Class 9 Mathematics Chapter 4 Quadratic Equations: Previous Year Questions & Solutions (2020–2025)

Reading Chapter 4 once teaches you what a quadratic equation is; solving 10 previous year questions teaches you *how examiners think*. Past papers reveal three critical patterns: (1) Examiners always test standard form (ax² + bx + c = 0, a ≠ 0) by disguising it as word problems or mixed-degree expressions. (2) Factorisation is preferred in 1-mark and 3-mark questions when roots are rational integers. (3) The discriminant (Δ = b² – 4ac) appears in at least one question per paper to test conceptual depth, not just calculation. By solving questions in the exact order and style they appear in official papers, you build pattern recognition. You'll instantly know: if a question asks 'without solving, find nature of roots', use discriminant immediately. If it says 'solve', check if factorisation works first before reaching for the formula. This exam-ready mindset only develops through deliberate practice, not passive reading. The questions below span the 2020–2025 question bank and reflect genuine CBSE difficulty levels.

One-mark questions test recall of definitions, standard form recognition, and the discriminant formula. These five are representative of the question types that cycle through papers. **Q1: Identify the quadratic equation.** (A) 2x² + 3x + 5 = 0 (B) x³ + 2x + 1 = 0 (C) 3x + 5 = 0 (D) x⁴ + x² – 1 = 0 *Answer: (A) 2x² + 3x + 5 = 0* A quadratic must have degree exactly 2 with non-zero leading coefficient. Only (A) qualifies. **Q2: Find the value of discriminant for 3x² – 5x + 2 = 0.** Δ = b² – 4ac = (–5)² – 4(3)(2) = 25 – 24 = **1** **Q3: If x = 2 is a root of x² + kx – 6 = 0, find k.** Substitute: 4 + 2k – 6 = 0 → 2k = 2 → **k = 1** **Q4: Express 2x² = 3x + 4 in standard form.** Rearrange: **2x² – 3x – 4 = 0** (verify a ≠ 0) **Q5: State the nature of roots if discriminant = 0.** **Real and equal roots.** When Δ = 0, the quadratic touches the x-axis at exactly one point.

Three-mark questions demand two skills: either solve a quadratic using an appropriate method *or* prove a relationship using the formula. Below are five types that repeat across papers. **Q1: Solve 6x² – x – 1 = 0 by factorisation.** 6x² – x – 1 = 0 Find factors of 6 × (–1) = –6 that sum to –1: these are –3 and 2. 6x² – 3x + 2x – 1 = 0 3x(2x – 1) + 1(2x – 1) = 0 (3x + 1)(2x – 1) = 0 *x = –1/3 or x = 1/2* ✓ **Q2: Solve x² – 4x – 5 = 0 by completing the square.** x² – 4x = 5 x² – 4x + 4 = 5 + 4 (x – 2)² = 9 x – 2 = ±3 *x = 5 or x = –1* ✓ **Q3: Use the quadratic formula to solve 2x² – 3x – 2 = 0.** x = [3 ± √(9 + 16)] / 4 = [3 ± 5] / 4 *x = 2 or x = –1/2* ✓ **Q4: If the roots of ax² + bx + c = 0 are equal, prove that b² = 4ac.** For equal roots, discriminant = 0 b² – 4ac = 0 *b² = 4ac* ✓ (Direct application of discriminant condition) **Q5: Find the value of k such that x² + 2kx + 9 = 0 has equal roots.** For equal roots: (2k)² – 4(1)(9) = 0 4k² – 36 = 0 k² = 9 *k = ±3* ✓

Five-mark questions combine multiple steps: converting a word problem to standard form, solving via an optimal method, and interpreting the result. Here are three representative full solutions. **Q1: A rectangular garden has length 3 m more than its width. If the area is 28 m², find the dimensions.** *Solution:* Let width = x m. Then length = (x + 3) m. Area: x(x + 3) = 28 x² + 3x – 28 = 0 Factorise: (x + 7)(x – 4) = 0 x = –7 (rejected, width cannot be negative) or x = 4 **Width = 4 m, Length = 7 m** ✓ *Verification: 4 × 7 = 28* ✓ **Q2: The sum of a number and its reciprocal is 10/3. Find the number.** *Solution:* Let the number = x. x + 1/x = 10/3 Multiply by x: x² + 1 = (10/3)x 3x² – 10x + 3 = 0 Using the quadratic formula: x = [10 ± √(100 – 36)] / 6 = [10 ± 8] / 6 x = 3 or x = 1/3 **The number is 3 or 1/3** ✓ *Verification: 3 + 1/3 = 10/3* ✓ **Q3: Solve 4x² – 4ax + (a² – b²) = 0 using the quadratic formula. Express roots in terms of a and b.** *Solution:* Here: A = 4, B = –4a, C = a² – b² Discriminant: Δ = (–4a)² – 4(4)(a² – b²) = 16a² – 16a² + 16b² = 16b² x = [4a ± √(16b²)] / 8 = [4a ± 4b] / 8 **x = (a + b)/2 or x = (a – b)/2** ✓ *Note: Roots are real for all real values of a and b since Δ = 16b² ≥ 0.*

The newly rationalized CBSE Class 9 Mathematics syllabus (2024–25 onwards) has shifted emphasis slightly. Historical observation suggests three key changes affect Quadratic Equations: (1) *Increased conceptual rigour*: Examiners now link the discriminant directly to the parabola graph. A question might ask: 'If Δ < 0, sketch why the parabola does not intersect the x-axis.' This tests geometric intuition, not just algebra. (2) *Real-world applications weighted more*: Word problems dominate the 5-mark sections. Garden, speed-distance, and profit-cost scenarios are common. Pure algebraic 'solve this equation' questions appear less frequently. (3) *Completing the square resurfacing*: For several years, this method was underemphasized, but recent papers show renewed interest, especially in questions that ask 'solve without using the formula' or 'derive the quadratic formula using this method.' (4) *Integer coefficients preferred*: Questions with fractional or irrational coefficients in the original form have decreased slightly; most papers stick to 'nice' integers to keep focus on method, not arithmetic. Students preparing for 2026–27 should practise converting messy real-world statements to standard form and solving via all three methods—not just the quickest one.

Exam day demands a two-minute decision tree: *Read the question. If it says 'solve', apply this logic:* **Step 1: Check if the equation factors easily.** Rewrite in standard form ax² + bx + c = 0. If the discriminant Δ = b² – 4ac is a perfect square (0, 1, 4, 9, 16, 25…), factorisation will work cleanly. Example: 2x² – 5x + 3 = 0 has Δ = 25 – 24 = 1 (perfect square) → factorise. Result: (2x – 3)(x – 1) = 0. **Step 2: If Δ is not a perfect square, use the quadratic formula.** It always works. Write carefully: x = [–b ± √(b² – 4ac)] / 2a. Avoid sign errors; this is where marks are lost. **Step 3: Completing the square is fastest only if b is even and 'nice'.** Example: x² – 6x + 5 = 0 (b = –6, even) → (x – 3)² = 4 → x = 5 or 1. Avoid this method if b is odd or large. **If the question asks 'without solving, find nature of roots':** Calculate Δ only. If Δ > 0, roots are real and unequal. If Δ = 0, roots are real and equal. If Δ < 0, roots are imaginary (not part of CBSE Class 9, but stated for completeness). **For word problems:** First *define* the variable clearly. Write the equation. Solve. **Always reject negative/non-physical solutions** (time cannot be negative, count cannot be fractional unless specified). Verify your answer substitutes back into the original statement. This three-step word-problem protocol secures marks even if algebra is slightly messy.

**Standard Form:** ax² + bx + c = 0, where a ≠ 0 and a, b, c ∈ ℝ. **Discriminant:** Δ = b² – 4ac - If Δ > 0: Two distinct real roots - If Δ = 0: One repeated real root (or two equal roots) - If Δ < 0: No real roots **Quadratic Formula:** x = [–b ± √(b² – 4ac)] / 2a **Sum and Product of Roots:** If α and β are roots of ax² + bx + c = 0, then: - Sum: α + β = –b/a - Product: αβ = c/a **Factorisation Method:** If ax² + bx + c = 0 factorises as (px + q)(rx + s) = 0, then x = –q/p or x = –s/r. **Completing the Square:** Convert ax² + bx + c = 0 to (x + p)² = q form, then take square roots. This requires dividing through by a first if a ≠ 1. Mastering these five tools and knowing when to use each is the entire chapter. Past papers show examiners reward method clarity as much as the final answer.