
DURATION: minutes
LESSON TOPIC: Sum of geometric series
(1) Determine the first term of a GP
(2) Find the common ratio of a GP
(3) State the formula for the sum of a GP
(4)Use the formula to calculate the sum of a GP
INSTRUCTIONAL RESOURCES: formula for sum a GP written 0ncardboard sheet
MODE: Individual
TEACHER’S ROLE: the teacher ask the students to explain the concept of series as earlier discussed under
arithmetic series.
STUDENTS’ ROLE: the students explain the concept of series as the addition of the terms of the
sequence (GP) i.e. a + ar +ar2
+ ar3
+ . . . + arn-1

73
MODE: Whole
TEACHER’S ROLE: the teacher leads the students to deduce the sum of geometric series i.e.
The following expression represents a general geometric series where the terms are added.
. a + ar +ar2
+ ar3
+ . . . + arn-1
If s is the sum, then
S = . a + ar +ar2
+ ar3
+ . . . + arn-1
(1)
Multiplying both sides by r
rs=. a + ar +ar2
+ ar3
+ . . . + arn-1
(2)
Subtract (2) from (1)
S- sr = a - arn
S (1-r) = a(1- rn
)
S=
a(1−rn)
1−r
(3)
Or multiplying the numerator denominator by -1
S =
a(rn−1)
r−1
(4)
If r < 1, formula (3) is convenient.
If r>1, formula (4) is convenient
STUDENTS’ ROLE: the students participate in the deduction of the formula above.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve some problem.
Example 1:Find the sum of the GP 1+11+171+ . . . as far as the fifth term.
Solution:
a=1, r=11, n=5
Using S =
a(rn−1)
r−1
, since r>1
=
115−1
11−1
=
161051−1
10

74
= 16105
STUDENTS’ROLE: the students give the value of a, r, and state the formula to be used and why. They also
help in the simplification above.
TEACHER’S ROLE: the teacher guides the students to solve additional problem.
Example 2: find the sum of the GP 1+3+9+… +729.
Solution:
a=1, r= 3, un
= 729
3n-1
= 729
3n-1
= 36
n-1= 6
n = 1+6 = 7
Using s=
a(rn−1)
r−1
, since r>1
=
37−1
3−1
=
2187−1
2
= 1093
∴ the sum is 1093
Example 3: the third term of a GP is 12 and the fifth term is 48. Find the sum of the first eleven terms.
Solution:
ar2
= 12 (1)
ar4
= 48 (2)
Dividing (2) by (1)
ar4
ar2
=
48
12
r2
= 4
r2
= 22

75
r = 2
put r=2 in (1)
(2)2
a = 12
4a = 12
a=3
∴ a=3, r = 2, n = 11
Using,S =
a(rn−1)
r−1
, since r>1
=
3(211−1)
2−1
= 3(2048-1)
= 3(2047)
= 6141
STUDENTS’ ROLE: the students supply all the values needed in solving the problem and help the teacher
in simplifying.
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives the students similar problem to solve.
Questions:
Given the GP 54 +18 + … +
2
27
.
1. Find its first term
2. Find its common ratio
3. State the formula for sum of GP
4. Use the formula in (3) above to calculate the sum of the GP.
(2) calculate the sum of the GP 448 +224 +112 +… as far as the 6th
term.
STUDENTS’ ROLE: students solve the problem above.
ASSIGNMENT: the 2nd
term of a GP is 12 more than the first term, given that the common ratio is half of
the first term. Find third term and the sum of the first five terms.
REFERENCES (1) Newgeneral mathematics for sss book 2 by M.F. macrae et al pages 205–211
(3) Comprehensive mathematics for sss by D. N. adu pages 105 - 112

76
DATE:
CLASS:
TIME:
PERIOD:
DURATION: minutes
LESSON TOPIC: Sum of a GP to infinity
(1) Determine the first term of a GP
(2) Find the common ratio of a GP
(3) State the formula for the sum of a GP to infinity
(4) Calculate the sum of a GP to infinity
INSTRUCTIONAL RESOURCES: formula for sum a GP to infinity written 0ncardboard sheet
MODE: Individual
TEACHER’S ROLE: the teacher once again test the students’ ability to find the common ratio of a GP
especially if its involves fractions e.g.
(1) What is the common ratio in the GP 6+
6
10
+
6
100
+ . . .
(2) What is the common ratio in the GP 16 + 2 +
1
4
+ …
(3) What is the common ratio in the GP 1 +
1
2
+
1
4
+ …
STUDENTS’ ROLE: the students supplies the answers to the problem above
(1)
1
10
(2)
1
8
(3)
1
2

77
TEACHER’S ROLE: the teacher explains the concept of the sum of GP to infinity by asking the students to
determine the value of
1
2
n as n increases to infinity.
STUDENTS’ ROLE: the students substitute the values of n=1 to infinity and observe what happens to
1
2
n
as n increases to infinity.
MODE: whole class
TEACHER’S ROLE: the teacher leads then in turn explains that if the common ratio r, of a GP is a fraction
such that -1<r<+1, the value rn
in S=
a(1−rn)
1−r
becomes
S =
a
1−r
This formula gives the sum to infinity of a geometric series.
STUDENTS’ROLE: the students listen attentively and are made to state the formula above one after the
other.
TEACHER’S ROLE: the teacher guides the students to solve some problems.
EXAMPLE 1: find the sum to infinity of the GP 6 +
6
10
+
6
100
+ …
Solution:
a=6, r=
1
10
S =
a
1−r
S =
6
1−
1
10
=
6
9
10
=
60
9
= 6
2
3
STUDENTS’ ROLE: the students will be given a similar work to do.

78
(1) Given the GP 16 + 2 +
1
4
+…
(a) Determine the first term
(b) Find the common ratio
(c) State the formula for sum of GP to infinity of the GP above
SOLUTION:
(a) a=16
(b) r=
1
8
(c) S =
a
1−r
(d) S =
16
1−
1
8
=
16
7
8
=
16 x 8
7
=
128
7
= 18
2
7
.
Example 2: the sum of a GP to infinity is 80.find the second term if its first term is 48.
Solution:
Here we seek to find r the common ratio first.
a=48, sum = 80
S =
a
1−r
48
1−r
= 80
80(1-r) = 48
80 – 80r = 48
-80r = 48 – 80
-80r = -32
r=
32
80
r=
2
5

79
∴ The second term ar = 48 x
2
5
=
96
5
= 19
1
5
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives the students similar problem to solve.
QUESTIONS:
Given
1
4
+
1
8
+
1
16
+…
(a) Determine the first term of the GP
(b) Determine the common ratio
(c) State the formula for the sum of GP to infinity
(d) Calculate the sum to infinity of the GP
STUDENTS’ ROLE: students solve the problem above.
ASSIGNMENT: find the sum to infinity of the series
1
4
+
1
8
+
1
16
+…
REFERENCES (1) New general mathematics for sss book 2 by M.F. macrae et al pages 212–213.
DATE:
CLASS:
TIME:
PERIOD:

80
DURATION: minutes
UNIT TOPIC: Quadratic equation
LESSON TOPIC: Revision of perfect squares
(1) Determine the coefficient of a given letter
(2) find half of the coefficient in (ii) above
(3)Find the squares of the numbers
(4) Make a given quadratic equation a perfect square
(5) Factorize (5) above
INSTRUCTIONAL RESOURCES: step by step approaches on how to make an expression a perfect written
on a cardboard sheet
MODE: Individual
TEACHER’S ROLE: the teacher ask the students to list numbers whose square roots are whole numbers.
STUDENTS ROLE: the students list perfect squares like 4, 9, 16, 25, 36, etc.
TEACHER’S ROLE: the teacher leads the students to explain that a perfect square is a number or
expression that can be written as (x+a)2
or a number with two same factors.
STUDENTS’ ROLE: the students are made to list more perfect squares.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to explain further a perfect square especially of a
algebraic expression
Example 1: what must added to 𝑥2
+ 8𝑥 to a make it a perfect square?
SOLUTION:
Suppose 𝑥2
+ 8𝑥 + 𝑘 is a perfect square and is equal to (x +a)2
ie

81
𝑥2
+ 8𝑥 + 𝑘 = (x +a)2
𝑥2
+ 8𝑥 + 𝑘 = 𝑥2
+ 2𝑎 + 𝑎2
Comparing the coefficient of x
2a= 8
a=4
Comparing the constant term
k= 𝑎2
k = 16
∴ 16 must be added to make the expression a perfect square
Check: 𝑥2
+ 8𝑥 + 16= (x +4)2
The teacher explains that in practice, the quantity to be added is the square of half the coefficient of x or
whatever letter is used.
STUDENTS’ROLE: the students helps the teacher to find the values of a and k in above example
MODE: the entire class/ individual
TEACHER’S ROLE: the teacher guides the students to apply the concept above in the example below.
EXAMPLES2: what must be added to the following to make it a perfect square? Factorize the result.
(a) 𝑥2
- 6x (b) 𝑥2
+
2
3
𝑥 (c) 𝑥2
+ 10𝑥𝑦 (d) 𝑥2
−
5
2
𝑥
𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏:
(a) 𝑥2
- 6x
Coefficient of x = - 6
1
2
of x = - 3
(- 3)2
= 9
Hence 9 must be added
∴ 𝑥2
− 6𝑥 + 9 = (x – 3)2
(c) 𝑥2
+ 10𝑥𝑦

82
Coefficient of x is 10y
1
2
of 10y = 5y
(5y)2
= 25y2
Hence 25y2
must be added
∴ 𝑥2
+ 10𝑥𝑦 + 25y2
= (x + 5y)2
(b) 𝑥2
+
2
3
𝑥
Coefficient of x =
2
3
1
2
Of
2
3
=
1
3
(
1
3
)2
=
1
9
Hence
1
9
must be added
∴ 𝑥2
+
2
3
𝑥 +
1
9
= (x +
1
3
)2
(d) Coefficient of x = -
5
2
1
2
of -
5
2
= -
5
4
( -
5
4
)2
=
25
16
Hence
25
16
must be added
∴ 𝑥2
−
5
2
𝑥 +
25
16
= ( x -
5
4
)2
STUDENTS’ ROLE: the students will be given a similar work to do.
(1) What must be added to 𝑘2
− 1
1
3
𝑘 to make it a perfect square. Factorize the result.
SOLUTION:
Coefficient of k is - 1
1
3
1
2
𝑜f -
4
3
= -
2
3
( -
2
3
)2
=
4
9
Hence 𝑘2
− 1
1
3
𝑘 +
4
9
= ( k -
2
3
)2
STEP V: Evaluation

83
MODE: Individual
TEACHER’S ROLE: the students will be given more work to do.
Given the expression (a) 𝑐2
− 4𝑐(b) 𝑥2
+
5
3
𝑥 (c)𝑝2
– 1
2
3
p (d) y +5py
1. Determine the coefficient of c, x, p and y in (a) – (d) above
2. Find half of the coefficient obtained above
3. Square the result in (2) above
4. Add the result in (3) above to make it a complete quadratic equation and obtain a perfect
square of it
5. Factorize the result above
𝑺𝑻𝑼𝑫𝑬𝑵𝑻𝑺 𝑹𝑶𝑳𝑬: The students solve the problems above.
ASSIGNMENT: what must be added to the following to make it a perfect square?
(a) 𝑥2
+
5
6
xy (b) 𝑥2
− 0.4𝑥
REFERENCES: (1) Newgeneral mathematics for sss book 2 by M.F. macrae et al pages 36 -37 (2)
comprehensive mathematics by D.B Adu pages 70 -71
DATE:
CLASS:
TIME:
PERIOD:
DURATION: minutes
LESSON TOPIC: Completing the squares
(1) Make the coefficient of 𝑥2
a unity
(2) Rearrange a given quadratic equation by taking the constant term to the RHS

84
(3) Complete the squares in the LHS
(4)Factorize the LHS
(5) Simplify the RHS
(6) Solve to obtain two values of the unknown.
INSTRUCTIONAL RESOURCES: steps involved in solving quadratic equation by completing the squares
written on a cardboard sheet.
MODE: Individual/ whole class
TEACHER’S ROLE: the teacher ask the students to list the steps involved in making and expression a
perfect squares.
STUDENTS ROLE: the students respond to the teacher question. Thus: find the coefficient of the letter
involved, find half the coefficient above and find its square.
TEACHER’S ROLE: the teacher the steps involved in solving quadratic equation by completing the
squares.
unity
(2) Rearrange the quadratic equation by taking the constant term to the RHS
(3) Complete the squares on the LHS
(4)Factorize (2) above
(6) Find two values of the unknown
STUDENTS’ ROLE: the students repeat the steps involved in solving quadratic equation by completing
the squares as above.
MODE: whole class

85
TEACHER’S ROLE: the teacher leads the students to solve a problem
Example 1: solve 𝑝2
+ 4p – 21 =0 by completing the squares
Solution:
𝑝2
+ 4p – 21 =0
Rearrange
𝑝2
+ 4p = 21
Add 4 to both sides
𝑝2
+ 4p +4 = 21 + 5
(𝑝 + 2)2
= 25
Take the Square root both sides
P +2 = √25
P = - 2 ± 5
P = - 2 +5 or – 2 -5
P = 3 or -7
STUDENTS’ROLE: the students are given a similar problem to solve
Solve by completing the squares 2𝑥2
-3x – 5 =0
SOLUTION:
2𝑥2
-3x – 5 =0
2𝑥2
-3x = 5
2𝑥2
-
3
2
x =
5
2
Add
9
16
to both sides
𝑥2
-
3
2
x +
9
16
=
5
2
+
9
16
(𝑥 −
3
4
)2
=
40+9
16
(𝑥 −
3
4
)2
=
49
16
X -
3
4
= ±√
49
16

86
X =
3
4
±
7
4
X =
10
4
or-
4
4
X =
5
2
or - 1
TEACHER’S ROLE: the teacher guides the students to solve additional problem
EXAMPLES2: solve 4𝑦2
+ 5y -21 = 0 by completing the squares.
SOLUTION:
4𝑦2
+ 5y -21 = 0
Divide through by 4
4𝑦2
4
+
5y
4
–
21
4
=
0
4
𝑦2
+
5y
4
–
21
4
= 0
Rearrange
𝑦2
+
5y
4
=
21
4
Add
25
64
to both sides
𝑦2
+
5y
4
+
25
64
=
21
4
+
25
64
(y +
5
8
)2 =
361
64
y+
5
8
= ±√
361
64
y +
5
8
= ±
19
8
y= -
5
8
±
19
8
y=
−5+19
8
or
−5−19
8
y=
14
8
or
−24
8
y=
7
4
or - 3

87
STUDENTS’ ROLE: the students will be asked to solve
2
3
𝑥2
- 8x – 9 = 0
Solution:
Multiply through by
2
3
𝑥2
- 12x -
27
2
= 0
Add 36 to both sides
𝑥2
- 12x +36 =
27
2
+36
(x-6)2
=
27+72
2
(x – 6)2
=
99
2
Take the square root of both sides
X -6 = ±√
99
2
X = 6 ±√
99
2
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher again gives the students a similar work to do.
Given the quadratic equation 2𝑥2
- 5x - 4 = 0
1. Make the coefficient of 𝑥2
a unity
2. Rearrange the given equation
3. Complete the squares on the LHS
4. Factorize the RHS
5. Solve the equation 2𝑥2
-5x – 4= 0
𝑺𝑻𝑼𝑫𝑬𝑵𝑻𝑺 𝑹𝑶𝑳𝑬:The students solve the problems above.
ASSIGNMENT: Find the roots of the equation 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0, using completing the method.
REFERENCES : (1) New general mathematics for sss book 2 by M.F. macrae et al pages 37 -3 (2)
comprehensive mathematics by D.B Adu pages

88
DATE:
CLASS:
TIME:
PERIOD:
DURATION:
LESSON TOPIC: The use of formula from completing the square
(1) Identify the constants a, b, c in a 𝑥2
+ bx + c = 0
(2) State the quadratic formula correctly
(3) Substitute in the formula correctly
(4)Simplify to obtain the roots of an equation
INSTRUCTIONAL RESOURCES: Formula for solving quadratic equation written on a cardboard paper.

89
TEACHER’S ROLE: the teacher informs the students that the general quadratic equation a𝑥2
+ bx + c = 0
can be solved using a formula
STUDENTS ROLE: the teacher ask the students to identify the values of a, b, and c in the equation𝑥2
−
5𝑥 + 6 = 0, i.e.
a=2, b= -5, c=6
TEACHER’S ROLE: the teacher leads the students to deduce the quadratic formula from completing the
square.
𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0
𝑥2
+
𝑏
𝑎
𝑥 +
𝑐
𝑎
= 0 (dividing through by a)
𝑥2
+
𝑏
𝑎
𝑥 =
−𝑐
𝑎
(carrying
𝑐
𝑎
to the other side)
𝑥2
+
𝑏
𝑎
𝑥 + (
𝑏
2𝑎
)
2
=
−𝑐
𝑎
+ (
𝑏
2𝑎
)
2
… … (𝑓𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑖𝑛𝑔 𝐿𝐻𝑆)
(𝑥 +
𝑏
2𝑎
)
2
=
−4𝑎𝑐+ 𝑏2
4𝑎2
(𝑥 +
𝑏
2𝑎
)
2
=
𝑏2 − 4𝑎𝑐
4𝑎2
𝑥 +
𝑏
2𝑎
= √
𝑏2 − 4𝑎𝑐
4𝑎2
𝑥 =
−𝑏
2𝑎
±√
𝑏2 − 4𝑎𝑐
4𝑎2 =
−𝑏
2𝑎
±
√ 𝑏2 − 4𝑎𝑐
2𝑎
𝒙 =
−𝒃 ±√ 𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
STUDENTS’ ROLE: the students participate in deducing the formula above by supplying components like
the coefficient of x is
b
a
, half of
b
a
is
b
2a
and square of
b
2a
as
b2
4a2 etc.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve a problem using the formula
Example 1: solve 3𝑥2
− 5𝑥 − 2 = 0
Solution:

90
Here a= 3, b= -5, c= -2
X =
−𝑏±√𝑏2−4𝑎𝑐
2𝑎
X =
− (−5) ±√(−5)2 – 4(3)(−2)
2(3)
X=
5 ±√25+24
6
X =
5±√49
6
X =
5±7
6
Either,
X =
5+7
6
or
5−7
6
=
12
6
or
−2
6
= 2 or - -
1
3
STUDENTS’ROLE: the students will be given a similar problem to solve
Solve by formula𝑥2
+4x – 21 =0
Solution:
a=1, b = 4, c = - 21
x =
−4±√42−4(1)(−21)
2
x =
−4±√16+84
2
x =
−4±√100
2
x =
−4±10
2
Either
x =
−4+10
2
or
−4−10
2
=
6
2
or
−14
2
= 3 or - 7

91
Examples2: solve 4𝑥2
-25x +25 = 175 by formula
Solution:
Rewrite the equation 4𝑥2
- 25x +25 = 175 as
4𝑥2
- 25x – 150 = 0
a=4, b= -25, c= -150
x =
−(−25)±√(−25)2−4(4)(−150)
2(4)
x =
25±√625+2400
8
x =
25±√3025
8
x =
25 ±55
8
=
80
8
or -
15
4
STUDENTS’ ROLE: the students participate in the solution above.
STEP V: Evaluation
MODE: Individual
TEACHER’S ROLE: the teacher gives the students a similar work to do.
Given the equation3𝑥2
- 8x – 3 = 5
1. Identify the values of a, b, c
2. State the quadratic formula
3. Substitute (2) in (3)
4. Simplify to obtain the roots of the equation
Solve by formula the equation 3𝑥2
- 8x – 3 = 5
𝑆𝑇𝑈𝐷𝐸𝑁𝑇𝑆 𝑅𝑂𝐿𝐸: The students solve the problems above.
Correction
3𝑥2
- 8x – 3 = 5
3𝑥2
- 8x – 3 -5 = 0

92
3𝑥2
- 8x – 8 = 0
a=3 , b = -8, c = - 8
x =
−(−8)±√82−4(3)(−8)
2(3)
=
8±√64+96
6
X =
8+√160
6
or
8−√160
6
ASSIGNMENT: Use formula to solve 6𝑥2
+ 13𝑥 + 6 = 0
REFERENCES (1) New general mathematics for sss book 2 by M.F. macrae et al pages41-42 (2)

93
DATE:
CLASS:
TIME:
PERIOD:
DURATION: minutes
LESSON TOPIC: Completing the squares
a unity
(2) Rearrange a given quadratic equation by taking the constant term to the RHS
(3) Complete the squares in the LHS
(4)Factorize the LHS
(6) Solve to obtain two values of the unknown.
INSTRUCTIONAL RESOURCES: steps involved in solving quadratic equation by completing the squares
written on a cardboard sheet.
TEACHER’S ROLE: the teacher ask the students to list the steps involved in making and expression a
perfect squares.
STUDENTS ROLE: the students respond to the teacher question. Thus: find the coefficient of the letter
involved, find half the coefficient above and find its square.

94
TEACHER’S ROLE: the teacher the steps involved in solving quadratic equation by completing the
squares.
unity
(2) Rearrange the quadratic equation by taking the constant term to the RHS
(3) Complete the squares on the LHS
(4)Factorize (2) above
(5) Simp lify the RHS
(6) Find two values of the unknown
STUDENTS’ ROLE: the students repeat the steps involved in solving quadratic equation by completing
the squares as above.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve a problem
Example 1: solve 𝑝2
+ 4p – 21 =0 by completing the squares
Solution:
𝑝2
+ 4p – 21 =0
Rearrange
𝑝2
+ 4p = 21
Add 4 to both sides
𝑝2
+ 4p +4 = 21 + 5
(𝑝 + 2)2
= 25
Take the Square root both sides
P +2 = √25
P = - 2 ± 5
P = - 2 +5 or – 2 -5
P = 3 or -7
STUDENTS’ROLE: the students are given a similar problem to solve

95
Solve by completing the squares 2𝑥2
-3x – 5 =0
Solution:
2𝑥2
-3x – 5 =0
2𝑥2
-3x = 5
2𝑥2
-
3
2
x =
5
2
Add
9
16
to both sides
𝑥2
-
3
2
x +
9
16
=
5
2
+
9
16
(𝑥 −
3
4
)2
=
40+9
16
(𝑥 −
3
4
)2
=
49
16
X -
3
4
= ±√
49
16
X =
3
4
±
7
4
X =
10
4
or -
4
4
X =
5
2
or - 1
Examples2: solve 4𝑦2
+ 5y -21 = 0 by completing the squares.
Solution:
4𝑦2
+ 5y -21 = 0
Divide through by 4
4𝑦2
4
+
5y
4
–
21
4
=
0
4
𝑦2
+
5y
4
–
21
4
= 0
Rearrange
𝑦2
+
5y
4
=
21
4

96
Add
25
64
to both sides
𝑦2
+
5y
4
+
25
64
=
21
4
+
25
64
(y +
5
8
)2 =
361
64
y+
5
8
= ±√
361
64
y +
5
8
= ±
19
8
y= -
5
8
±
19
8
y=
−5+19
8
or
−5−19
8
y=
14
8
or
−24
8
y=
7
4
or - 3
STUDENTS’ ROLE: the students will be asked to solve
2
3
𝑥2
- 8x – 9 = 0
Solution:
Multiply through by
2
3
𝑥2
- 12x -
27
2
= 0
Add 36 to both sides
𝑥2
- 12x +36 =
27
2
+36
(x-6)2
=
27+72
2
(x – 6)2
=
99
2
Take the square root of both sides
X -6 = ±√
99
2
X = 6 ±√
99
2
STEP V: Evaluation
MODE: Individual

97
TEACHER’S ROLE: the teacher again gives the students a similar work to do.
Solve 𝑡2
- 6t + 9 = 0 by completing the square
Solution:
𝑡2
- 6t + 9 = 0
𝑡2
- 6t = - 9
Add 9 to both sides
𝑡2
- 6t + 9 = - 9 +9
(t -3)2
= 0
t-3 = ±0
t=3±0
t=3 twice
𝑆𝑇𝑈𝐷𝐸𝑁𝑇𝑆 𝑅𝑂𝐿𝐸: The students solve the problems above.
ASSIGNMENT: Find the roots of the equation 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0, using completing the method.
REFERENCES (1) New general mathematics for sss book 2 by M.F. macrae et al pages 37 -39 (2)
DATE:
CLASS:

98
TIME:
PERIOD:
DURATION:
LESSON TOPIC: word problem leading to quadratic equation and solution
(1) Use letters to represent numbers
(2) Form quadratic equation from a given word problem
(3) Solve the equation formed in (2) above
(4)Interpret the result obtained
INSTRUCTIONAL RESOURCES: step by step approaches in solving practical problems written on a
cardboard sheet.
MODE: whole class
TEACHER’S ROLE: the teacher ask the students to mention all the methods of solving quadratic equation
and the general quadratic equation 𝑎𝑥2
+ 𝑏𝑥 + 𝑐 = 0 and prepare to use any of the methods in the
problems to be solved in the course of the lesson
STUDENTS’ ROLE: the students mentions, factorization, completing the square, formula and graph as the
four ways of solving quadratic equation
TEACHER’S ROLE: the teacher gives the steps involved in solving word problems.
(1) Choose a letter to represent a number required
(2) The letter chosen must represent a number and not a quantity
(3) Change each statement in the question to statement containing the letter you have chosen.
(4) Link up all the parts to form an equation
STUDENTS’ ROLE: the students will be requested to list the steps involved in solving.

99
STEP III: DISCUSSION word problem as enumerated by the teacher as in above.
MODE: whole class
TEACHER’S ROLE: the teacher leads the students to solve a problem.
Example 1: the sum of two numbers is 35. Their product is 300. Find the numbers
Solution:
Let one of the number be x
=>The other will be 35 – x , since their sum is 35.
Also their product is 300, hence
X(35 –x) = 300
35x - 𝑥2
= 300
𝑥2
- 35x +300= 0
𝑥2
− 35𝑥 − 15𝑥 + 300 = 0
x2
− 20x − 15x + 300 = 0
X(x-20) – 15(x-15)
(x-20)(x-15) = 0
Either x-20=0 or x-15= 0
X= 20 or 15
∴ the numbers are 15 and 20
Example 2: the product of two consecutive odd numbers is 323. Find the numbers.
Solution:
Let the number be x and x+2
But x(x+2) = 323
x2
+2x – 323 = 0
x2
+ 19x − 17x − 323 = 0
X(x+19) -17(x +19) = 0
(x+19)(x-17) = 0
X+19 = 0 or x-17 = 0
X = -19 or 17

100
∴ The odd numbers are 17 and 19.
STUDENTS’ROLE: the students factorizes the quadratic equation formed in the two
TEACHER’S ROLE: the teacher guides the students to solve additional problems
Examples3: the sum of digits of a two – digit number is 12. The tens digit is the square of its unit digit.
Find the numbers.
Solution:
Let the two digit number be xy
But x +y = 12 ( 1st
sentence)
∴ y=12 – x ………..(1)
Also x = (12 – x)2
(second sentence)
X = 144 – 24x +x2
x2
− 25x + 144 = 0
x2
− 16x − 9x + 144 = 0
X(x-16) -9(x-16) = 0
(x -16)(x-9) = =0
Either x- 16 = =0 or x-9 = 0
X= 16 or 9
But x ≠ 16
Hence x = 9
Put x = 9 in C1) to find y
Y=12-9
= 3
∴ the two digit number is 93
STUDENTS’ ROLE: the students participate in the solution above.
STEP V: Evaluation
MODE: Individual

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