KSSM_Form_4_Additional_Mathematics_Notes

EXAMPLE 8

Solve the following simultaneous equations using the graphical representation method.

x² + y² = 20 Tip: To determine the
x+y=6 points to be plotted
on the graph, students
Solution: are encouraged to
construct a table.

For equation x² + y² = 20

x –4 –2 0 2 4
y ±2 ±4 ± √20 ±4 ±2

For equation x + y = 6

x –4 –2 0 2 4
y 10 8 6 4 2

Sketch a graph
based on the table
constructed.

Based on the graph, there are two points of intersection which is (2,4) and (4,2).
Thus, the solutions to the simultaneous equations are x = 2, y = 4 and x = 4, y = 2.

47

Further Exploration

Visualizing Linear Equations in Three Variables

https://www.youtube.com/watch?v=Wm27Y6hxbRs

Types of Linear Systems in Three Variables

https://www.youtube.com/watch?v=WAzUwzV1F3g

Solving Systems of Equations in Three Variables

https://www.youtube.com/watch?v=fjfmNfIgQ2Q

System of Linear Equations in Three Variables (Substitution method)

https://www.youtube.com/watch?v=pYrynnnqoEI

Three Variable Systems with Infinite or Null Solution Sets

https://www.youtube.com/watch?v=pOSxxRosKoQ

Solving Simultaneous Equation (One linear and one non-linear)

https://www.youtube.com/watch?v=cP9Zzzff2VE

48

Summative Exercises

3.1 Systems of Linear Equations in 3.2 Simultaneous Equations Involving
Three Variables One Linear Equation and One Non-
Linear Equation
1) Based on the following system of linear
equations 1) Solve the following simultaneous equations
x − 2y + z = 10 2x + y = 1
2x + 2y + 3z = 13 x2 + y2 = 1
–2x + 5y + z = –7
Find the value of x + y + z. 2) Solve the simultaneous equations
x2 – y = 14
2) Solve the following system of linear 2y – 4 = 12x
equations using the elimination method.
x + 2y – z = 2 3) Solve the simultaneous equation
2x – 3y + z = –1 x² − y² = 7
5x – y – 2z = –3 2y = 2 + x

3) Solve the following system of linear 4) Solve the following simultaneous equations
equations. using the graphical representation method
x+y+z=6 x + 2y – 4 = 0
x2 + y2 = 7 – xy
2 y + 5z = −7
2x + 5y − z = 27 5) A straight line with equation 4x + y – 5 = 0
intersects the curve 27x2 + 21xy = –2y2 at
4) Adam has 90 balls of four colours: yellow, points A and B. Find the coordinates of the
pink, blue, and black.60 of them are not pink, points A and B.
25 are black, and 67 are not yellow. What is
the number of yellow, pink and blue balls 6) ym
respectively? Fish pond

5) The standard equation of a circle is xm
x2+y2+Ax+By+C=0. Find the equation of
the circle that passes through the points (1,1), 10m Banana
(2, –4) and (5,5).
30 m
6) A total of RM50,000 is invested in three Ms Wong has a rectangular plot of land. She
funds paying 6%, 8%, and 10% simple grows bananas trees and rears fish in the
interest. The yearly interest is RM3,700. regions as shown in the diagram. Given that
Twice as much money is invested at 6% as the region planted with bananas is 460m2 and
invested at 10%. How much was invested in the perimeter of the rectangular fish pond is
each of the funds. 48m, find the value of x and of y.

49

Chapter 4 Indices, Surds and Logarithms

Chapter 4: Indices, Surds and Logarithms

4.1 Law of Indices

Index

Base

Some law of indices to recall:
1. = × × … × ×

n times of a

2. − = 1


3. 0 = 1

4. = ξ



5. × = +

6. ÷ = −

7. ሺ ሻ = ሺ ሻ =

8. ሺ ሻ =

9. ቀ ቁ =



4.1.1 Simplifying algebraic expressions involving indices

By using the law of indices from the above, algebraic expressions can be simplified. We can apply
techniques like grouping like terms together when we are simplifying them.

EXAMPLE 1 (ii) 7 3 4
2 −2
Simplify the following:
(i) ሺ4 −3 3ሻ−2

Solution:

(i) ሺ4 −3 3ሻ−2 = 4−2 −3×−2 3×−2

= 1 6 −6
42
= 1 6 −6
16

50

Chapter 4 Indices, Surds and Logarithms

(i) 7 3 4 = 7 ቀ 32ቁ ቀ −42ቁ
2 −2

= 7ሺ 3−2ሻ൫ 4−ሺ−2ሻ൯

= 7 6

EXAMPLE 2

Show that 7 +3 + 7 − 7 +2 is divisible by 5 for all positive integer .
Solution:
7 +3 + 7 − 7 +2 = 7 73 + 7 − 7 72

= 7 ሺ73 + 1 − 72ሻ
= 7 ሺ343 + 1 − 49ሻ
= 7 ሺ295ሻ
Since 295 is a multiple of 5, therefore 7 +3 + 7 − 7 +2 is divisible by 5 for all
positive integer .

4.1.2 Solving problems involving indices

All equations involving indices can be solved following the rule below:

If = , then = or if = , then = when > and ≠ .

EXAMPLE 3

Solve the following equations: (ii) 34 = 1
(i) 49ሺ7−4ሻ = 72+ 9 3 −3

Solution:

(i) 49ሺ7−4ሻ = 72+ Equate both side by comparing the indices
72ሺ7−4ሻ = 72+
7−2 = 72+

+ 2 = −2

= −4

(ii) 34 = 1
9 3 −3

34 = 1
32 3 −3

34−2 = 3−ሺ −3ሻ

−ሺ − 3ሻ = 2 Equate both side by comparing the indices

= 1

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Chapter 4 Indices, Surds and Logarithms

4.2 Law of Surds

4.2.1 Comparing rational numbers and irrational numbers, and relating surds
with rational numbers

Rational numbers are numbers that can be expressed in the form of in the simplest form, where


and are integers and ≠ 0.

Decimals that can be converted into fractions are rational numbers, such as 0.111 … = 1
9

Decimals that cannot be converted into fractions are irrational numbers.

Surds are numbers that are in the square root form ൫ξ ൯, where is any positive integer. Surds have
infinite decimal places and are non-recurring. The following are some examples:

ξ2 = 1.41421 … has infinite decimal places and non-recurring. Hence ξ2 is a surd.

ξ4 = 2 is an integer. Hence ξ4 is not a surd.

Besides, ξ is the surd of order , which means the raise to the 1 th power.


In short, when a number cannot be simplified by eliminating the root, it is classified as a surd.

Recurring decimals are decimals that have a number, or a group of numbers repeated infinitely in the
decimal places. They are all rational numbers. For example:

0.262626 … can be written as fraction ቀ2996ቁ, and can also be denoted by 0. 2̇ 6̇ or 0. 26.

EXAMPLE 4

Convert the following recurring decimals into fractions:

(i) 0.535353 … (ii) 0. 128

Solution: Multiply the decimals by
10 raise to how many
(i) Let = 0.535353 … numbers in the group of
100 = 53.535353 … the recurring decimals (in
this case 2 because 53 is
100 − = 53.535353 … − 0.535353 … repeated).
99 = 53
= 53 Subtract the two equations
of 100 and .
99

(ii) Let = 0. 128
1000 = 128. 128

1000 − = 128. 128 − 0. 128
999 = 128
= 128

999

52

Chapter 4 Indices, Surds and Logarithms

4.2.2 Making and verifying conjectures on ξ × ξ and ξ ÷ ξ

For > 0 and > 0,

ξ × ξ = ξ By using law of indices


ξ ÷ ξ = √

EXAMPLE 5

Simplify the following surds:

(i) ξ6 (ii) ξ7×ξ6
ξ24 ξ3

Solution:

(i) ξ6 = √6
ξ24
24

= √1

4

= 1
2

(ii) ξ7×ξ6 = √7×6
ξ3
3

= ξ14

4.2.3 Simplifying expression involving surds

For some surds ξ , may divided by numbers of factor , then the surds can be simplified. For
example:

ξ72 can be written as ξ6 × 6 × 2, the number 72 has 2 factors of 6. It can be further simplified.

ξ72 = ξ6 × 6ξ2
= 6ξ2

Besides, expressions involving surds can also be simplify by basic algebraic manipulations and
operations such as addition, subtraction or grouping likes term(surds).

EXAMPLE 6

Simplify the following expressions: (ii) ൫5 + 3ξ3൯൫2 − 2ξ3൯
(i) ξ150 − ξ24

Solution:

(i) ξ150 − ξ24 = ξ25 × 6 − ξ4 × 6
= 5ξ6 − 2ξ6
= 3ξ6

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Chapter 4 Indices, Surds and Logarithms

(ii) ൫5 + 3ξ3൯൫2 − 2ξ3൯ = 10 − 10ξ3 + 6ξ3 − 6൫ξ3൯2
= 10 − 4ξ3 − 18
= −4ξ3 − 8

4.2.4 Rationalising the denominators for the expressions involving surds

Fractions in the form of 1 can be rationalised and be simplified to eliminate the surds from the
ξ ± ξ

denominator. To rationalise means to multiply conjugate surds to the numerator and denominator.

Conjugate surds are the surds with similar value but with the opposite operator, the following tables
show the conjugate surds of each surd.

Surds Conjugate Surds
1 1

ξ + ξ ξ − ξ
1 1

ξ ξ
1 1

ξ − ξ ξ + ξ

EXAMPLE 7

Rationalise the following:

(i) 1 (ii) 1
4ξ7 3ξ5−2ξ3

Solution:

(i) 1 = 1 × 4ξ7
4ξ7 4ξ7 4ξ7

= 4ξ7 Multiply conjugate
surds to both numerator
4ξ7×4ξ7 and denominator.

= 4ξ7

112

= ξ7
28

(ii) 1= 1× 3ξ5+2ξ3
3ξ5+2ξ3
3ξ5−2ξ3 3ξ5−2ξ3

= 3ξ5+2ξ3

൫3ξ5−2ξ3൯൫3ξ5+2ξ3൯

= 3ξ5+2ξ3
൫3ξ5൯2−൫2ξ3൯2

= 3ξ5+2ξ3

45−12

= 3ξ5+2ξ3
33

54

Chapter 4 Indices, Surds and Logarithms

4.2.5 Solving problems involving surds

While solving problems involving surds, apply the law of indices, law of surds and techniques to
simplify and rationalise surds.

EXAMPLE 8

Solve − 5ξ + 6 = 0 for all real .

Solution:

− 5ξ + 6 = 0

൫ξ − 2൯൫ξ − 3൯ = 0

ξ − 2 = 0 or ξ − 3 = 0

ξ = 2 or ξ = 3
= 4 or = 9

4.3 Law of Logarithms

4.3.1 Relating the equations in index form with logarithmic form and determine
the logarithmic value of a number

For index form equation = , a function logarithm is introduced to help solve this type of equation.
= ⟺ log = where > 0 and ≠ 1

Since 0 = 1 and 1 = , apply it into the logarithm function we get:

log 1 = 0 and log = 1

If we were to find the reverse of a result from the logarithmic function, we use antilogarithm or
antilog for short.

Also, if a base is not mentioned in the logarithmic function, it is understood as base 10.

log10 = ⟺ antilog =

In short, Number Index

= ⟺ log =

Base

Suppose index form equation as a function:
ሺ ሻ = ,

−1ሺ ሻ = log

55

Chapter 4 Indices, Surds and Logarithms

EXAMPLE 9

Express 35 = 243 in logarithmic form.
Solution:

35 = 243
log3 243 = 5
Express log7 343 = 3 in index form.
Solution:
log7 343 = 3
73 = 343

EXAMPLE 10 (ii) log10 100
(ii) log7 = 2
1. Find the value of the following:
(i) log3 27

Solution:

(i) Let log3 27 =
3 = 27
3 = 33
= 3

Thus, log3 27 = 3

(ii) Let log10 100 =
10 = 100
10 = 102
= 2

Thus, log10 100 = 2

2. Solve the following:
(i) log2 = 4

Solution:

(i) log2 = 4
24 =
= 16

(ii) log7 = 2
72 =
= 49

56

Chapter 4 Indices, Surds and Logarithms (ii) antilog 0.2346

EXAMPLE 11

Find the value of the following:
(i) antilog 3
Solution:
(i) antilog 3 = 1000
(ii) antilog 0.2346 = 1.7163

4.3.2 Proving the laws of logarithms

The basic laws of logarithm are as follows:
If , , are positive and ≠ 1, then

(i) log + log = log

(ii) log − log = log

(iii)log = log for any real number .

The above formulae can be proven by applying the law of indices:

Let = and = that , , are positive and ≠ 1, we have = log and = log .

(i) = ×
= +

log = +
log = log + log

(ii) =


= −



log = −


log = log − log


(iii) = ሺ ሻ

=

log =
log = log

57

Chapter 4 Indices, Surds and Logarithms

EXAMPLE 12

Given log2 5 = 2.322 and log2 9 = 3.170, determine the value of:

(i) log2 45 (ii) log2 9
5

Solution:

(i) log2 45 = log2 5 × 9
= log2 5 + log2 9
= 2.322 + 3.170

= 5.492

(ii) log2 9 = log2 9 − log2 5
5

= 3.170 − 2.322

= 0.848

EXAMPLE 13

Without using calculator, find the value of:

(i) log5 10 + log5 15 − log5 6
(ii) 2 log3 6 − log3 4
(iii) 3 log2 3ξ4

Solution:

(i) log5 10 + log5 15 − log5 6 = log5 10×15
6

= log5 25

= log5 52

= 2 log5 5

=2

(ii) 2 log3 6 − log3 4 + log3 9 = log3 62 − log3 4 + log3 9

= log3 36×9
4

= log3 81

= log3 34

= 4 log3 3

=4

(iii) 3 log2 3ξ4 = log2൫3ξ4൯3
= log2 22
= 2 log2 2
=2

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Chapter 4 Indices, Surds and Logarithms

4.3.3 Simplifying algebraic expressions using the law of logarithms

By using the law of logarithms as stated in 4.3.2, we can simplify algebraic expressions:

EXAMPLE 14

1. Simplify the following expressions into single logarithms:

(i) log5 2 + log5 (ii) log + log 2 − 4 log

Solution:

(i) log5 2 + log5 = log5ሺ 2 × ሻ
= log5 2

(ii) log + log 2 − 4 log = log + log 2 − log 4
= log ቀ 42ቁ
= log ቀ 23ቁ

2. Given = log2 5, = log2 7 and = log2 3, write the following

expressions in terms of , and/or .

(i) log2 35

(ii) log2 75
7

(iii) log2 21
ξ5

Solution:

(i) log2 35 = log2 5 × 7
= log2 5 + log2 7
=p+q

(ii) log2 75 = log2 ቀ25×3ቁ
7
7

= log2 ቀ52×3ቁ

7

= log2 52 + log2 3 − log2 7

= 2 log2 5 + log2 3 − log2 7

= 2 + −

(ii) log2 21 = log2 ቀ7×3ቁ
ξ5
ξ5

= log2 7 + log2 3 − log2 5ቀ21ቁ

= log2 7 + log2 3 − 1 log2 5
2

= + − 1

2

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Chapter 4 Indices, Surds and Logarithms

4.3.4 Proving the relationship of log = log and determining the logarithm of
log

a number

Suppose we have , and are positive numbers, ≠ 1 and ≠ 1, also let log = , then = .

log = log

log = log

= log
log

Therefore, we get

log = log
log

The following summarises the above proof:

If , and are positive numbers, ≠ and ≠ , then

=


Additionally, if we have = , then

log = log = log = 1
log log log

Often, we change the base of logarithms to base 10 or base conventionally to evaluate the value of
the logarithms. is a mathematical constant with the value of non-recurring decimal 2.71828 …
Logarithms with base are called the natural logarithms and are written as log or ln .

EXAMPLE 15

Given = log3 , express the following expressions in term of m by

changing the bases of the logarithmic functions:

(i) log27 2 (ii) log 81 3

Solution:

(i) log27 2 = log3 2
log3 27

= 2 log3
log3 33

= 3 2 3
log3

= 2

3

(ii) log 81 3 = log3 81 3
log3

= log3 34+log3 3

log3

= 4 log3 3+3 log3



= 4+3



60

Chapter 4 Indices, Surds and Logarithms

4.3.5 Solving problems involving the law of logarithms

By applying the law of logarithms, we can solve problems involving indices that cannot be solved by
the way of equating = or = .

EXAMPLE 16 (ii) 32 − 3 +1 + 2 = 0

Solve the following equations:
(i) 3 +1 = 7 −2

Solution:

(i) 3 +1 = 7 −2
log 3 +1 = log 7 −2

ሺ + 1ሻ log 3 = ሺ − 2ሻ log 7
log 3 + log 3 = log 7 − 2 log 7
log 3 − log 7 = −2 log 7 − log 3
ሺlog 3 − log 7ሻ = −2 log 7 − log 3

= −2 log 7−log 3

log 3−log 7

= 5.890

(ii) 32 − 3 +1 + 2 = 0

ሺ3 ሻ2 − 3ሺ3 ሻ + 2 = 0

ሺ3 − 2ሻሺ3 − 1ሻ = 0

3 = 2 or 3 = 1

log 3 = log 2 or log 3 = log 1

log 3 = log 2 or log 3 = 0

= log 2 or = 0
log 3

= 0.631 or = 0

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Chapter 4 Indices, Surds and Logarithms

4.4 Application of Indices, Surds and Logarithms

4.4.1 Solving problems involving indices, surds and logarithms

EXAMPLE 17

(i) A colony of bacteria weighs ሺ ሻ = 100 + 30.02 gram after time (in
minutes). Find the time needed for the bacteria’s mass to reach 180g.

Solution:

100 + 30.02 = 180

30.02 = 80

log 30.02 = log 80

0.02 log 3 = log 80

0.02 = log 80
log 3

= 199.43 minutes

(ii) The total number of people infected by a highly infectious virus follow
the equation ሺ ሻ = 1500 + 0.9 after time (in days). Determine the
minimum number of days needed for the virus to infect more than 9500
people.

Solution:

1500 + 0.9 > 9500
0.9 > 8000

ln 0.9 > ln 8000
0.9 ln > ln 8000

> ln 8000

0.9

> 9.986

We round the answer the nearest integer larger than 9.986, the minimum
number of days needed for the virus to infect more than 9500 people is 10
days.

62

Chapter 4 Indices, Surds and Logarithms

Summative Exercises

4.1 Law of Indices 3. Given that
2 log3( − ) = 1 + log3 + log3
1. Simplify the expressions below:
Show that 2 + 2 = 5 .
13 4. (i) Simplify the expression:

(4 2) log2( + 2) + 5 log4 − 3 log2 √
(ii) Hence, solve the equation:
(i) 2√ log2( + 2) + 5 log4 − 3 log2 √ = 3
3 +1+9 5. Find the value of:
(ii) 3 −1+1
log2 3 × log3 4 × log4 5 × log5 6
2. Show that:
(i) 3 + 3 +1 + 3 +3 is divisible by 31. 4.4 Application of Indices, Surds
(ii) 7 + 7 +2 + 72 + 1 is divisible by 25. and Logarithms

3. Solve the following equations: 1. The diagram below shows 3 circles.
(i) 3 −3 = 9
(ii) 25(5 +1) = 52 −1 All circles touch each other and lie on a
common tangent of PQ. Circle A has a radius of
4.2 Law of Surds 12cm, and circle B has a radius of 6cm. Find the
radius of the circle C.
1. Convert the following recurring decimals 2. Given a car’s value will deplete over years,
the car’s value after t years is RM80000 (89) .
into fractions: Find:
(i) 0. 1̅̅̅4̅ (i) the value of the car after 5 years.
(ii) 0. 1̅ (ii) number of years for the car to drop below
(iii) 0.25̅̅̅6̅ RM35000 for the first time.

2. Rationalise the following expressions: 63

(i) 3
2√2
1
(ii) 3√7−2√2

(iii) 2√3−3√5
2√3+3√5

4.3 Law of Logarithms

1. Solve the following equations:

(i) log2(log3(2 + 1)) = 2
(ii) log64(log2(3 − 2)) = log25 3√5
(iii) 3log3(2 −3) = 15

2. Determine the value(s) of in the following

equations:

(i) log9(2 + 12) = log3( + 2)
2
(ii) log7 2 = log2(4 − 1)

(iii) ln 3 +4 = 25log5 √5

CHAPTER 5 PROGRESSIONS

Chapter 5 Progressions

5.1 Arithmetic Progression

5.1.1 What is an Arithmetic Progression?

An arithmetic progression is a sequence of numbers with a common difference between any two
consecutive terms.
The difference between any two consecutive terms in an arithmetic progression is called the common
difference, indicated by the symbol d.
Let’s say we have an arithmetic progression as follow:

120, 128/, 136, 144, …

To calculate d, we can apply the formula

=
where represents the 优 term, n = 1, 2, 3, 4, …

For this case,

= 120 = 128 = 136
= = =
= 128 = 136 = 144
=8 =8 =8

We can see that = = , thus this is an arithmetic progression.
(Arithmetic progression always has the same difference between any two consecutive terms.)

There are three cases for the values of d.

1. d > 0
The terms in the arithmetic progression are increasing constantly,
Eg: 20, 40, 60, 80, 100, …

2. d < 0
The terms in the arithmetic progression are decreasing constantly,
Eg: 100, 80, 60, 40, 20, …

3. d = 0
The terms in the arithmetic progression are neither increasing nor decreasing.
Eg: 20, 20, 20, 20, 20, …

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CHAPTER 5 PROGRESSIONS

EXAMPLE 1

Given the following sequences, determine whether they are arithmetic sequences.

(a) 80,83,86,89, … (b) 865, 859, 853, 846, …

Solution: 80 (b) = 859 865
(a) = 83 83 =6
86 = 853 859
=3 =6
= 86 = 846 853
=3 =7
= 89
=3

Since there is a common difference = Since ≠ , it is not an arithmetic
= , it is an arithmetic sequence. sequence.

Now, let’s explore the properties of arithmetic progression a little bit more.

Suppose we have an arithmetic progression a, b, c, …, we can find d by either subtracting a from b or
subtracting b from c. Thus,

b a=c b

From the following equation, we get the following relationship for any three consecutive terms
(a, b, c) in an arithmetic progression.

2b = c + a

5.1.2 How to find the value of the 优 term?

Assuming that a means the first term of the arithmetic progression and d means the common
difference, we can see the following pattern in the arithmetic progression.

12,19,26,33, …

The first term 12 = 12 + 0 x 7 =a+0xd
The second term 19 = 12 + 1 x 7 =a+1xd
The third term 26 = 12 + 2 x 7 =a+2xd
The fourth term 33 = 12 + 3 x 7 =a+3xd

Hence, we can conclude that = a + (n 1)d

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CHAPTER 5 PROGRESSIONS

where represents the 优 term,
a represents the first term,
n represents the number of terms,
d represents the common difference.

EXAMPLE 2

Given the arithmetic sequence 63,69,75, …,
(a) find the 优 term,
(b) find the smallest term which is greater than 100 for this sequence, and
(c) find the value of n if it is known that the 优 term has the value 165.

Solution: 63
First term, a = 63

Common difference, d = 69
=6

(a) = a + (n 1)d 1)6
The 优 term, t = 63 + (25
= 63 + 144

= 207

(b) > 100

a + (n 1)d > 100

63 + (n 1)6 > 100

57 + 6n > 100

100 7

n> 6
n > 7.167

The smallest integer n after 7.167 is 8, hence the smallest term which is greater than 100 in
this sequence is the 优 term, .

= a + (n 1)d 1)6
The 优 term, = 63 + (8

= 63 + 42

= 105

(c) = a + (n 1)d

165 = 63 + (n 1)6

165 = 57 + 6n

16 7

n= 6
= 18

5.1.3 How to find the sum of the first n terms?

Let’s say we have a finite arithmetic progression given by
a , a + d , a + 2d , a + 3d , a + 4d , … , l d , l

66

CHAPTER 5 PROGRESSIONS

where a represents the first term,
d represents the common difference,
l represents the last term.

Assuming the arithmetic progression has n terms, we can calculate the sum of the first n terms, ,

Sum of the first n terms, = a + a + d + a + 2d + a + 3d + a + 4d + … + l d + l ○1

We can also write in the following way, ○2
Sum of the first n terms, = l + l d + l 2d + l 3d + l 4d + … + a + d + a ○3

○1 + ○2 : 2 = n(a + l)

= (a + l)

From ○3 , we can deduce the sum of the first n terms, as follow:

Sum of the first n terms, = (a + )

= [a + a + (n 1)d] ○4
= [2a + (n 1)d]

From above, we get two formulae,

= (a + l)

= [2a + (n 1)d]

where represents sum of the first n terms in a finite arithmetic progression,
n represents number of terms,
a represents the first term,
l represents the last term,
d represents the common difference.

EXAMPLE 3

Given the arithmetic sequence 28, 41, 54, …,
(a) find the sum of the first 14 terms.
(b) find the sum of the terms from the fifth term to the tenth term, and
(c) find the smallest value of n such that the sum of the first n terms is more than 500.

67

CHAPTER 5 PROGRESSIONS

Solution: 28
First term, a = 28

Common difference, d = 41
= 13

(a) = [2a + (n 1)d] = 14[2(28) + (14 1)13]

Sum of the first 14 terms,

(b) Sum of terms from = 7(225)
= 1575
优 term to 10 优 term = 1)13] 4[2(28) + (4 1)13]

[10

= 2(28) + (10

= 5(173) 2(95)
= 865 190
= 675

(c) > 500

[2a + (n 1)d] > 500

[2(28) + (n 1)13] > 500

(43 + 13n) > 500

13 + 43n 1000 > 0
n < 10.58, n > 7.27
Since will be always positive, n < 10.58 is rejected, hence the smallest value of n such
that the sum of the first n terms is more than 500 is 8.

5.1.4 What are the applications of arithmetic progression?

So far, we have gone thru what actually arithmetic progression is and also know how to find the 优
term as well as the sum of the first n terms. However, do you even know what the applications of
arithmetic progression in our real life are? Here are some examples of the applications.

In some of the cinemas and symposium halls or even stadiums, the arrangement of the seats forms an
arithmetic progression where the number of seats increases constantly from the first row to the last
row. For instance, there are 8 seats for the first row, 9 seats for the second row, 10 seats for the second
row and so on. The design makes the cinemas, halls and stadiums to have a stylish appearance.

68

CHAPTER 5 PROGRESSIONS

With the passage of time, our asset like cars and houses may depreciate by a fixed amount, k per year.
The value of the asset each year can be well represented by an arithmetic progression p, p k, p
2k, … where p represents the initial value of the asset.

Do you know Old Faithful? It is a natural geyser situated at the Yellow Stone National Park that
produces long eruptions which can be predicted using the knowledge of arithmetic progression. The
time between eruptions is believed to be based on the length of previous eruption.

If an eruption lasts for 1 minutes, then the next eruption will occur in approximately 46 minutes.
If an eruption lasts for 2 minutes, then the next eruption will occur in approximately 58 minutes.
If an eruption lasts for 3 minutes, then the next eruption will occur in approximately 70 minutes.
From the information above, we can clearly see that the time for the next eruption follows an
arithmetic progression 46, 58, 70, …. with a common difference of 12.
Try to predict how many minutes after the previous eruption will the next eruption occur if the
previous eruption lasts for 5 minutes? Yes, it is the fifth term of the progression and it is 94 minutes.

69

CHAPTER 5 PROGRESSIONS

1

Have you ever used ladder before to fix the lamps or perhaps clean the fan? Actually, the lengths of
the rungs of a ladder decrease uniformly from the bottom to the top. Take a simple example, the
bottom rung of a ladder has a length of 40 cm and the lengths of the rungs decline constantly by 3 cm
until the top rung. There are five rungs for the ladder, thus we can express the length of the rungs
from the bottom to the top in an arithmetic progression 37 cm, 34 cm, 31 cm, 28 cm, 25 cm.

Rung

Apart from the examples shown, the application of arithmetic progression still can be observed in
many other aspects of our daily life. Have you figured them out?

1 https://en.m.wikipedia.org/wiki/Old_Faithful

70

CHAPTER 5 PROGRESSIONS

5.2 Geometric Progression

5.2.1 What is a Geometric Progression?

A geometric progression is a sequence of numbers in which any term after the first term is obtained
by multiplying the previous term by a non-zero constant.
The non-zero constant is called the common ratio, indicated by the symbol .
Let’s say we have a geometric progression as follow:

60, 180, 540, 1620, …

To calculate , we can apply the formula

=
where represents the 优 term, n = 1, 2, 3, 4, …

For this case,

== =

= = =

10 40 16 0

= 60 =1 0 = 40
=3 =3 =3

We can see that = = , thus this is a geometric progression.
(Geometric progression always has same non-zero common ratio between any two consecutive terms.)

There are five cases for the values of .

1. > 1
The terms in the geometric progression are increasing exponentially if the first term is
positive and vice versa.
Eg 1: 20, 40, 80, 160, 320, … (Increase exponentially)
Eg 2: 20, 40, 80, 160, 320, … (Decrease exponentially)

2. < 1
The terms in the geometric progression increases in their absolute values and alternate in
signs.
Eg: 20, 40, 80, 160, 320, …

3. = 1
The terms in the geometric progression stay the same.
Eg: 20, 20, 20, 20, 20, …

71

CHAPTER 5 PROGRESSIONS

4. = 1
It forms a sequence of numbers in same magnitude with alternating signs.
Eg: 20, 20, 20, 20, 20, …

5. 1 < < 1 where 0
The terms of the geometric sequence show exponential decay to 0.
Eg: 20, 4, 0.8, 0.16, 0.032, …

EXAMPLE 4

Given the following sequences, determine whether they are geometric sequences.

(a) 25, 100, 400, 1600, … (b) 3600, 600, 100, 20, …

Solution: 600

100 (b) = 3600

(a) = 1

=4 =6

400 100

= 100 = 600

=4 1

= 1600 =6
400
0
=4
= 100
Since = = , it is a geometric sequence.
1

=

Since , it is not a geometric

sequence.

Now, let’s explore the properties of geometric progression a little bit more.
Suppose we have a geometric progression a, b, c, we can find by either dividing a from b or
dividing b from c. Thus,

=

From the following equation, we get the following relationship for any three consecutive terms
(a, b, c) in a geometric progression.

= ac

72

CHAPTER 5 PROGRESSIONS

5.2.2 How to find the value of the 优 term?

Assuming that a means the first term of the geometric progression and r means the common ratio,
we can see the following pattern in the geometric progression.

10,20,40,80, …

The first term, 10 = 10 x 0 =ax
The second term, 20 = 10 x 1 =ax
The third term, =ax
The fourth term, 40 = 10 x =ax
80 = 10 x 3

Hence, we can conclude that

=a

where a represents the first term, 1,

r represents the common ratio, where r
represents the 优 term.

EXAMPLE 5

Given the geometric sequence 0.5, 1, 2, 4, …,
(a) find the 10 优 term,
(b) find the smallest term which is greater than 100 for this sequence, and
(c) find the value of n if it is known that the 优 term of the sequence has the value 1024.

Solution:
First term, a = 0.5

1

Common ratio, r = 0
=2

(a) = a = 0.5 10 1
The 10 优 term,

= 0.5(512)

= 256

(b) > 100

a > 100
(0.5) 1 > 100
1 > 200
ln( 1) > ln 200

(n 1) ln 2 > ln 200

ln 00

n 1 > ln

73

CHAPTER 5 PROGRESSIONS

5.2.3 How to find the sum of the first n terms?

Let’s say we have a geometric progression given by
a , ar , a , a , … , a , a

where a represents the first term, 1,
r represents the common ratio, where r
n represents the number of terms.

From the geometric progression, we can calculate the sum of the first n terms, ,

= a + ar + a + a + … + a + a ○1

We multiply with the common ratio, r, ○2
r = ar + a + a + a + … + a + a

○1 ○2 ,

(1 r) = a a
=

74

CHAPTER 5 PROGRESSIONS

=

○2 ○1 ,

(r 1) = a a

=

= −
−

Hence, we can conclude that

=

−
=−

where represents sum of the first n terms in a finite geometric progression,
a represents the first term,
r represents the common ratio where r 1,
n represents the number of terms.

We can use either of these formulae to calculate . Two of the formulae will give the same results.
Note that,

= 1 . −
1 =−

EXAMPLE 6

Consider the geometric sequence 13, 39, 117, …, 85293. Find the sum of the terms.

Solution:

First term, a = 13

3h

Common ratio, r = 13
=3

First, since we know that the last term of the progression is 85923, we apply the formula
= a to find the value of n.

85293 = 13 3 1

h3 31
31
13 =
6561 =

3= 3 1

8=n 1

n=9

75

CHAPTER 5 PROGRESSIONS

Now, we know that n is 9, so we can apply the formula = to find the sum of
terms, .

Sum of terms, 13 3h 1
31
= 1h6 4
4
13

=

= 63973

5.2.4 Can we calculate sum to infinity of the geometric progression?

As we have discussed earlier, there are five cases totally for the value of common ratio, r as follows:

1. > 1 0
2. < 1
3. = 1
4. = 1
5. 1 < < 1 where

We shall investigate whether or not we can calculate sum to infinity of the geometric progression for
every case one by one.

When > 1 or < 1, the terms in the geometric progression keep getting larger and larger in
magnitude, the sequence of terms diverges towards infinity, hence we are unable to find the sum to
infinity for this case.

3, 9, 27, 81, 243, …

When = 1, all the terms in the geometric progression will be the same, the series diverges, hence we
are unable to find the sum to infinity for this case.

5, 5, 5, 5, 5, …

When = 1, it forms a sequence of two values in same magnitude with alternate signs. The sum to
infinity of this kind of geometric progression oscillates between two values which are 0 and the first
term of the sequence, hence we are unable to find the sum to infinity for this case.

S=0

4, 4, 4, 4, 4, …

S=4

When 1 < < 1, the value of the term in the geometric progression approaches 0 (negligible) when
the number of terms approaches infinity. The series is said to converge to 0, hence we are able to find
the sum of infinity for this case.

100, 10, 1, 0.1, 0.01, 0.001, …

76

CHAPTER 5 PROGRESSIONS
Let’s have a visual look for this case.

2

The areas of the coloured squares follow a geometric progression
11 1 1
4 , 16 , 64 , … with a common ratio of 4.
The areas of the squares become smaller and approach zero.

You can also have visual look on infinite geometric progression via this video:
https://youtu.be/-y1Ob0K63hc

Now, we shall explore how to calculate the sum to infinity of geometric progression with the
restriction that 1 < < 1.

We have learnt that = −
or = − in our previous subtopic. From the formulae,

lim = lim − = When n approaches ,
= − approaches 0

=

Thus, we deduce a new formula here.

=
where represents the “sum” of an infinite geometric progression,

a represents the first term,
r represents the common ratio where 1 < r < 1, r 0.

2 https://en.m.wikipedia.org/wiki/Geometric_series

77

CHAPTER 5 PROGRESSIONS

EXAMPLE 7

Given a geometric progression 0.9, 0.81, 0.729, …, find the sum to infinity.

Solution:
First term, a = 0.9

01

Common ratio, r = 0 h
= 0.9

Sum to infinity, =

0h

=1 0h

0h

=01
=9

EXAMPLE 8

Express the recurring decimal 0.65454… in the simplest fraction form.

Solution:

0.65454… = 0.6 + (0.054 + 0.00054 + 0.0000054 + … )

= 0.6 +

3 00 4 An infinite geometric progression 0.054, 0.00054, 0.0000054
with = 0.054 and = 0.01
= + 1 0 01
=
3 00 4

= + 0 hh

33

=+

36

=

5.2.5 What are the applications of geometric progression?

Geometric progression is an important field of mathematics and have crucial applications in various
fields like physics, biology, finance, etc. Here are some of the examples of the applications in
different fields.

Mr. Lee, an entrepreneur, has invested RM5000 in a fixed deposit plan with an interest rate at 3.7%
yearly. His investment each subsequent year after that follows a geometric progression
RM5000(1.037), RM5000 1 037 , RM5000 1 037 3 , …. By using knowledge of geometric
progression, he can easily estimate the amount of savings he will have after n years, .

78

CHAPTER 5 PROGRESSIONS

1st year 2nd year 3rd year

Normally, bacteria will replicate by binary fission, a process by which a bacterium split into two. As a
result, the amount of the bacteria follows a geometric progression as the population of bacteria
doubles every generation time. For most of the bacteria, the generation time will be within 20 to 60
minutes under optimum conditions. Thus, the population of bacteria after a given time period, , the
initial population of the bacteria, 0 and number of divisions are related by the following equation:

Doesn’t it seem alike to the formula = = 0x
1?

Now, let’s have an example to see how geometric progression is applied for this case. Supposed a

bacterial cell can split into two in 25 minutes, given initial population of the bacteria is 100, what will

be the population of the bacteria after 5 hours, ? Since the number of divisions, n that will happen
in 5 hours will be 12 (300 minutes/25 minutes), we can get 409600 (100 x 1 ) bacteria after 5 hours.

The number of the bacteria after every division is illustrated as follow:

200, 400, 800, 1600, 3200, 6400, …

79

CHAPTER 5 PROGRESSIONS

Geometric progression is applied in fractal geometry. In mathematics, a fractal is a self-similar
geometric figure in which they exhibit similar pattern at increasingly small scales and also at different
levels of views. Koch snowflake is one of the examples for the fractal space.

3The Koch snowflake is formed in a sequence of stage. First, start with an equilateral triangle. On the

middle third of each side, draw an equilateral triangle of 1 of the length of the previously drawn
3

triangle, bending outwards and then erase the original middle third of that side. You now should have

a 12-sided polygon with 6 vertices. Continue the process for each of the 12 sides again. You now

should have a 36-sided polygon with 18 vertices. Then, continue the process until you get a beautiful

snowflake.

There are other examples of fractal geometries as well.

You may check how geometric progression is applied in fractal geometry via these videos:
https://youtu.be/rqP_bvOoJak
https://youtu.be/ocswsR6qFll

3 https://en.m.wikipedia.org/wiki/Koch_snowflake

80

CHAPTER 5 PROGRESSIONS

Now, consider an infinite geometric progression 0 1 3 4 where the values of x are greater

than 1 and less than 1 but not equal to zero. This geometric progression has 1 as its first term and x
as its common ratio. Since the values of x are restricted between 1 and 1, we can calculate the finite

sum of this progression, . The calculations are as below:

=

01 3 4 + … = 11
3 4 +… = 11
11

A new relation is formed from the progression. This relation is very useful to derive other useful

formulae to be applied in the field of Calculus as well as Physics.

For example, if we differentiate the following relation, we can get the following formula:

t1 1 = 1 1 for 1 1 where 0

Arithmetic Progression (Part 1)

https://youtu.be/hEU3byoqyVY

Arithmetic Progression (Part 2)

https://youtu.be/Q98NyR50mHw

Geometric Progression (Part 1)

https://youtu.be/3AYb5tqPle4

Geometric Progression (Part 2)

https://youtu.be/RLZaGJ9nK2U

Arithmetic and Geometric Progression Formulae

https://youtu.be/gua96ju_FBk

81

CHAPTER 5 PROGRESSIONS

Summative Exercises

5.1 Arithmetic Progression 5.2 Geometric Progression

1) Given an arithmetic progression in which the first 1) Given the geometric sequence 20, 40, 80, …
term is 1038, the third term is 964 and the last (a) find the 10 优 term,
term is 594. Find the number of terms for the (b) if the sequence only applies to the numbers
arithmetic progression. less than 10000, find the last term,
(c) find sum of the terms from t term to
2) Harry has RM x in his piggy bank. He saves RM12 7 优 term.
on the first day, RM15 on the second day, RM18
on the third day and follows this pattern for the 2) Puan Kiah invests RM 5000 in her fixed deposit
subsequent days. On the tenth day, he saves his saving account with an annual interest rate of 3%.
money as usual and finds that his piggy bank has How many years minimum does it take to have
a total of RM605 inside. Find the value of x. more than RM 6000 if she does not withdraw the
money throughout the duration?
3) It is known that the middle two terms of an
arithmetic progression add up to 268 and the third 3) A square paper with length of 22 cm is folded in
term is 101. Given the progression has 16 terms, half. The action is repeated for another 5 times.
find the last term. Calculate the area of the folded paper at the end.

4) Rashid draws a circle with a radius of 3 cm at 4) Given the first term is 2000, the last term is
first. Then, he increases the radius to 6 cm and 1180.98 and there are six terms for a geometric
draw the new circle besides the previous circle. sequence. Find the sum of the first four terms.
He keeps drawing by following the arithmetic
sequence until he had drawn 7 circles in total. 5) A ball is dropped from a height of 24 cm. It
After that, he draws a rectangle to inscribe the rebounds to a height of 21 cm and continues to
circles. Find the area of the rectangle. rebound to 0.875 of its height continuously.
Calculate the total distance it travels before it
5) The sum of the first five terms and the first eight comes to a rest.
16070
terms for an arithmetic progression are 3 6) Given 7 cylinders arranged in a row in which their
radius and height follow geometric progression.
and 3 6. Find the tenth term. The radius and height are incrementing by 2 times
and 1.4 times respectively. The first cylinder has
6) Given an arithmetic sequence 78, 82, 86, … a radius of 1 cm and height of 5 cm. Calculate the
(a) find the sum of the 10 terms after the 优 term, volume of the last cylinder.
(b) find the smallest term which is more than 200,
(c) if the last second term of the sequence is 158, 7) Express the recurring decimal 5.3888…. in its
find the sum of the last five terms. simplest fraction form.

82

CHAPTER 1 FUNCTIONS

SOLUTIONS

Chapter 1 Functions

1.1 Functions

1) (a) ( ) = 2 + 4

4 is the object. Therefore,

(4) = 42 + 4 Whenever see “x” in the function ( ) = 2 + 4, substitute = 4

= 16 + 4

= 20

(b) 4 is the image. Therefore,

( ) = 4

2 + 4 = 4 Since ( ) = 2 + 4

2 = 0

= 0

2) (a) Function is undefined when denominator = 0

− 4 = 0

= 4 Since =
= 4

(b) (2) = 2 2 is mapped onto itself under

2 + 3
2−4 =2
2 + 3
−2 = 2
2 + 3 = −4

2 = −7
7

= − 2

83

CHAPTER 1 FUNCTIONS

3) Step 1: You need to have 4 points in order to construct a graph.

The four points are: x f (x)
(a) left endpoint of the graph, left endpoint → = −4 −4 1
(b) vertex,
(c) y-intercept, −3 0 vertex → ( ) = 0
(d) right endpoint of the graph. y-intercept → = 0 0 3
right endpoint → = 1 1 4

(a) When = −4, (c) When = 0,
(−4) = |−4 + 3| (0) = |0 + 3|
= | − 1| = |3|
=1 =3

(b) When ( ) = 0 (d) When = 1,
| + 3| = 0 (1) = |1 + 3|
+ 3 = 0 = |4|
= −3 =4

Step 2: You need to plot the points onto a graph paper and indicate the x-axis and y-axis.
Step 3: Connect the points in the shape of V with vertex as the turning point.

( )

(1,4)

(0,3)

( ) = | + 3|

(-4,1) Range of the function → 0 ≤ f(x) ≤ 4
0 and 4 are included in the range of function

(-3,0) x

84

CHAPTER 1 FUNCTIONS

1.2 Composite Functions

1) In order to find ( ), you need to have the function ( ) and ( ).
Now, the question has given the function ( ).
Therefore you need to find the function ( ).
( ) = 2 2 + 5
( ) = 2 + 1
[ ( )] =2 ( ) + 1
2 2 + 5 = 2 ( ) + 1
2 ( ) = 2 2 + 4
( ) = 2 + 2
( ) = [ ( )]
= (2 + 1)
= (2 + 1)2 + 2
= 4 2 + 4 + 1 + 2
= 4 2 + 4 + 3

( ) = 2
4 2 + 4 + 3 = 2
4 2 + 4 + 1 = 0
(2 + 1)(2 + 1) = 0

1
= − 2

85

CHAPTER 1 FUNCTIONS

2) 1 +
( ) = 1 − , ≠ 1

(a) 2( ) = ( )

1 +
= (1 − )

= 1 + 1 +
1 −
1 −
1 +

1 −

(1 − ) + (1 + )

= 1− + )
(1 − ) − (1

1 −

2

= 1 −
−2

1 −

2 1 −
= 1 − × −2
1
= − , ≠ 0

(b) 3( ) = 2 ( )

= 2 1 +
(1 − )
1
= − 1 +

1 −

1 −
= − 1 +

− 1
= + 1 , ≠ −1

ALTERNATIVE METHOD

3( ) = 2( )

1
= (− )

= 1 + (− 1 )
1 − (− 1 )

− 1

= 1
+



− 1
= + 1 , ≠ −1

86

CHAPTER 1 FUNCTIONS

2) (c) 6( ) = 2 2 2( ) 1

= 2 2 (− 1 − 1
)
−

= 2 (− 11)

− +

= 2( )

1
= − , ≠ 0

ALTERNATIVE METHOD

6( ) = 3 3( )

= 3 − 1
( + 1)

= − 1 − 1
+ 1
+ 1
− 1

+ 1

( − 1) − ( + 1)

= ( − + 1 + 1)
1) + (

+ 1

−2

= + 1
2

+ 1

−2 + 1
= + 1 × 2

1
= − , ≠ 0

3) (a) ( ) = 5 (Roland could walk 5x distance in x hours)
(b) ( ) = 2

The furthest that Roland could r Spot where Roland is missing
walk is by walking through a r
straight line. The area that we
could possibly find Roland is
the area of the circle whereby
the radius of circle is r miles.

(c) ( ) = (5 )
= (5 )2
= 25 2

87

CHAPTER 1 FUNCTIONS

4) (a) In order to write the function for the height h of water after t seconds, we need to have the initial

height of water first.

= 2ℎ

1540 = 22 × 72 × ℎ
7

1540 × 7
ℎ = 22 × 72

ℎ = 10

ℎ( ) = 10 +

(b) (ℎ) = 2ℎ

= 22 × 72 × ℎ
7
= 154ℎ

(c) ℎ( ) = (10 + )

= 154(10 + )

= 1540 + 154

(d) When = 10,

ℎ(10) = 1540 + 154(10)

= 3080 3

1.3 Inverse Functions

1) 2
( ) = − 2

( ) = 3 − 2

(a) 2
= − 2

− 2 = 2

= 2 + 2

2 + 2
=

−1 ( ) = 2 + 2


−1 ( ) = 2 + 2 , ≠ 0


(b) = 3 − 2

2 = 3 −

3 −
= 2

−1( ) = 3 −
2
3 −
−1( ) = 2

88

CHAPTER 1 FUNCTIONS

1) (c) −1 −1( ) = −1 2 + 2
( )

3 − 2 + 2

=
2

1 3 − (2 + 2)
=2×


− 2
= 2 , ≠ 0

(d) In order to find ( )−1( ), we need to find ( ) first.
( ) = (3 − 2 )

2
= 3 − 2 − 2

21
= 1 − 2 , ≠ 2

2
= 1 − 2

− 2 = 2
−2 = 2 −
2 −
= −2

= − 2
( )−1( ) =
2− 2
2

( )−1( ) = − 2 , ≠ 0
2

89

CHAPTER 1 FUNCTIONS

(e) ( ) = ( 2 2)
−

2
= 3 − 2 ( − 2)

3( − 2) − 4
= − 2

3 − 10
= − 2

3 − 10
= − 2
− 2 = 3 − 10

− 3 = 2 − 10

( − 3) = 2 − 10

2 − 10
= − 3

( )−1( ) = 2 − 10
− 3

( )−1( ) = 2 − 10 , ≠ 3
− 3

CONCLUSION −2
2
−1 −1( ) = ( )−1( ) =

Therefore, −1 −1( ) = ( )−1( )

2) ( ) = − 4
(a) 2( ) = ( )
= ( − 4)
= ( − 4) − 4

= − 8

= − 4
= + 4

−1( ) = + 4
−1( ) = + 4

90

CHAPTER 1 FUNCTIONS
2) (b) = − 8

= + 8
( 2)−1( ) = + 8
( 2)−1( ) = + 8
= ( −1)2( )

= −1 −1( )
= −1( + 4)
= ( + 4) + 4
= + 8
= ( 2)−1( )
=
Therefore ( −1)2( ) = ( 2)−1( ).
(c) −1 ( ) = −1(5 + 2)

= (5 + 2) + 4
− 5 = 2 + 4

−4 = 6
3

= − 2

91

Chapter 2 Quadratic Functions

2.1 Quadratic Equations and Inequalities

1. 3 ( − 5) = 2 − 1

3 * − 15 = 2 − 1

3 * − 17 + 1 = 0

3 . * − 17 + 13/ = 0
3
12−213734* 12−213734*
3 0 * − 17 + − + 315 = 0
3

3 6. − 167/* − 237678 = 0

3 . − 167/* − 277 = 0
12
167/*
3 . − = 277
12
167/*
. − = 277
36

− 17 = ±:23767
6

= ±:23767 + 17
6

= 0.059 = 5.607

2. Given = 5 = 9.

, + = 5 + 9 = 14
, = 5(9) = 45
* − ( ) + ( ) = 0
* − 14 + 45 = 0

3. 2( − 5)* = 4( + 7)

2( * − 10 + 25) = 4 + 28

2 * − 20 + 100 = 4 + 28

2 * − 24 + 72 = 0

* − 12 + 36 = 0

= 1, = −12, = 36

+ = − (−12) = 12 = 36 = 36
1 1
Roots: 2 , 2

= 2 + 2 = 2( + ) = 2(12) = 24

= 2 (2 ) = 4( ) = 4(36) = 144

92

* − 24 + 144 = 0

4. Roots: −3 P
*
1 5
= −3 + 2 = − 2

= −3 .12/ = − 3
2
* − .− 52/ + .− 32/ = 0
5 3
* + 2 − 2 = 0

2 * + 5 − 3 = 0

Compare 2 * + 5 − 3 = 0 and 2 * + ( + 1) + = 2

+ 1 = 5

= 4

− 2 = −3

= −5

5. * − 4 + 3 > 0
( − 1)( − 3) > 0
When ( − 1)( − 3) = 0, = 1 = 3.



++
1− 3

Since ( − 1)( − 3) > 0, thus the range of values of is determined on the graph above the
-axis. Hence, the range of values of is < 1 > 3.



93

2.2 Types of Roots of Quadratic Equations

1. ( − 2) = 5
* − 2 − 5 = 0

= 1, = −2, = −5
* − 4 = (−2)* − 4(1)(−5) = 24 (> 0)
Since the discriminant is > 0. Thus, ( − 2) = 5 has two real roots.

( + 5) = 2 − 14
* + 5 − 2 + 14 = 0
* + 3 + 14 = 0

= 1, = 3, = 14
* − 4 = (3)* − 4(1)(14) = −47(< 0)
Since the discriminant is < 0. Thus, ( + 5) = 2 − 14 has no real root.

2. * + 2ℎ + 4 =
* + (2ℎ − 1) + 4 = 0

= 1, = 2ℎ − 1, = 4
* − 4 = 0
(2ℎ − 1)* − 4(1)(4) = 0
(2ℎ − 1)* − 16 = 0

2ℎ − 1 = ±√16
2ℎ − 1 = 4 2ℎ − 1 = −4

2ℎ = 5 2ℎ = −3

ℎ = 5 ℎ = − 3
2 2

3. * + 5 = 4

* − 4 + 5 = 0

= 1, = −4 , = 5

* − 4 = 0

(−4 )* − 4(1)(5 ) = 0

16 * − 20 = 0

16 * = 20

* = 5
4
1
= ± 2 √5

4. * + + = 0 1

= 1, = , = 2
* − 4 = 16
* − 4(1)( ) = 16

− = −4

= − 4
( − 4)* − 4 = 16

94

* − 8 + 16 − 4 = 16
* − 12 = 0
( − 0)( − 12) = 0
= 12 = 0 (ignored)
= − 4
= 12 − 4
= 8

2.3 Quadratic Functions

1.
( )

( ) = − * + + 6

( ) ( ) = − * − + 6

( ) ( ) = − 1 * + + 6
2


( ) ( ) = − * + + 1

2. ( ) = 0
2 * + − 5 = 0

= 2, = 1, = −5
* − 4 = (1)* − 4(2)(−5) = 41(> 0)
Thus, the quadratic function ( ) = 2 * + − 5 has two real roots. Since > 0, hence the

graph is a parabola which passes through minimum point and intersects two points on -axis.

95

3. ( ) = * + 4 − 6

= , = 4, = −6

* − 4 < 0

(4)* − 4( )(−6) < 0

16 + 24 < 0

24 < −16

< − 2
3

4. ( ) = 5 * − ( + 4) − 2
( ) = 5 * − * − 4 − 2
( ) = (5 − ) * − 4 − 2

= 5 − , = −4, = −2
* − 4 > 0
(−4)* − 4(5 − )(−2) > 0

16 + 40 − 8 > 0

56 > 8
< 7

96