Class 9 Mathematics Chapter 8: Algebraic Expressions and Identities – Important Questions with Complete Solutions

Algebraic Expressions and Identities has been a consistent topic in CBSE Class 9 board papers, typically accounting for 8–12 marks across different question formats. The 2024-25 rationalized curriculum emphasizes conceptual clarity over rote memorization, meaning examiners now favour questions that test application of identities in real-world contexts and multi-step algebraic manipulation. Recent trends show: • 1-mark questions focus on identifying expressions, degree of polynomials, and quick identity recognition. • 2-mark questions demand simplification using identities or combining like terms strategically. • 3-mark questions combine two or more identities or require algebraic proof. • 5-mark questions may integrate geometry (e.g., expressing area as an algebraic expression) or challenge students to derive or verify identities. The board also increasingly tests your ability to recognize when to apply which identity. For example, recognizing that x² − 9 = (x+3)(x−3) and not expanding unnecessarily saves time. Students who practise these patterns daily—with instant feedback on conceptual gaps—score 15–18 marks in this chapter consistently. That's why targeted, pattern-based practice is more effective than random problem-solving.

These questions test quick recall, identity recognition, and basic simplification. Each carries 1 mark and requires no working in exams—but always verify your answer mentally. **Q1.** If (a+b)² = a² + 2ab + b², then (3x+2)² equals: (A) 9x² + 4 (B) 9x² + 12x + 4 (C) 9x + 12x + 4 (D) 6x² + 12x + 4 **Answer:** (B) 9x² + 12x + 4 *Explanation:* Apply (a+b)² with a = 3x, b = 2: (3x)² + 2(3x)(2) + 2² = 9x² + 12x + 4. **Q2.** (a+b)(a−b) simplifies to: (A) a² − 2ab + b² (B) a² + b² (C) a² − b² (D) 2a² − b² **Answer:** (C) a² − b² *Explanation:* This is the difference of squares identity. Expanding: a(a−b) + b(a−b) = a² − ab + ab − b² = a² − b². **Q3.** The expression 5x²y − 3xy² + 2x²y + xy² simplifies to: (A) 7x²y − 2xy² (B) 9x²y − 4xy² (C) 7x²y − 4xy² (D) 5x²y − xy² **Answer:** (A) 7x²y − 2xy² *Explanation:* Combine like terms: (5x²y + 2x²y) + (−3xy² + xy²) = 7x²y − 2xy². **Q4.** If we use the identity (a−b)² = a² − 2ab + b², then (5m−3)² equals: (A) 25m² − 30m + 9 (B) 25m² − 15m + 9 (C) 10m² − 15m + 9 (D) 25m² + 30m + 9 **Answer:** (A) 25m² − 30m + 9 *Explanation:* With a = 5m, b = 3: (5m)² − 2(5m)(3) + 3² = 25m² − 30m + 9. **Q5.** Multiply: (2a + 3)(2a − 3) = (A) 4a² + 9 (B) 4a² − 6a − 9 (C) 4a² − 9 (D) 2a² + 3 **Answer:** (C) 4a² − 9 *Explanation:* Use (a+b)(a−b) = a² − b² with a = 2a, b = 3: (2a)² − 3² = 4a² − 9.

These questions test simplification and application of single identities or combining like terms. Show your working clearly. **Q1.** Simplify: (x+4)² − (x−4)² **Solution:** Using (a+b)² = a² + 2ab + b² and (a−b)² = a² − 2ab + b²: (x+4)² = x² + 8x + 16 (x−4)² = x² − 8x + 16 (x+4)² − (x−4)² = (x² + 8x + 16) − (x² − 8x + 16) = x² + 8x + 16 − x² + 8x − 16 = 16x **Alternative (faster):** Use a² − b² = (a+b)(a−b): (x+4)² − (x−4)² = [(x+4) + (x−4)][(x+4) − (x−4)] = (2x)(8) = 16x **Q2.** Multiply: (3a + 2b)(3a − 2b) **Solution:** Using (x+y)(x−y) = x² − y² with x = 3a, y = 2b: (3a + 2b)(3a − 2b) = (3a)² − (2b)² = 9a² − 4b² **Q3.** Simplify: 5(p²+3p) − 3(p² − 2p) + 2 **Solution:** Distribute: 5p² + 15p − 3p² + 6p + 2 Combine like terms: (5p² − 3p²) + (15p + 6p) + 2 = 2p² + 21p + 2 **Q4.** If (2x+5)² = ax² + bx + c, find a, b, and c. **Solution:** Expand (2x+5)² using (a+b)²: (2x+5)² = (2x)² + 2(2x)(5) + 5² = 4x² + 20x + 25 Comparing with ax² + bx + c: a = 4, b = 20, c = 25 **Q5.** Simplify: (m+3)(m+4) − (m+1)(m+6) **Solution:** Expand each product: (m+3)(m+4) = m² + 4m + 3m + 12 = m² + 7m + 12 (m+1)(m+6) = m² + 6m + m + 6 = m² + 7m + 6 Subtract: (m² + 7m + 12) − (m² + 7m + 6) = 12 − 6 = 6

These questions combine multiple identities or require proof and algebraic justification. **Q1.** Prove that (a+b)² − (a−b)² = 4ab **Solution:** LHS = (a+b)² − (a−b)² = (a² + 2ab + b²) − (a² − 2ab + b²) = a² + 2ab + b² − a² + 2ab − b² = 4ab = RHS Hence proved. **Alternative approach:** Use difference of squares: (a+b)² − (a−b)² = [(a+b) + (a−b)][(a+b) − (a−b)] = (2a)(2b) = 4ab **Q2.** Simplify: (2x+3)² + (2x−3)² − 2(4x² − 9) **Solution:** Expand (2x+3)²: (2x+3)² = 4x² + 12x + 9 Expand (2x−3)²: (2x−3)² = 4x² − 12x + 9 Add them: (4x² + 12x + 9) + (4x² − 12x + 9) = 8x² + 18 Expand 2(4x² − 9): 2(4x² − 9) = 8x² − 18 Final simplification: 8x² + 18 − (8x² − 18) = 8x² + 18 − 8x² + 18 = 36 **Q3.** If x = 2 and y = 3, find the value of (3x+2y)² − (3x−2y)² using identity without direct substitution. **Solution:** Use (a+b)² − (a−b)² = 4ab, where a = 3x, b = 2y: (3x+2y)² − (3x−2y)² = 4(3x)(2y) = 24xy Now substitute x = 2, y = 3: 24(2)(3) = 24 × 6 = 144 **Q4.** Multiply and simplify: (x+2)(x−3)(x+4) **Solution:** First multiply (x+2)(x−3): (x+2)(x−3) = x² − 3x + 2x − 6 = x² − x − 6 Now multiply (x² − x − 6)(x+4): = x²(x+4) − x(x+4) − 6(x+4) = x³ + 4x² − x² − 4x − 6x − 24 = x³ + 3x² − 10x − 24

These questions integrate multiple identities, require detailed working, and test deeper conceptual understanding. **Q1.** Prove that (a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca. Then, if a = 1, b = 2, c = 3, verify the result. **Solution:** **Proof:** (a+b+c)² = (a+b+c) × (a+b+c) = a(a+b+c) + b(a+b+c) + c(a+b+c) = a² + ab + ac + ba + b² + bc + ca + cb + c² = a² + b² + c² + ab + ab + bc + bc + ac + ac = a² + b² + c² + 2ab + 2bc + 2ca Hence proved. **Verification with a = 1, b = 2, c = 3:** LHS = (1+2+3)² = 6² = 36 RHS = 1² + 2² + 3² + 2(1)(2) + 2(2)(3) + 2(3)(1) = 1 + 4 + 9 + 4 + 12 + 6 = 36 LHS = RHS, verified. **Q2.** Simplify: [(x+1/x)² − (x−1/x)²] / [4] and then find its value when x = 2. **Solution:** Let a = x + 1/x and b = x − 1/x. We need: (a² − b²) / 4 Using a² − b² = (a+b)(a−b): a + b = (x + 1/x) + (x − 1/x) = 2x a − b = (x + 1/x) − (x − 1/x) = 2/x (a+b)(a−b) = 2x × 2/x = 4 So (a² − b²) / 4 = 4/4 = 1 The expression equals 1 for any non-zero x. When x = 2: Answer = 1 **Q3.** Prove that (x+y)² + (x−y)² = 2(x² + y²). Using this, if x² + y² = 13 and xy = 6, find (x+y)². **Solution:** **Proof:** LHS = (x+y)² + (x−y)² = (x² + 2xy + y²) + (x² − 2xy + y²) = 2x² + 2y² = 2(x² + y²) = RHS Hence proved. **Finding (x+y)²:** From the identity: (x+y)² + (x−y)² = 2(x² + y²) Rearrange: (x+y)² = 2(x² + y²) − (x−y)² Alternatively, expand (x+y)² directly: (x+y)² = x² + 2xy + y² = (x² + y²) + 2xy = 13 + 2(6) = 13 + 12 = 25 **Q4.** If x = √5 + √3 and y = √5 − √3, find x² + y² and xy. **Solution:** **Finding xy:** xy = (√5 + √3)(√5 − √3) Use (a+b)(a−b) = a² − b²: xy = (√5)² − (√3)² = 5 − 3 = 2 **Finding x² + y²:** Using (x+y)² + (x−y)² = 2(x² + y²): First, find x + y and x − y: x + y = (√5 + √3) + (√5 − √3) = 2√5 x − y = (√5 + √3) − (√5 − √3) = 2√3 (x+y)² + (x−y)² = (2√5)² + (2√3)² = 4(5) + 4(3) = 20 + 12 = 32 Using (x+y)² + (x−y)² = 2(x² + y²): 2(x² + y²) = 32 x² + y² = 16 Alternative: x² + y² = (x+y)² − 2xy = (2√5)² − 2(2) = 20 − 4 = 16 ✓

**Case-Study Question:** A rectangular garden has dimensions (2x+3) metres by (2x−3) metres. The gardener wants to expand it to a square garden with side (2x+3) metres. **(i)** Find the original area of the rectangular garden in terms of x. **Solution:** Area of rectangle = length × breadth = (2x+3) × (2x−3) Use (a+b)(a−b) = a² − b²: = (2x)² − 3² = 4x² − 9 m² **(ii)** Find the area of the new square garden in terms of x. **Solution:** Area of square = side² = (2x+3)² = (2x)² + 2(2x)(3) + 3² = 4x² + 12x + 9 m² **(iii)** Find the increase in area when expanding from rectangle to square. **Solution:** Increase in area = Square area − Rectangular area = (4x² + 12x + 9) − (4x² − 9) = 4x² + 12x + 9 − 4x² + 9 = 12x + 18 = 6(2x + 3) m² Interpretation: The increase is proportional to the length (2x+3) with factor 6. **(iv)** If x = 4 metres, calculate the numerical increase in area. **Solution:** Increase = 6(2x+3) = 6(2(4)+3) = 6(8+3) = 6(11) = 66 m² Original rectangular area = 4(4)² − 9 = 64 − 9 = 55 m² New square area = 4(4)² + 12(4) + 9 = 64 + 48 + 9 = 121 m² Increase = 121 − 55 = 66 m² ✓

Mastering Chapter 8 requires not just understanding identities once, but recognizing when and how to apply them in varied contexts. Traditional studying lacks this adaptive repetition. CBSETUTOR.ai's AI-powered tutor changes this: **Daily Pattern Recognition:** The AI generates randomized questions on (a+b)², (a−b)², and (a+b)(a−b) with varying coefficients and variables. You solve 5–7 problems daily, building neural recognition so you instantly spot which identity applies. **Instant Conceptual Feedback:** If you expand (3x+4)² incorrectly as 9x² + 16 (missing the middle term), the AI doesn't just mark it wrong—it highlights where the identity 2ab went and shows why 2(3x)(4) = 24x is essential. **Spaced Repetition Scheduling:** The tutor tracks which question types you struggle with (e.g., negative coefficients, combining identities) and resurfaces them at optimal intervals. Studies show this increases retention by 40%. **Real Exam Simulations:** Practice full 1-mark, 2-mark, 3-mark, and 5-mark sections under timed conditions. The AI adjusts difficulty based on your performance—harder questions appear only after you master basics. **Multi-Modal Explanations:** Watch a 2-minute video on why (a+b)² ≠ a² + b², solve a similar problem, then tackle a board-style question. This layered approach locks concepts in place. **Progress Analytics:** See exactly which identities you're strongest in and which need more drills. Many students discover they excel at proof questions but stumble on substitution—the AI then personalizes practice. Start a 3-day free trial at cbsetutor.ai and watch your Chapter 8 confidence grow. No credit card needed—access all 18 questions and AI explanations immediately.