Quadratic Equations & Inequalities
Practice official JEE Main Quadratic Equations chapter-wise PYQs (2019-2024). Includes Newton Sums shortcuts, location of roots, timed test & step-by-step solutions.
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4-Step Solution & Tricks
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🎯 High-Yield Syllabus Breakdown
These high-frequency concepts account for over 90% of all questions asked in Quadratic Equations & Inequalities:
- Roots & Coefficients Relations (Vieta Formulas)
- Newton Sums Formula for αⁿ + βⁿ
- Location of Roots (Discriminant & Vertex Conditions)
- Common Roots of Quadratic Equations
- Transformation of Roots & Modulus Inequalities
⚡ Essential Formulas & Speed Shortcuts
Formulas and time-saving shortcuts to solve Quadratic Equations questions in under 45 seconds:
Newton Sums: If a·α² + b·α + c = 0 and Sₙ = αⁿ + βⁿ, then a·Sₙ + b·Sₙ₋₁ + c·Sₙ₋₂ = 0 directly.
Both roots positive: D ≥ 0, -b/(2a) > 0, and c/a > 0.
Both roots greater than k: D ≥ 0, -b/(2a) > k, and a·f(k) > 0.
📖 Sample Solved Exam Questions
Preview authentic questions from this chapter with full step-by-step solutions:
💡 View Step-by-Step Solution & Answer
Correct Answer: 3
- Step 1: Since α and β are roots of x² - 6x - 2 = 0, they satisfy α² - 6α - 2 = 0 and β² - 6β - 2 = 0.
- Step 2: Multiply the first equation by α⁸ and the second by β⁸ to get: α¹⁰ - 6α⁹ - 2α⁸ = 0 and β¹⁰ - 6β⁹ - 2β⁸ = 0.
- Step 3: Subtracting the two gives (α¹⁰ - β¹⁰) - 6(α⁹ - β⁹) - 2(α⁸ - β⁸) = 0, which means a₁₀ - 2a₈ = 6a₉.
- Step 4: Dividing both sides by 2a₉ gives (a₁₀ - 2a₈) / (2a₉) = 6a₉ / (2a₉) = 3.
⚡ Speed Tip: Newton's Sum Shortcut: For ax² + bx + c = 0, a·aₙ + b·aₙ₋₁ + c·aₙ₋₂ = 0 directly gives a₁₀ - 2a₈ = 6a₉.
💡 View Step-by-Step Solution & Answer
Correct Answer: Arithmetic Progression (AP)
- Step 1: Let the roots be α and β. Then α + β = -b/a and αβ = c/a.
- Step 2: Given α + β = 1/α² + 1/β² = (α² + β²)/(αβ)² = ((α+β)² - 2αβ)/(αβ)².
- Step 3: Substitute the root formulas: -b/a = [(-b/a)² - 2(c/a)] / (c/a)² = (b² - 2ac) / c².
- Step 4: Cross multiply: -b c² = a b² - 2 a² c ⇒ 2 a² c = a b² + b c², which means ca², ab², bc² (or bc², ca², ab²) are in AP.
⚡ Speed Tip: Standard symmetric identity: Whenever 2(term2) = term1 + term3, the sequence is always in AP.
💡 View Step-by-Step Solution & Answer
Correct Answer: 1
- Step 1: Let t = eˣ > 0 for all real x. The equation becomes t⁴ + t³ - 4t² + t + 1 = 0.
- Step 2: Since t ≠ 0, divide by t²: (t² + 1/t²) + (t + 1/t) - 4 = 0.
- Step 3: Put u = t + 1/t ≥ 2. Then t² + 1/t² = u² - 2, giving (u² - 2) + u - 4 = 0 ⇒ u² + u - 6 = 0.
- Step 4: Factorizing (u + 3)(u - 2) = 0 gives u = 2 or u = -3. Since u ≥ 2, only u = 2 is admissible, which gives t + 1/t = 2 ⇒ t = 1 ⇒ eˣ = 1 ⇒ x = 0 (exactly 1 real root).
⚡ Speed Tip: AM-GM Inequality: t + 1/t ≥ 2 for all t > 0. Immediately reject roots u < 2 without further calculation.
❓ Frequently Asked Questions (FAQs)
How many questions come from Quadratic Equations in JEE Main?
Typically 1 to 2 direct questions appear in every JEE Main shift, and quadratic concepts are heavily used in calculus and algebra.
What is the fastest shortcut for Sₙ = αⁿ + βⁿ problems?
Use Newton Sums: a·Sₙ + b·Sₙ₋₁ + c·Sₙ₋₂ = 0. This avoids expanding large powers like α¹⁰ + β¹⁰.