CBSE 10th Grade Mathematics > Chapter: Quadratic EquationsChapter: Quadratic EquationsPrevious Question 1 of 161 / 16NextReveal all steps (a) (x + 5)² = x(x + 10) Expand LHS: x² + 10x + 25 = x² + 10x Simplify: x² + 10x + 25 - x² - 10x = 0 25 = 0 (This is a contradiction) ∴ This is neither quadratic nor linear (no solution) (b) (2x - 3)(x + 4) = 2x² + 8 Expand LHS: 2x² + 8x - 3x - 12 = 2x² + 8 Simplify: 2x² + 5x - 12 = 2x² + 8 2x² + 5x - 12 - 2x² - 8 = 0 5x - 20 = 0 ∴ This is a linear equation (c) x³ + 2x² - x³ + 5x - 3 = 0 Simplify: 2x² + 5x - 3 = 0 ∴ This is a quadratic equation Standard form: 2x² + 5x - 3 = 0 a = 2, b = 5, c = -3 (a) √2x² = 3x - √5 √2x² - 3x + √5 = 0 ∴ a = √2, b = -3, c = √5 (b) (1/3)x² + (2/5)x = 1 (1/3)x² + (2/5)x - 1 = 0 Multiply by 15 (LCM of 3 and 5): 5x² + 6x - 15 = 0 ∴ a = 5, b = 6, c = -15 Or in fractional form: a = 1/3, b = 2/5, c = -1 (c) (x - √3)² = 2√3x x² - 2√3x + 3 = 2√3x x² - 2√3x + 3 - 2√3x = 0 x² - 4√3x + 3 = 0 ∴ a = 1, b = -4√3, c = 3 (a) x² + (√5 + 1)x + √5 = 0 We need factors of √5 that add to (√5 + 1) Try: (x + √5)(x + 1) = x² + x + √5x + √5 = x² + (1 + √5)x + √5 ∴ (x + √5)(x + 1) = 0 x = -√5 or x = -1 (b) 2x² - (1 + 2√2)x + √2 = 0 We need to split the middle term Product = 2√2, Sum = -(1 + 2√2) Try: -1 and -2√2 -1 + (-2√2) = -(1 + 2√2) (-1)(-2√2) = 2√2 2x² - x - 2√2x + √2 = 0 x(2x - 1) - √2(2x - 1) = 0 (x - √2)(2x - 1) = 0 x = √2 or x = 1/2 Given: √3x² + 10x + 7√3 = 0 a = √3, b = 10, c = 7√3 Using quadratic formula: x = (-b ± √(b² - 4ac)) / (2a) Calculate discriminant: D = b² - 4ac D = (10)² - 4(√3)(7√3) D = 100 - 4 × 7 × 3 D = 100 - 84 D = 16 Apply formula: x = (-10 ± √16) / (2√3) x = (-10 ± 4) / (2√3) x = (-10 + 4)/(2√3) or x = (-10 - 4)/(2√3) x = -6/(2√3) or x = -14/(2√3) x = -3/√3 or x = -7/√3 Rationalize: x = -3√3/3 or x = -7√3/3 x = -√3 or x = -7√3/3 (a) √2x² + 5x + 3√2 = 0 a = √2, b = 5, c = 3√2 D = b² - 4ac D = (5)² - 4(√2)(3√2) D = 25 - 4 × 3 × 2 D = 25 - 24 D = 1 Since D > 0, two distinct real roots (b) (1/2)x² - (√3/2)x + (1/8) = 0 a = 1/2, b = -√3/2, c = 1/8 D = b² - 4ac D = (-√3/2)² - 4(1/2)(1/8) D = 3/4 - 4/16 D = 3/4 - 1/4 D = 2/4 = 1/2 Since D > 0, two distinct real roots | (c) πx² + 2x + π = 0 a = π, b = 2, c = π D = b² - 4ac D = (2)² - 4(π)(π) D = 4 - 4π² D = 4(1 - π²) Since π ≈ 3.14, π² ≈ 9.87 1 - π² ≈ 1 - 9.87 = -8.87 < 0 Since D < 0, no real roots Given: (k - 2)x² + 2(k - 2)x + 2 = 0 Case 1: If k = 2, the equation becomes: 0·x² + 0·x + 2 = 0 2 = 0 (impossible, not a valid equation) So k ≠ 2 Case 2: For k ≠ 2, it's a quadratic equation a = k - 2, b = 2(k - 2), c = 2 For equal roots: D = 0 b² - 4ac = 0 [2(k - 2)]² - 4(k - 2)(2) = 0 4(k - 2)² - 8(k - 2) = 0 Factor out 4(k - 2): 4(k - 2)[(k - 2) - 2] = 0 4(k - 2)(k - 4) = 0 Since k ≠ 2: k - 4 = 0 k = 4 ∴ k = 4 Given: 2x² - 5x + 3 = 0 a = 2, b = -5, c = 3 α + β = -b/a = 5/2 αβ = c/a = 3/2 (a) α³ + β³ = (α + β)³ - 3αβ(α + β) = (5/2)³ - 3(3/2)(5/2) = 125/8 - 45/4 = 125/8 - 90/8 = 35/8 (b) 1/α² + 1/β² = (α² + β²)/(αβ)² First find α² + β²: α² + β² = (α + β)² - 2αβ = (5/2)² - 2(3/2) = 25/4 - 3 = 25/4 - 12/4 = 13/4 Now: 1/α² + 1/β² = (13/4)/(3/2)² = (13/4)/(9/4) = 13/9 Given: One root is 2 + √3 For rational coefficients, if 2 + √3 is a root, then 2 - √3 must also be a root (conjugate pair) Let α = 2 + √3 and β = 2 - √3 Sum of roots: α + β = (2 + √3) + (2 - √3) = 4 Product of roots: αβ = (2 + √3)(2 - √3) = 4 - 3 = 1 Quadratic equation: x² - (sum)x + (product) = 0 x² - 4x + 1 = 0 Verification for x = 2 + √3: (2 + √3)² - 4(2 + √3) + 1 = 4 + 4√3 + 3 - 8 - 4√3 + 1 = 7 + 4√3 - 8 - 4√3 + 1 = 0 ∴ The equation is x² - 4x + 1 = 0 Roots are 2 + √3 and 2 - √3 Given equation: px² + qx + r = 0 Let the roots be α and 1/α From the sum-product relations: Sum of roots: α + 1/α = -q/p ... (i) Product of roots: α × (1/α) = r/p ... (ii) From equation (ii): α × (1/α) = r/p 1 = r/p p = r ∴ r = p (Hence proved) Given equation: (a² + b²)x² - 2(ac + bd)x + (c² + d²) = 0 Coefficients: A = a² + b² B = -2(ac + bd) C = c² + d² For equal roots, D = 0 B² - 4AC = 0 B² = 4AC [-2(ac + bd)]² = 4(a² + b²)(c² + d²) 4(ac + bd)² = 4(a² + b²)(c² + d²) (ac + bd)² = (a² + b²)(c² + d²) Expand LHS: a²c² + 2abcd + b²d² = (a² + b²)(c² + d²) Expand RHS: a²c² + 2abcd + b²d² = a²c² + a²d² + b²c² + b²d² Simplify: a²c² + 2abcd + b²d² = a²c² + a²d² + b²c² + b²d² 2abcd = a²d² + b²c² 0 = a²d² - 2abcd + b²c² 0 = (ad - bc)² ad - bc = 0 ad = bc a/b = c/d ∴ a/b = c/d (Hence proved) Let the first digit be x and last digit be y Original number = 100x + 70 + y Number after interchange = 100y + 70 + x Given: New number = Original number + 99 100y + 70 + x = 100x + 70 + y + 99 100y + x = 100x + y + 99 99y = 99x + 99 99y - 99x = 99 y - x = 1 ... (equation 1) Also given: xy = 18 ... (equation 2) From equation 1: y = x + 1 Substitute in equation 2: x(x + 1) = 18 x² + x = 18 x² + x - 18 = 0 Factorising: (x + 9)(x - 2) = 0 x = -9 or x = 2 Since x is a digit, x = 2 From equation 1: y = 2 + 1 = 3 ∴ Original number = 100(2) + 70 + 3 = 273 Let son's present age = x years Father's present age = (45 - x) years [since sum = 45] Five years ago: Son's age = (x - 5) years Father's age = (45 - x - 5) = (40 - x) years Given: Product 5 years ago = 124 (x - 5)(40 - x) = 124 40x - x² - 200 + 5x = 124 -x² + 45x - 200 = 124 -x² + 45x - 324 = 0 Multiply by -1: x² - 45x + 324 = 0 Factorize: x² - 36x - 9x + 324 = 0 x(x - 36) - 9(x - 36) = 0 (x - 9)(x - 36) = 0 x = 9 or x = 36 If x = 9: Son = 9 years, Father = 45 - 9 = 36 years If x = 36: Son = 36 years, Father = 45 - 36 = 9 years (illogical) Verification for x = 9: 5 years ago: Son = 4, Father = 31 Product: 4 × 31 = 124 ∴ Son's present age = 9 years ∴ Father's present age = 36 years Let speed of boat in still water = x km/h Speed of stream = 4 km/h Speed upstream = (x - 4) km/h Speed downstream = (x + 4) km/h Distance = 32 km each way Time = Distance/Speed Time upstream = 32/(x - 4) hours Time downstream = 32/(x + 4) hours Total time = 6 hours 32/(x - 4) + 32/(x + 4) = 6 Taking LCM: [32(x + 4) + 32(x - 4)] / [(x - 4)(x + 4)] = 6 [32x + 128 + 32x - 128] / (x² - 16) = 6 64x / (x² - 16) = 6 64x = 6(x² - 16) 64x = 6x² - 96 6x² - 64x - 96 = 0 Dividing by 2: 3x² - 32x - 48 = 0 Using factorisation: 3x² - 36x + 4x - 48 = 0 3x(x - 12) + 4(x - 12) = 0 (3x + 4)(x - 12) = 0 x = -4/3 or x = 12 Since speed cannot be negative, x = 12 ∴ Speed of boat in still water = 12 km/h Let the smaller pipe fill the tank in x hours Then larger pipe fills in (x - 3) hours Time together = 6 hours 40 minutes = 6⅔ hours = 20/3 hours In 1 hour: Smaller pipe fills = 1/x of tank Larger pipe fills = 1/(x - 3) of tank Together they fill = 1/(20/3) = 3/20 of tank Equation: 1/x + 1/(x - 3) = 3/20 Taking LCM: (x - 3 + x) / [x(x - 3)] = 3/20 (2x - 3) / (x² - 3x) = 3/20 20(2x - 3) = 3(x² - 3x) 40x - 60 = 3x² - 9x 3x² - 9x - 40x + 60 = 0 3x² - 49x + 60 = 0 Using quadratic formula: x = [49 ± √(49² - 4(3)(60))] / (2 × 3) x = [49 ± √(2401 - 720)] / 6 x = [49 ± √1681] / 6 x = [49 ± 41] / 6 x = 90/6 or x = 8/6 x = 15 or x = 4/3 If x = 4/3, then x - 3 = -5/3 (negative, rejected) ∴ x = 15 hours ∴ Smaller pipe takes 15 hours ∴ Larger pipe takes 12 hours Let one side = x cm Then other side = (x + 5) cm [since difference is 5] Hypotenuse = 25 cm By Pythagoras theorem: x² + (x + 5)² = 25² x² + x² + 10x + 25 = 625 2x² + 10x + 25 = 625 2x² + 10x - 600 = 0 Dividing by 2: x² + 5x - 300 = 0 Using factorisation: x² + 20x - 15x - 300 = 0 x(x + 20) - 15(x + 20) = 0 (x - 15)(x + 20) = 0 x = 15 or x = -20 Since length cannot be negative, x = 15 cm Other side = 15 + 5 = 20 cm Verification: 15² + 20² = 225 + 400 = 625 = 25² ∴ The three sides are: 15 cm, 20 cm, and 25 cm Given: P = -x² + 16x - 48 (a) For break-even, P = 0 -x² + 16x - 48 = 0 Multiply by -1: x² - 16x + 48 = 0 Using factorisation: x² - 12x - 4x + 48 = 0 x(x - 12) - 4(x - 12) = 0 (x - 4)(x - 12) = 0 x = 4 or x = 12 ∴ Break-even at 4 thousand units or 12 thousand units i.e., 4,000 units or 12,000 units (b) For profit, P > 0 -x² + 16x - 48 > 0 x² - 16x + 48 < 0 (x - 4)(x - 12) < 0 This inequality holds when: 4 < x < 12 ∴ Company makes profit when selling between 4,000 and 12,000 units (exclusive)
