Exercise 4.3 - Q2
Textbook Exercise Problem
Find the roots of the quadratic equation: 2x² - 7x + 3 = 0 using the quadratic formula.
Step-by-Step Verified Solution:
- Step 1: Compare with ax² + bx + c = 0: a = 2, b = -7, c = 3.
- Step 2: Compute Discriminant: D = b² - 4ac = (-7)² - 4(2)(3) = 49 - 24 = 25.
- Step 3: Since D = 25 > 0, two distinct real roots exist. √D = √25 = 5.
- Step 4: x = [-(-7) ± 5] / [2(2)] = (7 ± 5) / 4.
- Step 5: x1 = (7 + 5) / 4 = 12 / 4 = 3; x2 = (7 - 5) / 4 = 2 / 4 = 1/2.
Final Answer: x = 3 and x = 1/2
Key Concept / Exam Strategy: Quadratic formula guarantees finding roots even when middle-term splitting is non-obvious.
Generated Exercise #1
Foundation
Question: Given that HCF(306, 657) = 9, find LCM(306, 657).
Core Concept Tested: HCF and LCM product relationship for two integers
Mentor Hint: LCM(a, b) = (a × b) / HCF(a, b).
Step-by-Step Solution:
- Step 1: Formula: HCF × LCM = a × b.
- Step 2: 9 × LCM = 306 × 657.
- Step 3: LCM = (306 × 657) / 9 = 34 × 657 = 22,338.
Verified Answer: LCM = 22,338
Generated Exercise #2
Intermediate
Question: Find the value of k for which the quadratic equation 2x² + kx + 3 = 0 has two equal real roots.
Core Concept Tested: Discriminant condition D = 0 for equal roots
Mentor Hint: Set b² - 4ac = 0.
Step-by-Step Solution:
- Step 1: Here a = 2, b = k, c = 3.
- Step 2: For equal roots, D = 0 => k² - 4(2)(3) = 0.
- Step 3: k² - 24 = 0 => k² = 24 => k = ±√24 = ±2√6.
Verified Answer: k = ±2√6
Generated Exercise #3
Intermediate
Question: The sum of two numbers is 15, and the sum of their reciprocals is 3/10. Find the numbers.
Core Concept Tested: Translating reciprocal constraints into quadratic equations
Mentor Hint: Let numbers be x and 15 - x. 1/x + 1/(15-x) = 3/10.
Step-by-Step Solution:
- Step 1: 1/x + 1/(15 - x) = 3/10 => 15 / [x(15 - x)] = 3/10.
- Step 2: 5 / [15x - x²] = 1/10 => 15x - x² = 50 => x² - 15x + 50 = 0.
- Step 3: (x - 10)(x - 5) = 0 => x = 10 or x = 5.
Verified Answer: The numbers are 10 and 5
Generated Exercise #4
JEE-NEET Foundation
Question: If α and β are the roots of 2x² - 5x + 7 = 0, form a new quadratic equation whose roots are 2α + 3β and 3α + 2β.
Core Concept Tested: Symmetric root transformations using Vieta formulas
Mentor Hint: Sum of roots α + β = 5/2, αβ = 7/2.
Step-by-Step Solution:
- Step 1: New sum S = (2α + 3β) + (3α + 2β) = 5(α + β) = 5(5/2) = 25/2.
- Step 2: New product P = (2α + 3β)(3α + 2β) = 6α² + 13αβ + 6β² = 6(α + β)² + αβ.
- Step 3: P = 6(25/4) + 7/2 = 75/2 + 7/2 = 82/2 = 41.
- Step 4: Equation: x² - Sx + P = 0 => x² - (25/2)x + 41 = 0 => 2x² - 25x + 82 = 0.
Verified Answer: 2x² - 25x + 82 = 0
Generated Exercise #5
Advanced HOTS
Question: Solve for x: √(x + 5) + √(x + 12) = √(2x + 41).
Core Concept Tested: Irrational radical equation solving via systematic squaring
Mentor Hint: Square both sides, isolate the radical cross-term, and square again.
Step-by-Step Solution:
- Step 1: Squaring both sides: (x + 5) + (x + 12) + 2√((x+5)(x+12)) = 2x + 41.
- Step 2: 2x + 17 + 2√(x² + 17x + 60) = 2x + 41 => 2√(x² + 17x + 60) = 24 => √(x² + 17x + 60) = 12.
- Step 3: Square again: x² + 17x + 60 = 144 => x² + 17x - 84 = 0.
- Step 4: (x + 21)(x - 4) = 0 => x = 4 (x = -21 gives extraneous negative inside root).
- Step 5: Verify x = 4: √9 + √16 = 3 + 4 = 7 = √49. Valid.
Verified Answer: x = 4