Exploring Algebraic Identities - Class 9 Maths Chapter 4 Notes

Section 4.1

Introduction: The Magic of Consecutive Square Patterns

An Algebraic Identity is a universal mathematical equality that holds true for every possible value assigned to its variables. Let us begin by observing a surprising numerical mystery!

Example 1: The Three Consecutive Square Paradox

Take any 3 consecutive square numbers. Add the smallest and largest squares, then subtract twice the middle square:

Set 1 (1, 4, 9): $ (1 + 9) - (2 \times 4) = 10 - 8 = \mathbf{2} $
Set 2 (9, 16, 25): $ (9 + 25) - (2 \times 16) = 34 - 32 = \mathbf{2} $
Set 3 (25, 36, 49): $ (25 + 49) - (2 \times 36) = 74 - 72 = \mathbf{2} $

Why does this always yield 2 regardless of which three consecutive squares you choose? We will use algebra to prove this universal truth in Section 4.3!

Section 4.2

Visualising Identities: Geometric Proof of $ (a + b)^2 $

Take a line segment of length $ (a + b) $ units and construct a square on it. We can partition this outer square into two smaller squares of area $ a^2 $ and $ b^2 $, and two identical rectangles of area $ ab $:

$ (a + b)^2 = a^2 + 2ab + b^2 $

Figure 4.1: Geometric Partitioning of a Square of side (a + b) into a², b², and two ab Rectangles

Identity vs. Equation: What's the Difference?
• Equation: $ x^2 - 1 = 24 $ is true ONLY for specific values ($ x = 5 $ or $ x = -5 $).
• Identity: $ (a + b)^2 = a^2 + 2ab + b^2 $ is true for EVERY value of $ a $ and $ b $ (positive, negative, zero, or rational numbers).

Solved Conceptual Example 1: Expansion & Fast Mental Math

(A) Expand $ (5x + 2y)^2 $:
Apply $ (a + b)^2 $ with $ a = 5x $ and $ b = 2y $:
\[ (5x + 2y)^2 = (5x)^2 + 2(5x)(2y) + (2y)^2 = \mathbf{25x^2 + 20xy + 4y^2} \]

(B) Calculate $ 43^2 $ mentally using identities:
Write $ 43 = (40 + 3) $:
\[ 43^2 = (40 + 3)^2 = 40^2 + 2(40)(3) + 3^2 = 1600 + 240 + 9 = \mathbf{1849} \]

Section 4.3

Factorisation & The Identity $ (a - b)^2 $

Identities work in reverse! If an algebraic trinomial matches $ a^2 + 2ab + b^2 $, we can factor it directly into $ (a + b)^2 $.

$ (a - b)^2 = a^2 - 2ab + b^2 $

Figure 4.2: Visual Deduction of (a - b)² by Subtracting Rectangular Regions from a²

Mathematical Proof of the Consecutive Square Paradox!

Let three consecutive integers be $ (n - 1) $, $ n $, and $ (n + 1) $. Their squares are $ (n - 1)^2 $, $ n^2 $, and $ (n + 1)^2 $.
Sum of smallest and largest squares: \[ (n - 1)^2 + (n + 1)^2 = (n^2 - 2n + 1) + (n^2 + 2n + 1) = 2n^2 + 2 \] Subtracting twice the middle square ($ 2n^2 $): \[ (2n^2 + 2) - 2n^2 = \mathbf{2} \] This proves why the result is ALWAYS 2 for ANY set of 3 consecutive square numbers!

Solved Conceptual Example 2: Common Factor Extraction & Factorisation

Problem: Factor completely $ 50p^2 + 60pq + 18q^2 $.

Solution:
Step 1: Notice 2 is a common factor among all terms:
\[ 50p^2 + 60pq + 18q^2 = 2(25p^2 + 30pq + 9q^2) \]
Step 2: Recognize inside the bracket: $ 25p^2 = (5p)^2 $, $ 9q^2 = (3q)^2 $, and $ 30pq = 2(5p)(3q) $.
Step 3: Apply $ a^2 + 2ab + b^2 = (a + b)^2 $:
\[ 25p^2 + 30pq + 9q^2 = (5p + 3q)^2 \]
Thus, the complete factorisation is $ 2(5p + 3q)^2 $.

Section 4.4

Trinomial Square $ (a + b + c)^2 $ & Śhrīdharāchārya's Method

When squaring a sum of three terms $ (a + b + c)^2 $, we substitute $ d = b + c $ and expand to get:

$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca $

Figure 4.3: Geometric Representation of Trinomial Expansion (a + b + c)²

Śhrīdharāchārya's Fast Squaring Method (750 CE)

Using the identity $ a^2 - b^2 = (a + b)(a - b) \implies a^2 = (a + b)(a - b) + b^2 $, ancient Indian mathematician Śhrīdharāchārya devised a genius shortcut to square numbers ending in 5 or near round numbers!

Example: Calculate $ 55^2 $ in 3 Seconds!
Choose $ b = 5 $:
\[ 55^2 = (55 + 5)(55 - 5) + 5^2 = (60 \times 50) + 25 = 3000 + 25 = \mathbf{3025} \]

Solved Conceptual Example 3: Trinomial Expansion

Expand $ (3x - 2y + 4z)^2 $:
Treat terms as $ a = 3x $, $ b = -2y $, $ c = 4z $:
\[ (3x - 2y + 4z)^2 = (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x) \]
\[ = \mathbf{9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx} \]

Practice Exercise 4.1 Expand expressions and calculate squares of round numbers using $(a + b)^2$.

Solve Exercise 4.1

Practice Exercise 4.2 Factor perfect squares and evaluate squares using $(a - b)^2$.

Solve Exercise 4.2

Practice Exercise 4.3 Trinomial expansions $(a + b + c)^2$, identity verification, and general square calculations.

Solve Exercise 4.3

Section 4.5 & 4.6

Factorisation Using Algebra Tiles & Splitting Middle Term

Multiplying two binomials gives: $ (x + a)(x + b) = x^2 + (a + b)x + ab $. Visualizing this with Algebra Tiles involves creating a rectangle with an $ x^2 $-tile, $ (a+b) $ $ x $-tiles, and $ ab $ unit tiles.

Figure 4.4: Algebra Tiles Rectangular Array for x² + 7x + 12 = (x + 3)(x + 4)

Method of Splitting the Middle Term (Without Tiles)

To factor $ x^2 + Bx + C $, we find two numbers $ a $ and $ b $ such that:

$ a + b = B \quad \text{and} \quad a \times b = C $

Solved Conceptual Example 4: Splitting Middle Term with Negative Signs

Factor $ x^2 - 5x + 6 $:
We need $ a + b = -5 $ and $ ab = 6 $.
Factors of 6: $ (-2) \times (-3) = 6 $ and $ (-2) + (-3) = -5 $.
Split middle term: $ x^2 - 2x - 3x + 6 = x(x - 2) - 3(x - 2) = \mathbf{(x - 2)(x - 3)} $.

Section 4.7

Volumetric Cubes: Binomial Cubes & Sum/Difference of Cubes

Expanding into 3-dimensional volume: A cube of edge $ (a + b) $ splits into 2 smaller cubes ($ a^3, b^3 $) and 6 cuboids ($ 3a^2b $ and $ 3ab^2 $):

$ (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a + b) $
$ (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a - b) $

Sum and Difference of Cubes

$ x^3 + y^3 = (x + y)(x^2 - xy + y^2) $
$ x^3 - y^3 = (x - y)(x^2 + xy + y^2) $

The Grand Master Identity

$ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Solved Conceptual Example 5: Master Cubic Identity Application

Problem: Given $ x + y + z = 10 $, $ xyz = 25 $, and $ x^2 + y^2 + z^2 = 38 $. Find $ x^3 + y^3 + z^3 $.

Solution:
Step 1: Find $ (xy + yz + zx) $ using $ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) $:
\[ 100 = 38 + 2(xy + yz + zx) \implies 2(xy + yz + zx) = 62 \implies xy + yz + zx = 31 \]
Step 2: Substitute into the Master Identity:
\[ (10)(38 - 31) = x^3 + y^3 + z^3 - 3(25) \]
\[ 10 \times 7 = x^3 + y^3 + z^3 - 75 \implies 70 + 75 = \mathbf{145} \]
Thus, $ x^3 + y^3 + z^3 = \mathbf{145} $.

Section 4.8

Simplifying Rational Expressions & Real-World Problems

Rational algebraic expressions are simplified by factoring numerator and denominator and cancelling non-zero common factors.

Solved Conceptual Example 6: Real-World Pool Dimension Problem

Problem: A rectangular swimming pool has breadth 4 metres less than its length and an area of 96 sq. metres. Find its dimensions.

Solution:
Let length = $ x $ metres, Breadth = $ (x - 4) $ metres.
Area equation: $ x(x - 4) = 96 \implies x^2 - 4x - 96 = 0 $.
Split middle term: $ -12 \times 8 = -96 $ and $ -12 + 8 = -4 $.
\[ x^2 - 12x + 8x - 96 = 0 \implies x(x - 12) + 8(x - 12) = 0 \implies (x - 12)(x + 8) = 0 \]

Since length cannot be negative, $ x = 12 $.
Length = 12 metres, Breadth = 8 metres.

Practice Exercise 4.4 Complete factorisation identities, find products mentally, and factorise trinomials.

Solve Exercise 4.4

Practice Exercise 4.5 Factor and simplify complex rational expressions using standard and master cubic identities.

Solve Exercise 4.5

Interactive Suite

Interactive Explorer Widgets

Select Identity Type:

Value A (or N):

Value B (if applicable):

Q1. What is the value when you add the smallest and largest of any 3 consecutive square numbers and subtract twice the middle square?

A) 0

B) 2

C) 4

Q2. Who introduced the fast squaring method N² = (N+b)(N-b) + b² in 750 CE?

A) Śhrīdharāchārya

B) Āryabhaṭa

C) Brahmagupta

Master Formula Sheet

Summary of All 11 Essential Identities

1. $ (x + y)^2 = x^2 + 2xy + y^2 $
2. $ (x - y)^2 = x^2 - 2xy + y^2 $
3. $ (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx $
4. $ (x + y)(x - y) = x^2 - y^2 $
5. $ (x + a)(x + b) = x^2 + (a + b)x + ab $
6. $ (ax + b)(cx + d) = acx^2 + (ad + bc)x + bd $
7. $ (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 $
8. $ (x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 $
9. $ x^3 + y^3 = (x + y)(x^2 - xy + y^2) $
10. $ x^3 - y^3 = (x - y)(x^2 + xy + y^2) $
11. $ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $

Exploring Algebraic Identities — Class 9 Maths Chapter 4 Notes