Class 9th Mathematics

Chapters covered:

1. Number System

2. Polynomials

3. Coordinate Geometry

4. Linear Equations in two variables

CBSE Class 9 Mathematics

Course Content: Chapters 1-4

Comprehensive Study Guide with Problem-Solving Strategies

Chapter 1: Number Systems

1.1 Introduction to Number Systems

What are Number Systems?

A number system is a way of organizing and classifying different types of numbers. Think of it as a family tree of numbers, where each type of number has its own special properties and uses.

The Number Family:

Natural Numbers (N): The counting numbers: 1, 2, 3, 4, 5, ...

Whole Numbers (W): Natural numbers plus zero: 0, 1, 2, 3, 4, ...

Integers (Z): All positive and negative whole numbers: ..., -3, -2, -1, 0, 1, 2, 3, ...

Rational Numbers (Q): Numbers that can be expressed as pq where p and q are integers and q0

Irrational Numbers: Numbers that cannot be expressed as a simple fraction

Real Numbers (R): All rational and irrational numbers combined

Key Concept: Rational Numbers

A number r is called a rational number if it can be written in the form pq, where p and q are integers and q0.

Examples:

35 is rational

−25=−251 is rational

0=01 is rational

Every integer is a rational number!

Why is q0? Division by zero is undefined in mathematics. If we allowed q=0, we would break the fundamental rules of arithmetic.

Solved Problem 1.1: Identifying Number Types

Problem: Determine whether the following statements are true or false. Give reasons.

Every whole number is a natural number.

Every integer is a rational number.

Every rational number is an integer.

Solution:

False. Zero is a whole number but not a natural number. Natural numbers start from 1, while whole numbers include 0.

True. Every integer m can be expressed as m1, which is in the form pq. Therefore, every integer is a rational number.

False. The number 35 is a rational number, but it is not an integer because it lies between 0 and 1.

Solved Problem 1.2: Finding Rational Numbers Between Two Numbers

Problem: Find five rational numbers between 1 and 2.

Solution Method 1 (Midpoint Method):

To find a rational number between r and s, we can calculate r+s2.

Between 1 and 2: 1+22=32=1.5

Between 1 and 1.5: 1+1.52=2.52=1.25=54

Between 1.5 and 2: 1.5+22=3.52=1.75=74

Between 1 and 1.25: 1+1.252=1.125=98

Between 1.75 and 2: 1.75+22=1.875=158

Solution Method 2 (Equivalent Fractions):

Convert to fractions with the same denominator:

1=66 and 2=126

Five numbers between them: 76,86,96,106,116

Important Insight: There are infinitely many rational numbers between any two rational numbers!

1.2 Irrational Numbers

What Makes a Number Irrational?

A number s is called irrational if it cannot be written in the form pq, where p and q are integers and q0.

Common Examples of Irrational Numbers:

2=1.41421356...

3=1.73205080...

=3.14159265...

0.10110111011110... (non-terminating, non-repeating pattern)

Historical Note

The ancient Greek mathematician Pythagoras and his followers discovered irrational numbers around 400 BC. Legend says that Hippasus of Croton discovered that 2 is irrational, which contradicted the Pythagorean belief that all numbers could be expressed as ratios. This discovery was considered so shocking that, according to myth, Hippasus met an unfortunate end!

Locating Irrational Numbers on the Number Line

Constructing 2 on the Number Line:

Draw a square OABC with each side 1 unit long.

By Pythagoras theorem, diagonal OB = 12+12=2

Using a compass with center O and radius OB, draw an arc intersecting the number line at point P.

Point P represents 2 on the number line.

Solved Problem 1.3: Constructing Square Roots

Problem: Show how to represent 3 on the number line.

Solution:

Start with the construction of 2 (as described above).

At point B, construct a perpendicular BD of 1 unit length.

Using Pythagoras theorem: OD=OB2+BD2=(2)2+12=2+1=3

With compass center O and radius OD, draw an arc intersecting the number line at Q.

Point Q represents 3.

General Pattern: Once you have located n−1, you can locate n by constructing a perpendicular of unit length and applying the Pythagorean theorem.

1.3 Real Numbers and Their Decimal Expansions

Types of Decimal Expansions

Every real number has a decimal expansion, which falls into one of these categories:

Type

Example

Number Type

Terminating

0.875=78

Rational

Non-terminating recurring

0.333...=0.3=13

Rational

Non-terminating non-recurring

0.10110111011110...

Irrational

Table 1: Classification of decimal expansions

Key Theorem

For Rational Numbers:

The decimal expansion of a rational number is either terminating or non-terminating recurring.

For Irrational Numbers:

The decimal expansion of an irrational number is non-terminating non-recurring.

Solved Problem 1.4: Finding Decimal Expansions

Problem: Find the decimal expansions of 103, 78, and 17.

Solution:

For 103:

Performing long division:

103=3.333...=3.3

The remainder is always 1, so the digit 3 repeats forever. This is non-terminating recurring.

For 78:

78=0.875

The remainder becomes 0 after three steps. This is a terminating decimal.

For 17:

17=0.142857142857...=0.142857

The block of digits 142857 repeats. This is non-terminating recurring with a repeating block of 6 digits.

Solved Problem 1.5: Converting Recurring Decimals to Fractions

Problem: Express 0.3 in the form pq.

Solution:

Let x=0.333...

Multiply both sides by 10:

10x=3.333...

Notice that 3.333...=3+0.333...=3+x

Therefore:

10x=3+x

9x=3

x=39=13

Verification: 13=0.333... ✓

Solved Problem 1.6: Converting Repeating Blocks

Problem: Express 1.27 in the form pq.

Solution:

Let x=1.272727...

Since two digits repeat, multiply by 100:

100x=127.272727...

Notice that 127.272727...=126+1.272727...=126+x

Therefore:

100x=126+x

99x=126

x=12699=1411

Key Strategy: When n digits repeat, multiply by 10n.

1.4 Operations on Real Numbers

Properties of Operations

Real numbers satisfy several important properties:

Commutative Property:

Addition: a+b=b+a

Multiplication: ab=ba

Associative Property:

Addition: (a+b)+c=a+(b+c)

Multiplication: (ab)c=a(bc)

Distributive Property:

a(b+c)=ab+ac

Operations with Irrational Numbers

Important Rules:

Sum or difference of a rational and irrational number is irrational.

Example: 2+3 is irrational

Product or quotient of a non-zero rational and irrational number is irrational.

Example: 23 is irrational

Sum, difference, product, or quotient of two irrational numbers may be rational or irrational.

6+(−6)=0 (rational)

2+3 (irrational)

Solved Problem 1.7: Operating with Surds

Problem: Simplify: (22+53)+(2−33)

Solution:

Combine like terms:

=22+2+53−33

=32+23

This is irrational (sum of two irrational numbers).

Solved Problem 1.8: Multiplying Surds

Problem: Multiply 65 by 25.

Solution:

6525=(62)(55)

=125=60

Note: aa=a for any positive real number a.

Rationalizing the Denominator

What is Rationalization?

Converting a denominator containing an irrational number into a rational number.

Key Identity: (a+b)(a−b)=a2−b2

Solved Problem 1.9: Basic Rationalization

Problem: Rationalize the denominator of 12.

Solution:

Multiply numerator and denominator by 2:

12=1222=22

Solved Problem 1.10: Rationalizing Binomial Denominators

Problem: Rationalize the denominator of 12+3.

Solution:

Multiply by the conjugate 2−32−3:

12+3=12+32−32−3

=2−3(2)2−(3)2=2−34−3=2−3

Key Strategy: For denominator (a+b), multiply by a−ba−b.

1.5 Laws of Exponents for Real Numbers

Basic Laws of Exponents

For positive real numbers a and b, and rational numbers m and n:

aman=am+n

(am)n=amn

aman=am−n

ambm=(ab)m

a0=1 (for a0)

a−n=1an

a1n=na

amn=nam=(na)m

Solved Problem 1.11: Applying Exponent Laws

Problem: Simplify: 223213

Solution:

Using Law 1: aman=am+n

223213=223+13=233=21=2

Solved Problem 1.12: Fractional Exponents

Problem: Simplify: (64)12 and (32)15

Solution:

For (64)12:

6412=64=8

For (32)15:

3215=532=525=2

Practice Problems: Chapter 1

Set A: Number Systems Basics

Is zero a rational number? Can you write it in the form pq?

Find six rational numbers between 3 and 4.

State whether every natural number is a whole number, and vice versa.

Which of these are irrational: 25, 5, 227, ?

Set B: Decimal Expansions

Write the decimal expansion of 111 and identify if it terminates or repeats.

Express 0.47 in the form pq.

Find an irrational number between 17 and 27.

Classify as rational or irrational: 225, 7.478478..., 1.101001000100001...

Set C: Operations and Rationalization

Simplify: (3+7)(2+5)

Rationalize: 17, 53−5

Simplify: 55

Show that 27 is irrational.

Set D: Exponents

Find: (64)12, (125)13

Simplify: 22/321/521/3

Evaluate: (9)32

Simplify: (712812)

Chapter 2: Polynomials

2.1 Introduction to Polynomials

What is a Polynomial?

A polynomial is an algebraic expression consisting of variables and coefficients, involving only addition, subtraction, multiplication, and non-negative integer exponents.

General Form:

p(x)=anxn+an−1xn−1+...+a2x2+a1x+a0

where a0,a1,a2,...,an are constants (coefficients) and an0.

Important Terminology

Coefficient: The numerical factor of a term.

In 3x2, the coefficient is 3

In −5x, the coefficient is -5

Degree: The highest power of the variable in the polynomial.

Degree of x3+2x−5 is 3

Degree of 7 (constant) is 0

Terms: The parts of a polynomial separated by + or - signs.

x2+3x−5 has three terms: x2, 3x, and −5

Types of Polynomials

Based on Number of Terms:

Monomial: One term (e.g., 5x3, −2y)

Binomial: Two terms (e.g., x+1, y2−4)

Trinomial: Three terms (e.g., x2+5x+6)

Based on Degree:

Linear Polynomial: Degree 1 (e.g., 2x+3)

Quadratic Polynomial: Degree 2 (e.g., x2−5x+6)

Cubic Polynomial: Degree 3 (e.g., x3+2x2−x+1)

What is NOT a Polynomial?

These expressions are NOT polynomials:

1x (negative exponent: x−1)

x (fractional exponent: x1/2)

3y+y2 (contains negative exponent)

Solved Problem 2.1: Identifying Polynomials

Problem: Which of the following are polynomials?

4x2−3x+7

y2+2

3t+tt

x2−2x

Solution:

Polynomial. All exponents are whole numbers.

Polynomial. Can be written as y2+2y0. The coefficient 2 is allowed.

Not a polynomial. 3t=3t1/2, which has fractional exponent.

Not a polynomial. 2x=2x−1, which has negative exponent.

Solved Problem 2.2: Finding Degree and Coefficients

Problem: For p(x)=5x3−4x2+7x−8, find:

The degree

The coefficient of x2

The constant term

The coefficient of x3

Solution:

Degree = 3 (highest power of x)

Coefficient of x2 = -4

Constant term = -8 (term with x0)

Coefficient of x3 = 5

2.2 Zeros of a Polynomial

What is a Zero?

A zero of a polynomial p(x) is a value c such that p(c)=0.

Example: For p(x)=x−2:

p(2)=2−2=0, so 2 is a zero of p(x)

p(0)=0−2=−20, so 0 is not a zero

Finding Zeros

To find the zero of a linear polynomial ax+b:

Set the polynomial equal to zero: ax+b=0

Solve for x: x=−ba

Important Facts About Zeros

A linear polynomial has exactly one zero

A quadratic polynomial can have at most two zeros

A cubic polynomial can have at most three zeros

A polynomial of degree n can have at most n zeros

The zero polynomial has infinitely many zeros (every real number)

A non-zero constant polynomial has no zeros

Solved Problem 2.3: Finding Polynomial Values

Problem: Find p(0), p(1), and p(−2) for p(x)=x2−5x+6.

Solution:

For p(0):

p(0)=(0)2−5(0)+6=0−0+6=6

For p(1):

p(1)=(1)2−5(1)+6=1−5+6=2

For p(−2):

p(−2)=(−2)2−5(−2)+6=4+10+6=20

Solved Problem 2.4: Verifying Zeros

Problem: Verify whether 2 and -3 are zeros of p(x)=x2+x−6.

Solution:

For x=2:

p(2)=(2)2+2−6=4+2−6=0

Therefore, 2 is a zero of p(x). ✓

For x=−3:

p(−3)=(−3)2+(−3)−6=9−3−6=0

Therefore, -3 is also a zero of p(x). ✓

2.3 Remainder Theorem and Factor Theorem

Remainder Theorem

When a polynomial p(x) is divided by (x−a), the remainder is p(a).

Mathematically: p(x)=(x−a)q(x)+p(a)

where q(x) is the quotient.

Factor Theorem

(x−a) is a factor of p(x) if and only if p(a)=0.

Two-way relationship:

If p(a)=0, then (x−a) is a factor of p(x)

If (x−a) is a factor of p(x), then p(a)=0

Solved Problem 2.5: Using Remainder Theorem

Problem: Find the remainder when p(x)=x3−3x2+5x−6 is divided by (x−2).

Solution:

By Remainder Theorem, remainder = p(2)

p(2)=(2)3−3(2)2+5(2)−6

=8−12+10−6

=0

The remainder is 0, which means (x−2) is a factor of p(x).

Solved Problem 2.6: Using Factor Theorem

Problem: Determine whether (x+1) is a factor of p(x)=4x3+3x2−4x−3.

Solution:

For (x+1)=(x−(−1)) to be a factor, we need p(−1)=0.

p(−1)=4(−1)3+3(−1)2−4(−1)−3

=−4+3+4−3

=0

Since p(−1)=0, (x+1) is a factor of p(x). ✓

Solved Problem 2.7: Finding Unknown Coefficient

Problem: Find the value of k if (x−1) is a factor of p(x)=4x3−3x2−4x+k.

Solution:

If (x−1) is a factor, then p(1)=0.

p(1)=4(1)3−3(1)2−4(1)+k=0

4−3−4+k=0

−3+k=0

k=3

2.4 Factorization of Polynomials

Methods of Factorization

Method 1: Common Factor Method

Factor out the greatest common factor (GCF).

Method 2: Splitting the Middle Term

Used for quadratic polynomials.

Method 3: Using Factor Theorem

Find one zero, then use it to find other factors.

Solved Problem 2.8: Factoring by Common Factor

Problem: Factorize 6x2+9x.

Solution:

The GCF of 6x2 and 9x is 3x.

6x2+9x=3x(2x+3)

Solved Problem 2.9: Splitting the Middle Term

Problem: Factorize 6x2+17x+5.

Solution:

Step 1: Find two numbers whose product = 65=30 and sum = 17.

These numbers are 2 and 15.

Step 2: Split the middle term:

6x2+17x+5=6x2+2x+15x+5

Step 3: Group and factor:

=2x(3x+1)+5(3x+1)

=(3x+1)(2x+5)

Verification: (3x+1)(2x+5)=6x2+15x+2x+5=6x2+17x+5 ✓

Solved Problem 2.10: Factoring Cubic Polynomials

Problem: Factorize x3−23x2+142x−120.

Solution:

Step 1: Try to find one zero. Test factors of 120: ±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12...

Let p(x)=x3−23x2+142x−120

p(1)=1−23+142−120=0

So (x−1) is a factor!

Step 2: Divide or use grouping:

x3−23x2+142x−120=(x−1)(x2−22x+120)

Step 3: Factor the quadratic x2−22x+120:

Find two numbers whose product = 120 and sum = -22.

These are -10 and -12.

x2−22x+120=(x−10)(x−12)

Final Answer:

x3−23x2+142x−120=(x−1)(x−10)(x−12)

2.5 Algebraic Identities

Standard Identities

Identity I: (x+y)2=x2+2xy+y2

Identity II: (x−y)2=x2−2xy+y2

Identity III: x2−y2=(x+y)(x−y)

Identity IV: (x+a)(x+b)=x2+(a+b)x+ab

Identity V: (x+y+z)2=x2+y2+z2+2xy+2yz+2zx

Identity VI: (x+y)3=x3+y3+3xy(x+y)

Identity VII: (x−y)3=x3−y3−3xy(x−y)

Identity VIII: x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)

Solved Problem 2.11: Expanding Squares

Problem: Expand (2x+3y)2.

Solution:

Using Identity I with x=2x and y=3y:

(2x+3y)2=(2x)2+2(2x)(3y)+(3y)2

=4x2+12xy+9y2

Solved Problem 2.12: Difference of Squares

Problem: Factorize 49a2−25b2.

Solution:

Recognize this as x2−y2 where x=7a and y=5b:

49a2−25b2=(7a)2−(5b)2

=(7a+5b)(7a−5b)

Solved Problem 2.13: Perfect Square Trinomial

Problem: Factorize 4x2+20x+25.

Solution:

Check if this is a perfect square: (ax)2+2(ax)(b)+b2

4x2=(2x)2

25=52

20x=22x5

Yes! This is (2x+5)2.

Answer: 4x2+20x+25=(2x+5)2

Solved Problem 2.14: Expanding Cubes

Problem: Expand (2a+3b)3.

Solution:

Using Identity VI with x=2a and y=3b:

(2a+3b)3=(2a)3+(3b)3+3(2a)(3b)(2a+3b)

=8a3+27b3+18ab(2a+3b)

=8a3+27b3+36a2b+54ab2

Solved Problem 2.15: Sum of Cubes

Problem: Factorize 8x3+27y3.

Solution:

Recognize as a3+b3 where a=2x and b=3y.

Using the formula a3+b3=(a+b)(a2−ab+b2):

8x3+27y3=(2x)3+(3y)3

=(2x+3y)[(2x)2−(2x)(3y)+(3y)2]

=(2x+3y)(4x2−6xy+9y2)

Practice Problems: Chapter 2

Set A: Polynomial Basics

Which of the following are polynomials: x2+3x+5, 2x+x, x+2?

Find the degree and coefficients of 3x4−5x2+7x−2.

Write a quadratic polynomial with zeros 2 and -3.

Classify as monomial, binomial, or trinomial: 5x3, x2−4, 2y2+3y+1.

Set B: Zeros and Theorems

Find the zeros of p(x)=3x−6.

Verify if 1 is a zero of x2−2x+1.

Use the Remainder Theorem to find the remainder when x3−2x2+x−1 is divided by (x−1).

Find k if (x−2) is a factor of x3−4x2+x+k.

Set C: Factorization

Factorize: x2+7x+12

Factorize: 2x2−7x+3

Factorize: x3−6x2+11x−6 (Hint: x=1 is a zero)

Factorize: 6x2+5x−6

Set D: Algebraic Identities

Expand: (3x+4)2

Factorize: x2−16

Expand: (x−2)3

Factorize: x3+8 (Hint: 8=23)

Simplify: (x+2)(x−2)

Expand: (2a+3b+c)2

Chapter 3: Coordinate Geometry

3.1 Introduction to Coordinate Geometry

Why Coordinate Geometry?

Imagine trying to describe the exact location of your house to someone who has never been there. You might say "near the park" or "on Main Street," but these descriptions are vague. What if you could give precise numerical coordinates, like GPS does?

Coordinate Geometry (also called Cartesian Geometry) allows us to describe the exact position of any point in a plane using two numbers.

Real-Life Applications

GPS Navigation: Your location is described using latitude and longitude (coordinates!)

Maps: Grid references like "C5" use coordinate principles

Computer Graphics: Every pixel on your screen has x and y coordinates

Architecture: Building plans use coordinate systems for precision

Historical Background

René Descartes (1596-1650), the French mathematician and philosopher, invented this system. Legend says he developed the idea while lying in bed, watching a fly on the ceiling, and wondering how to describe its position!

In honor of Descartes, this system is also called the Cartesian system.

3.2 The Cartesian System

Components of the Cartesian Plane

The Two Axes:

x-axis: The horizontal number line

y-axis: The vertical number line

Origin (O): The point where the axes intersect, with coordinates (0, 0)

The Four Quadrants:

The axes divide the plane into four regions called quadrants, numbered counterclockwise:

Quadrant

Signs

Example

I

(+,+)

(3,4)

II

(−,+)

(−2,5)

III

(−,−)

(−4,−3)

IV

(+,−)

(5,−2)

Table 2: Quadrant characteristics

Understanding Coordinates

Every point in the plane is represented by an ordered pair (x,y):

x-coordinate (abscissa): Distance from y-axis (horizontal distance)

y-coordinate (ordinate): Distance from x-axis (vertical distance)

Important: (3,5)(5,3). The order matters!

Special Points

Points on the x-axis: Have the form (x,0)

Examples: (3,0), (−5,0), (7.5,0)

Points on the y-axis: Have the form (0,y)

Examples: (0,4), (0,−2), (0,6.5)

The Origin: (0,0) - the only point on both axes

Solved Problem 3.1: Identifying Coordinates

Problem: Given the points A, B, C, and D on the coordinate plane:

Point A is 4 units right of the origin and 3 units up

Point B is 2 units left of the origin and 5 units up

Point C is 3 units left and 2 units down from the origin

Point D is 5 units right and 4 units down from the origin

Write the coordinates of each point and identify which quadrant each is in.

Solution:

Point A: (4,3) - Quadrant I (both positive)

Point B: (−2,5) - Quadrant II (x negative, y positive)

Point C: (−3,−2) - Quadrant III (both negative)

Point D: (5,−4) - Quadrant IV (x positive, y negative)

3.3 Plotting Points on the Cartesian Plane

Step-by-Step Plotting Process

To plot a point (x,y):

Start at the origin (0, 0)

Move x units along the x-axis

If x>0: move right

If x<0: move left

If x=0: stay on y-axis

From there, move y units parallel to the y-axis

If y>0: move up

If y<0: move down

If y=0: stay on x-axis

Mark the point!

Solved Problem 3.2: Plotting Multiple Points

Problem: Plot the following points on a coordinate plane and join them in order: A(1,1), B(4,1), C(4,4), D(1,4). What shape do you get?

Solution:

Plotting Process:

Point A(1,1): 1 unit right, 1 unit up

Point B(4,1): 4 units right, 1 unit up

Point C(4,4): 4 units right, 4 units up

Point D(1,4): 1 unit right, 4 units up

When joined in order A→B→C→D→A, these points form a square with side length 3 units.

Solved Problem 3.3: Reading Coordinates from a Graph

Problem: A point P is located 3 units left of the origin and 5 units below the origin. What are its coordinates? In which quadrant does it lie?

Solution:

Moving 3 units left: x=−3

Moving 5 units below: y=−5

Coordinates: P(−3,−5)

Quadrant: III (both coordinates negative)

Solved Problem 3.4: Points on Axes

Problem: Where do the following points lie?

A(5,0)

B(0,−3)

C(−2,0)

D(0,0)

Solution:

A(5,0) lies on the positive x-axis (y-coordinate is 0)

B(0,−3) lies on the negative y-axis (x-coordinate is 0)

C(−2,0) lies on the negative x-axis (y-coordinate is 0)

D(0,0) is the origin

Distance Formula (Preview)

While not in the current syllabus, you'll soon learn that the distance between two points (x1,y1) and (x2,y2) is:

d=(x2−x1)2+(y2−y1