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Chapter 4 Master Notes

Exploring Algebraic Identities

Integrated Concept → Formula → Application → Practice Questions (100/100 Kit)

Topic 1: Equations vs. Algebraic Identities & Pattern Discovery

Concept & Core Rules
$$\mathbf{\text{Consecutive Squares Identity: } (n - 1)^2 + (n + 1)^2 - 2n^2 = 2}$$

Immediate Solved Application

Example 1.1: Distinguishing Equations & Identities NCERT

Question: State whether (i) $3x + 5 = 20$ and (ii) $(a + 2)^2 = a^2 + 4a + 4$ are equations or identities.

Solution:

(i) $3x + 5 = 20 \implies 3x = 15 \implies x = 5$. Since it is true ONLY when $x = 5$, it is an Algebraic Equation.

(ii) $(a + 2)^2 = a^2 + 4a + 4$. LHS $= (a+2)^2 = a^2 + 4a + 4 =$ RHS for any value of $a$ (e.g. at $a=0, 1, -3$). Thus, it is an Algebraic Identity.

Topic 1: Test Your Understanding
  1. Evaluate $(29)^2 + (31)^2 - 2(30)^2$ without calculating individual squares. [Hint: Apply $(n-1)^2 + (n+1)^2 - 2n^2 = 2 \implies \mathbf{2}$]
  2. Is $(x - 3)(x + 3) = x^2 - 9$ an equation or an identity? Explain. [Answer: Identity]

Topic 2: Identity 1 — Square of Sum $(x + y)^2 = x^2 + 2xy + y^2$

$$\mathbf{(x + y)^2 = x^2 + 2xy + y^2}$$ $$\text{Comparison Property: } \mathbf{(a + b)^2 > a^2 + b^2} \quad \text{for all } a, b > 0 \quad (\text{Difference is } 2ab)$$
\(a^2\)
\(ab\)
\(ab\)
\(b^2\)
Fig 4.1: Area Model of $(a + b)^2 = a^2 + ab + ab + b^2$

Immediate Solved Application

Example 2.1: Binomial Expansion & Rapid Squaring NCERT

Question: (i) Expand $(5x + 2y)^2$.    (ii) Compute $105^2$ using Identity 1.

Solution:

(i) $(5x + 2y)^2 = (5x)^2 + 2(5x)(2y) + (2y)^2 = \mathbf{25x^2 + 20xy + 4y^2}$.

(ii) $105^2 = (100 + 5)^2 = 100^2 + 2(100)(5) + 5^2 = 10000 + 1000 + 25 = \mathbf{11025}$.

Example 2.2: Reverse Factorisation RS Aggarwal

Question: Factorise $36x^2 + 60xy + 25y^2$.

Solution:

Rewrite terms as perfect squares and double product: $(6x)^2 + 2(6x)(5y) + (5y)^2$.

Comparing with $a^2 + 2ab + b^2 = (a + b)^2$ where $a = 6x, b = 5y$:

$$= \mathbf{(6x + 5y)^2 = (6x + 5y)(6x + 5y)}$$
Topic 2: Test Your Understanding
  1. Expand $\left(\frac{3}{4}x + \frac{2}{3}y\right)^2$. [Ans: $\frac{9}{16}x^2 + xy + \frac{4}{9}y^2$]
  2. Evaluate $205^2$ using Identity 1. [Ans: $42025$]
  3. Factorise $49p^2 + 70pq + 25q^2$. [Ans: $(7p + 5q)^2$]

Topic 3: Identity 2 — Square of Difference $(x - y)^2 = x^2 - 2xy + y^2$

$$\mathbf{(x - y)^2 = x^2 - 2xy + y^2}$$

Immediate Solved Application

Example 3.1: Binomial Expansion & Rapid Evaluation NCERT

Question: (i) Expand $(4a - 3b)^2$.    (ii) Evaluate $97^2$ using Identity 2.

Solution:

(i) $(4a - 3b)^2 = (4a)^2 - 2(4a)(3b) + (3b)^2 = \mathbf{16a^2 - 24ab + 9b^2}$.

(ii) $97^2 = (100 - 3)^2 = 100^2 - 2(100)(3) + 3^2 = 10000 - 600 + 9 = \mathbf{9409}$.

Topic 3: Test Your Understanding
  1. Expand $\left(2x - \frac{1}{2x}\right)^2$. [Ans: $4x^2 - 2 + \frac{1}{4x^2}$]
  2. Evaluate $198^2$ using $(200 - 2)^2$. [Ans: $39204$]
  3. Factorise $64x^2 - 48xy + 9y^2$. [Ans: $(8x - 3y)^2$]

Topic 4: Identity 3 — Difference of Two Squares & Śhrīdharāchārya Method

$$\mathbf{x^2 - y^2 = (x + y)(x - y)}$$ $$\text{Śhrīdharāchārya's Rapid Squaring Formula (750 CE): } \mathbf{a^2 = (a + b)(a - b) + b^2}$$

Immediate Solved Application

Example 4.1: Product Evaluation & Śhrīdharāchārya Squaring NCERT

Question: (i) Evaluate $105 \times 95$.    (ii) Compute $55^2$ using Śhrīdharāchārya's formula.

Solution:

(i) $105 \times 95 = (100 + 5)(100 - 5) = 100^2 - 5^2 = 10000 - 25 = \mathbf{9975}$.

(ii) $55^2 = (55 + 5)(55 - 5) + 5^2 = (60 \times 50) + 25 = 3000 + 25 = \mathbf{3025}$.

Topic 4: Test Your Understanding
  1. Evaluate $104 \times 96$ using Identity 3. [Ans: $9984$]
  2. Compute $65^2$ using Śhrīdharāchārya's formula $(70 \times 60) + 25$. [Ans: $4225$]
  3. Factorise $\frac{a^2}{9} - \frac{b^2}{16}$. [Ans: $\left(\frac{a}{3} + \frac{b}{4}\right)\left(\frac{a}{3} - \frac{b}{4}\right)$]

Topic 5: Identity 4 — Product of General Linear Factors $(x + a)(x + b)$

$$\mathbf{(x + a)(x + b) = x^2 + (a + b)x + ab}$$ $$\text{General Form: } (px + a)(qx + b) = pq \cdot x^2 + (pb + qa)x + ab$$

Immediate Solved Application

Example 5.1: Product & Middle-Term Factorisation NCERT

Question: (i) Evaluate $103 \times 107$.    (ii) Factorise $x^2 + 11x + 30$ by splitting middle term.

Solution:

(i) $103 \times 107 = (100 + 3)(100 + 7) = 100^2 + (3 + 7)100 + (3 \times 7) = 10000 + 1000 + 21 = \mathbf{11021}$.

(ii) For $x^2 + 11x + 30$, find numbers adding to $11$ and multiplying to $30 \implies 5, 6$.
$x^2 + 5x + 6x + 30 = x(x + 5) + 6(x + 5) = \mathbf{(x + 5)(x + 6)}$.

Topic 5: Test Your Understanding
  1. Evaluate $95 \times 96$ using $(100 - 5)(100 - 4)$. [Ans: $9120$]
  2. Factorise $x^2 - 7x + 12$ by middle-term splitting. [Ans: $(x - 3)(x - 4)$]

Topic 6: Identity 5 — Trinomial Square $(x + y + z)^2$

$$\mathbf{(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx}$$
Sign Variations

Immediate Solved Application

Example 6.1: Trinomial Expansion & Factorisation RS Aggarwal

Question: (i) Expand $(3x - 2y + 4z)^2$.    (ii) Factorise $4x^2 + 9y^2 + 16z^2 + 12xy - 24yz - 16zx$.

Solution:

(i) $(3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x) = \mathbf{9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24zx}$.

(ii) Negative terms involve $z \implies z$-term is $-4z$.
$(2x)^2 + (3y)^2 + (-4z)^2 + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x) = \mathbf{(2x + 3y - 4z)^2}$.

Topic 6: Test Your Understanding
  1. Expand $(-2x + 5y - 3z)^2$. [Ans: $4x^2 + 25y^2 + 9z^2 - 20xy - 30yz + 12zx$]
  2. Evaluate $119^2 = (100 + 10 + 9)^2$ using Identity 5. [Ans: $14161$]

Topic 7: Identities 6 & 7 — Binomial Cubes $(x + y)^3$ and $(x - y)^3$

$$\mathbf{(x + y)^3 = x^3 + y^3 + 3xy(x + y) = x^3 + 3x^2y + 3xy^2 + y^3}$$ $$\mathbf{(x - y)^3 = x^3 - y^3 - 3xy(x - y) = x^3 - 3x^2y + 3xy^2 - y^3}$$

Immediate Solved Application

Example 7.1: Cubic Binomial Expansion NCERT

Question: (i) Expand $(2x + 3y)^3$.    (ii) Evaluate $104^3$.

Solution:

(i) $(2x)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2 + (3y)^3 = \mathbf{8x^3 + 36x^2y + 54xy^2 + 27y^3}$.

(ii) $104^3 = (100 + 4)^3 = 100^3 + 4^3 + 3(100)(4)(100 + 4) = 1000000 + 64 + 1200(104) = 1000064 + 124800 = \mathbf{1124864}$.

Topic 7: Test Your Understanding
  1. Expand $(3a - 2b)^3$. [Ans: $27a^3 - 54a^2b + 36ab^2 - 8b^3$]
  2. Evaluate $99^3$ using $(100 - 1)^3$. [Ans: $970299$]

Topic 8: Identities 8 & 9 — Sum & Difference of Cubes

$$\mathbf{x^3 + y^3 = (x + y)(x^2 - xy + y^2)}$$ $$\mathbf{x^3 - y^3 = (x - y)(x^2 + xy + y^2)}$$

Immediate Solved Application

Example 8.1: Cubic Factorisation NCERT

Question: Factorise (i) $27a^3 + 64b^3$    (ii) $125x^3 - 8y^3$.

Solution:

(i) $(3a)^3 + (4b)^3 = (3a + 4b)[(3a)^2 - (3a)(4b) + (4b)^2] = \mathbf{(3a + 4b)(9a^2 - 12ab + 16b^2)}$.

(ii) $(5x)^3 - (2y)^3 = (5x - 2y)[(5x)^2 + (5x)(2y) + (2y)^2] = \mathbf{(5x - 2y)(25x^2 + 10xy + 4y^2)}$.

Topic 8: Test Your Understanding
  1. Factorise $8m^3 + 27n^3$. [Ans: $(2m + 3n)(4m^2 - 6mn + 9n^2)$]
  2. Factorise $216x^3 - 125y^3$. [Ans: $(6x - 5y)(36x^2 + 30xy + 25y^2)$]

Topic 9: Identity 10 — Sum of Three Cubes & Conditional Rule

$$\mathbf{x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)}$$ $$\text{Conditional Rule: If } \mathbf{x + y + z = 0} \implies \mathbf{x^3 + y^3 + z^3 = 3xyz}$$

Immediate Solved Application

Example 9.1: Conditional Evaluation & HOTS Proof HOTS

Question: (i) Evaluate $(-12)^3 + 7^3 + 5^3$.    (ii) If $a + b + c = 5$ and $ab + bc + ca = 10$, prove $a^3 + b^3 + c^3 - 3abc = -25$.

Solution:

(i) $x = -12, y = 7, z = 5$. Sum $= -12 + 7 + 5 = 0 \implies x^3 + y^3 + z^3 = 3(-12)(7)(5) = \mathbf{-1260}$.

(ii) $(a + b + c)^2 = a^2 + b^2 + c^2 + 2(10) \implies 25 = a^2 + b^2 + c^2 + 20 \implies a^2 + b^2 + c^2 = 5$.
$a^3 + b^3 + c^3 - 3abc = (5)(5 - 10) = 5(-5) = \mathbf{-25}$.

Topic 9: Test Your Understanding
  1. Evaluate $28^3 + (-15)^3 + (-13)^3$ without computing cubes. [Ans: $3(28)(-15)(-13) = 16380$]
  2. If $x = 2y + 6$, evaluate $x^3 - 8y^3 - 36xy - 216$. [Ans: $0$]

Topic 10: Simplifying Rational Algebraic Expressions

Simplification Strategy

To simplify $\frac{P(x)}{Q(x)}$: Factorize both $P(x)$ and $Q(x)$ completely using identities or middle-term splitting, then cancel common factors.

Immediate Solved Application

Example 10.1: Rational Reduction NCERT

Question: Simplify $\frac{x^2 - 7x + 12}{5x^2 + 5x - 100}$.

Solution:

$$\text{Numerator: } x^2 - 7x + 12 = (x - 3)(x - 4)$$ $$\text{Denominator: } 5(x^2 + x - 20) = 5(x - 4)(x + 5)$$ $$\text{Ratio} = \frac{(x - 3)(x - 4)}{5(x - 4)(x + 5)} = \mathbf{\frac{x - 3}{5(x + 5)}}$$
Topic 10: Test Your Understanding
  1. Simplify $\frac{4x^2 - 1}{4x^2 + 4x + 1}$. [Ans: $\frac{2x - 1}{2x + 1}$]
  2. Simplify $\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}$. [Ans: $\frac{3(p - 3q)}{p - 2q}$]

Topic 11: Real-Life Case Studies & Exam HOTS

Case Study 1: Playground Walking Path Area CASE STUDY

A square village playground has side $40\text{ metres}$. A uniform walking path of width $s\text{ metres}$ is constructed all around the playground outside its boundary.

Questions:
1. Write the side length of the outer square including the path.
2. Express the area of the walking path in terms of $s$.
3. Calculate the path area if width $s = 2\text{ metres}$.
4. If the path area is $336\text{ m}^2$, find the width $s$.

Solution:

1. Outer side $= 40 + 2s\text{ m}$.

2. $\text{Path Area} = (40 + 2s)^2 - 40^2 = 4s^2 + 160s\text{ m}^2$.

3. At $s = 2\text{ m}$: $\text{Area} = 4(2)^2 + 160(2) = 16 + 320 = \mathbf{336\text{ m}^2}$.

4. $4s^2 + 160s = 336 \implies s^2 + 40s - 84 = 0 \implies (s + 42)(s - 2) = 0 \implies \mathbf{s = 2\text{ metres}}$.

HOTS 1: Divisibility Proof of $n^3 - n$ by 6 HOTS

Question: Prove that $n^3 - n$ is always divisible by $6$ for any natural number $n \in \mathbb{N}$.

Solution:

Factorize: $n^3 - n = n(n^2 - 1) = (n - 1)n(n + 1)$. This is the product of three consecutive integers.

Product is divisible by $2 \times 3 = \mathbf{6}$.

12. 100/100 Master Revision Checklist

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