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Quadratic Equation Questions for IBPS Clerk Pre, SBI Clerk Pre and IBPS, RRB., Directions: In each of these questions, two equations (I) and (II) are given. You have to solve, both the equations and give answer., 1., , I. x3 – 4913 = 0, , II. y2 – 361 = 0, , A. if x < y, B. if x ≤ y, C. if x > y, E. if x = y or relationship between x and y can't be established, , 2., , I. x2 = 361, , II. y3 = 7269 + 731, , A. if x < y, B. if x > y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 3., , I. 15x2 + x – 6 = 0, , I. x3 – 2744 = 0, , I. x2 – 8x – 20 = 0, , I. 2x2 + 9x + 7 = 0, , C. if x ≥ y, , I. x2 – 7x + 12 = 0, , I. 2x2 + 15x + 28 = 0, , D. if x < y, , II. 3y2 – 11y + 10 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 8., , D. if x < y, , II. y2 + 4y + 4 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 7., , D. if x < y, , II. 3y2 – 60y + 297 = 0, , A. if x > y, B. if x ≤ y, E. if x = y or relationship between x and y can't be established, , 6., , D. if x < y, , II. y2 – 256 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 5., , D. if x ≤ y, , II. 5y2 – 23y + 12 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 4., , D. if x ≥ y, , D. if x < y, , II. 2y2 + 13y + 21 = 0, , A. if x > y, B. if x ≥ y, C. if x < y, E. if x = y or relationship between x and y can't be established, , D. if x ≤ y

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9., , I. x2 – 8x + 15 = 0, , II. y2 – 12y + 36 = 0, , A. if x > y, B. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 10., , I. x2 + 9x + 20 = 0, , D. if x ≤ y, , C. if x < y, , D. if x ≤ y, , II. y2 = 16, , A. if x > y, B. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 11., , C. if x < y, , I. x2 + (343)1/3 = 56, , II. (y)4/3 × (y)5/3 – 295 = 217, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 12., , I. 5x + 4y = 8, , II. 3x + 2y = 4, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 13., , I. x2 + 8 = 6x, , +√, , I. √, , =√, , I. x2 –, , (, , ), √, , =0, , II., , √, , –√ =, , I. 2x2 + 7x + 5 = 0, , I. 2x2 – 13x + 21 = 0, , D. if x < y, , II. 3y2 + 5y + 2 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 17., , D. if x < y, , √, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 16., , D. if x < y, , II. y2 – 212 = 364, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 15., , D. if x < y, , II. y2 + 15 = 8y, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 14., , D. if x < y, , D. if x < y, , II. 3y2 – 14y + 15 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x ≤ y or no relationship can be established between x and y., , D. if x < y

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18., , I. 2x2 – 13x + 18 = 0, , II. y2 – 7y + 12 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 19., , I. x2 + 6x + 9 = 0, , D. if x < y, , II. y2 – y – 20 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 20., , I. 3x2 – 10x + 8 = 0, , D. if x < y, , II. 2y2 – 19y + 35 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 21., , I. x2 – 3 = 2x, , II. y2 + 5y + 6 = 0, , A. if x > y, B. if x < y, E. if x = y or relationship between x and y can't be established, , 22., , I. x2 – 25x + 114 = 0, , I., , √, , +, , √, , =5√, , C. if x ≥ y, , II. y2 + √, , C. if x ≥ y, , I. x2 – 7 √3 x + 36 = 0, , A. if x > y, D. if x < y, , 25., , I. x2 = 361, , A. if x > y, D. if x ≥ y, , 26., , I. x2 + 5x + 6 = 0, , D. if x ≤ y, , =√, , A. if x > y, B. if x < y, D. if x = y or relationship between x and y can't be established, , 24., , D. if x ≤ y, , II. y2 – 10y + 24 = 0, , A. if x > y, B. if x < y, E. if x = y or relationship between x and y can't be established, , 23., , D. if x < y, , C. if x ≤ y, , II. y2 – 11 √3 y + 84 = 0, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , II. y3 = 7269 + 731, B. if x ≤ y, C. if x < y, E. if x = y or relationship between x and y can't be established, , II. y2 – 4y – 12 = 0, , A. if x > y, B. if x ≤ y, D. if x = y or relationship between x and y can't be established, , C. if x ≥ y, E. if x < y

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27., , I. 25x2 – 90x + 72 = 0, , II. y2 + 26y + 168 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 28., , I. 3x2 – 8x – 16 = 0, , II. 3y2 – 19y + 28 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 29., , I. 12x2 – 4x – 5 = 0, , I. 6x2 – 13x – 44 = 0, , I. 3x + 5y = 34.5, , I. 35x2 + 4x – 63 = 0, , I. x2 – 1089 = 0, , I. x2 – 4√7x + 21 = 0, , I. 3x2 – 8x – 16 = 0, , D. if x < y, , II. 2y2 – 8√5y – 50 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 35., , D. if x < y, , II. 3y2 – 363 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 34., , D. if x < y, , II. 7y2 – 4y – 20 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 33., , D. if x < y, , II. 4x – 9y = – 1, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 32., , D. if x ≥ y, , II. 4y2 – 17y – 42 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 31., , D. if x < y, , II. 8y2 – 4y – 4 = 0, , A. if x > y, B. if x < y, C. if x = y, E. if x ≤ y or no relationship can be established between x and y., , 30., , D. if x < y, , D. if x < y, , II. 3y2 – 19y + 28 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , D. if x < y

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36., , I. 3x2 – 5x – 12 = 0, , II. 2y2 + 15y + 25 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 37., , I. 12x2 – 4x – 5 = 0, , II. 8y2 – 4y – 4 = 0, , A. if x > y, B. if x < y, C. if x = y, E. if x ≤ y or no relationship can be established between x and y., , 38., , I. 2x + 3y = 77, , I. x2 – 4 (√2 + √5) x + 16√10 = 0, , A. if x > y, , D. if x ≥ y, , II. 3x + 5y = 124, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 39., , D. if x < y, , D. if x < y, , II. y2 – 5(√3 + 2√2) y + 50√6 = 0, , B. if x ≤ y, , C. if x ≥ y, , D. if x < y, , E. if x = y or relationship between x and y can't be established, , 40., , I. x2 – 4√3x + 9 = 0, , II. y2 – √3y – 18 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 41., , I. x2 – 9x + 20 = 0, , II. 2y2 – 15y + 28 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 42., , I. x2 – x – 20 = 0, , I. x2 – 9x + 18 = 0, , I. x2 – 9 = 0, , D. if x < y, , II. y2 – 9√2y + 36 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 44., , D. if x < y, , II. y2 + y – 30 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 43., , D. if x < y, , D. if x < y, , II. 2y2 + 13y + 21 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , D. if x < y

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45., , I. 5x2 + 11x – 12 = 0, , II. 4y2 – 13y – 12 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 46., , I. x2 + 16x + 63 = 0, , II. y2 + 13y + 42 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 47., , I. 2x2 + 3x – 20 = 0, , I. x2 – 13.5x + 38 = 0, , I. x2 + 11x + 30 = 0, , I. 4x2 – 216 = 0, , D. if x < y, , II. y2 + y – 20 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 50., , D. if x < y, , II. y2 – 1.5y – 10 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 49., , D. if x < y, , II. 2y2 + 15y + 28 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , 48., , D. if x < y, , D. if x < y, , II. 5y3 – 810√6 = 0, , A. if x > y, B. if x ≤ y, C. if x ≥ y, E. if x = y or relationship between x and y can't be established, , D. if x < y

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CORRECT OPTIONS:, 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , E, , A, , B, , E, , E, , E, , A, , D, , C, , D, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , 19, , 20, , D, , D, , E, , E, , D, , B, , C, , E, , E, , D, , 21, , 22, , 23, , 24, , 25, , 26, , 28, , 29, , 30, , A, , C, , D, , B, , A, , B, , 27, A, , E, , E, , E, , 31, , 32, , 33, , 34, , 35, , 36, , 37, , 38, , 39, , 40, , A, , E, , E, , E, , E, , A, , E, , D, , E, , E, , 41, , 42, , 43, , 44, , 45, , 46, , 47, , 48, , 49, , 50, , C, , E, , E, , C, , E, , B, , E, , C, , B, , B

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EXPLANATIONS:, , 1., , I. x3 – 4913 = 0, or, x3 = 4913, x = 17, II. y2 = 361, or, y = ± 19, While comparing the values of x and y, one root value of y lies between the root values of x, Hence, option E is correct., , 2., , I. x2 = 361, x = ± 19, II. y3 = 7269 + 731, y3 = 8000, y = 20, x<y, Hence, option A is correct., , 3., , I. 15x2 + x – 6 = 0, 15x2 + 10x – 9x – 6 = 0, 5x (3x + 2) – 3 (3x + 2) = 0, (5x – 3) (3x + 2) = 0, 3 2, x= ,–, 5 3, II. 5y2 – 23y + 12 = 0, 5y2 – 20y – 3y + 12 = 0, 5y (y – 4) – 3 (y – 4) = 0, (y – 4) (5y – 3) = 0, 3, y = 4,, 5, x≤y, Hence, option B is correct.

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4., , I. x3 – 2744 = 0, x3 = 2744, x = 14, II. y2 – 256 = 0, y2 = 256, y = ± 16, While comparing the values of x and y, one root value of x lies between the root values of y., Hence, option E is correct., , 5., , I. x2 – 8x – 20 = 0, ⇒ x2 – 10x + 2x – 20 = 0, ⇒ x (x – 10) + 2 (x – 10) = 0, ⇒ (x – 10) (x + 2) = 0, Then, x = + 10 or x = – 2, II. 3y2 – 60y + 297 = 0, ⇒ y2 – 20y + 99 = 0 [Dividing both sides by 3], ⇒ y2 – 11y – 9y + 99 = 0, ⇒ y (y – 11) – 9 (y – 11) = 0, ⇒ (y – 11) (y – 9) = 0, Then, y = + 11 or y = + 9, So, when x = + 10, x < y for y = + 11 and x > y for y = + 9, And when x = – 2, x < y for y = + 11 and x < y for y = + 9, ∴ So, we can observe that one root value of x lies between the root values of y. Therefore, the, relation between x and y can't be determined., Hence, option (E) is correct.

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6., , I. 2x2 + 9x + 7 = 0, or, 2x2 + 2x + 7x + 7 = 0, or, 2x(x + 1) + 7 (x + 1) = 0, or, (2x + 7) (x + 1) = 0, 7, ∴ x = – 1, –, 2, II. y2 + 4y + 4 = 0, or, y2 + 2y + 2y + 4 = 0, or, y (y + 2) + 2 (y + 2) = 0, or, (y + 2)(y + 2) = 0, ∴ y = – 2, –2, Hence, relationship can't be established between x and y., Therefore, Option E is correct., , 7., , I. x2 – 7x + 12 = 0, or, x2 – 4x – 3x + 12 = 0, or, x (x – 4) – 3 (x – 4) = 0, or, x (x – 4) – 3 (x – 4) = 0, or, (x – 4) (x – 3) = 0, ∴ x = 3, 4, II. 3y2 – 11y + 10 = 0, or, 3y2 – 6y – 5y + 10 = 0, or, 3y(y – 2) – 5(y – 2) = 0, or, (3y – 5)(y – 2) = 0, 5, ∴ y = 2,, 3, Hence, x > y, Hence, option A is correct.

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8., , I. 2x2 + 15x + 28 = 0, or, 2x2 + 8x + 7x + 28 = 0, or, 2x (x + 4) + 7 (x + 4) = 0, or, (2x + 7) (x + 4) = 0, 7, ∴ x = – 4, –, 2, II. 2y2 + 13y + 21 = 0, or, 2y2 + 6y + 7y + 21 = 0, or, 2y (y + 3) + 7 (y + 3) = 0, or, (2y + 7) (y + 3) = 0, 7, ∴ y = –3, –, 2, Hence, x ≤ y., Therefore, Option D is the correct answer., , 9., , I. x2 – 8x + 15 = 0, or, x2 – 5x – 3x + 15 = 0, or, x (x – 5) – 3 (x – 5) = 0, or, (x – 5) (x – 3) = 0, ∴ x = 5, 3, II. y2 – 12y + 36 = 0, or, y2 – 6y – 6y + 36 = 0, or, y(y – 6) – 6 (y – 6) = 0, or, (y – 6) (y – 6) = 0, ∴ y = 6, 6, Hence, x < y., Hence, option C is correct., , 10., , I. x2 + 9x + 20 = 0, or, x2 + 4x + 5x + 20 = 0, or, x (x + 4) + 5 (x + 4) = 0, or, (x + 4) (x + 5) = 0, x = – 4, – 5, II. y2 = 16, y=√ =±4, While comparing the x and y values, we got one value of x is equal to y and other values is less than the, root values of y., Hence, x ≤ y., Hence, option D is correct.

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11., , I. x2 + (343)1/3 = 56, x2 + 7 = 56, x2 = 49, ∴ x = √49 = ±7, II. (y)4/3 × (y)5/3 – 295 = 217, (y)3 = 217 + 295, (y)3 = 512 = (8)3, or, y = 8, Here, x < y, Hence, option D is correct., , 12., , 5x + 4y = 8, 3x + 2y = 4, , .......(i) × 3, .......(ii) × 5, , 15x + 12y = 24, 15x + 10y = 20, –, – –, 2y = 4, y=2, , .......(iii), .......(iv), , Putting the value of y in (i), we get, 5x + 8 = 8, 5x = 0, ∴x=0, Here, x < y, Hence, option D is correct.

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13., , I. x2 + 8 = 6x, x2 – 6x + 8 = 0, x2 – 4x – 2x + 8 = 0, x(x – 4) – 2(x – 4) = 0, (x – 2) (x – 4) = 0, ∴ x = 2, 4, II. y2 – 8y + 15 = 0, y2 – 5y – 3y + 15 = 0, y (y – 5) – 3 (y – 5) = 0, (y – 3) (y – 5 ) = 0, y = 3, 5, Here, while comparing the root values of x and y, we find that one root value of y lies between the, value of x. Therefore, no relationship between x and y can be established, Hence, option E is correct., , 14., , I. √, , +√, , 7+ √, (√, , =√, = 13, , )2 = (6)2, , x + 15 = 36, ∴ x = 36 – 15 = 21, II. y2 – 212 = 364, y2 = 364 + 212, y2 = 576, y = ± 24, Here, relationship between x and y cannot be established, Hence, option E is correct.

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15., (10)5/2, I. x –, =0, √x, 2, , x2+1/2 – (10)5/2 = 0, (x)5/2 = (10)5/2, x = 10, II., , 18, 7, – √y =, √y, √y, , 18 – y = 7, y = 11, Here, x < y, Hence, option D is correct., , 16., , I. 2x2 + 7x + 5 = 0, ⇒ 2x2 + 2x + 5x + 5 = 0, ⇒ 2x (x + 1) + 5 (x + 1) = 0, ⇒ (2x + 5) (x + 1) = 0, x = – 2.5, – 1, II. 3y2 + 5y + 2 = 0, ⇒ 3y2 + 3y + 2y + 2 = 0, ⇒ 3y (y + 1) + 2 (y + 1) = 0, ⇒ (3y + 2) (y + 1) = 0, y = – 0.66, – 1, For x = – 2.5 and y = – 0.66, – 1, , x<y, , For x = – 1 and y = – 0.66, – 1 x ≤ y, Hence x is either less than or equal to y., Hence, option B is correct.

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17., , I. x2 + 6x – 112 = 0, x2 + 14x – 8x – 112 = 0, x (x + 14) – 8(x + 14) = 0, (x + 14)(x – 8) = 0, x = 8, – 14, II. y2 + 22y + 112 = 0, y2 + 8y + 14y + 112 = 0, y(y + 8) + 14(y + 8) = 0, (y + 8)(y + 14) = 0, y = – 8, – 14, For, x = – 14 and y = – 8, x<y, For, x = – 14 and y = – 14, x=y, But for x = 8 and y = – 8 and – 14, x>y, Therefore, relationship can’t be established, Hence, option E is correct., , 18., , I. 2x2 – 13x + 18 = 0, ⇒ 2x2 – 4x – 9x + 18 = 0, ⇒ 2x (x – 2) – 9 (x – 2) = 0, ⇒ (2x – 9) (x – 2) = 0, x = 4.5, 2, II. y2 – 7y + 12 = 0, ⇒ y2 – 4y – 3y + 12 = 0, ⇒ y (y – 4) – 3 (y – 4) = 0, ⇒ (y – 3) (y – 4) = 0, y = 4, 3, For x = 4.5 and y = 4 ,3 x > y, For x = 2 and y = 4, 3, , x<y, , Hence, no relationship can be established, Hence, option E is correct.

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19., , I. x2 + 6x + 9 = 0, ⇒ x2 + 3x + 3x + 9 = 0, ⇒ x (x + 3) + 3 (x + 3) = 0, ⇒ (x + 3) (x + 3) = 0, x = – 3, – 3, II. y2 – y – 20 = 0, ⇒ y2 – 5y + 4y – 20 = 0, ⇒ y (y – 5) + 4 (y – 5) = 0, ⇒ (y + 4) (y – 5) = 0, y = – 4, 5, For x = – 3 and y = – 4,, For x = – 3 and y = 5,, , x>y, x<y, , Hence, no relationship can be established, Hence, option E is correct., , 20., , I. 3x2 – 10x + 8 = 0, ⇒ 3x2 – 6x – 4x + 8 = 0, ⇒ 3x (x – 2) – 4 (x – 2) = 0, ⇒ (3x – 4) (x – 2) = 0, x = 4/3, 2, II. 2y2 – 19y + 35 = 0, ⇒ 2y2 – 14y – 5y + 35 = 0, ⇒ 2y (y – 7) – 5 (y – 7) = 0, ⇒ (2y – 5) (y – 7) = 0, y = 2.5, 7, Hence, x < y, Hence, option D is correct.

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21., , I. x2 – 3 = 2x, ∴ x2 – 2x – 3 = 0, ∴ x2 – 3x + x – 3 = 0, ∴ (x + 1) (x – 3) = 0, ∴ x = 3 or x = −1, II. y2 + 5y + 6 = 0, ∴ y2 + 3y + 2y + 6 = 0, ∴ (y + 3) (y + 2) = 0, ∴ y = – 3 or y = – 2, For both values of x,, , x>y, , Hence, option A is correct, , 22., , I. x2 – 25x + 114 = 0, ∴ x2 – 19x – 6x + 114 = 0, ∴ (x – 6)(x – 19) = 0, ∴ x = 19 or x = 6, II. y2 – 10y + 24 = 0, ∴ y2 – 6y – 4y + 24 = 0, ∴ (y – 4)(y – 6) = 0, ∴ y = 6 or y = 4, When x = 19, x > y, When x = 6, x ≥ y, Hence, x ≥ y, Hence, option C is correct.

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23., , I., , √, , or,, , +, , √, , √, , = 5√, , =5√, , 10 = 5x, x=2, II. y2 + √, =√, 2, y + 16 = 25, y2 = 25 – 16, y2 = 9, y=±3, While comparing the values of x and y, one root value of y lies between the two root values of x, Hence, option D is correct., , 24., , I. x2 – 7 √3 x + 36 = 0, ⇒ x2 – 4 √3 x – 3 √3 x + 36 = 0, ⇒ x (x – 4 √3) – 3√3 (x – 4√3) = 0, ⇒ (x – 3√3) (x – 4√3) = 0, ∴ x = 3√3, 4√3, II. y2 – 11√3 y + 84 = 0, ⇒ y2 – 4√3 y – 7√3 y + 84 = 0, ⇒ y (y – 4√3)– 7√3 (y – 4√3) = 0, ⇒ (y – 7 √3) (y – 4√3) = 0, ∴ y = 7 √3, 4√3, Now, While comparing the root values of x and y, x, y, 3√3 < 4√3, 3√3 < 7√3, 4√3 = 4√3, 4√3 < 7√3, Here, x ≤ y, Hence, option (B) is correct.

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25., , I. x2 = 361, x = ± 19, II. y3 = 7269 + 731, y3 = 8000, y = 20, x<y, Hence, option A is correct., , 26., , I. x2 + 5x + 6 = 0, ⇒ x2 + 3x + 2x + 6 = 0, ⇒ x (x + 3) + 2 (x + 3) = 0, ⇒ (x + 2) (x + 3) = 0, ∴ x = – 2, – 3, II. y2 – 4y – 12 = 0, ⇒ y2 – 6y + 2y – 12 = 0, ⇒ y (y – 6) + 2 (y – 6) = 0, ⇒ (y + 2) (y – 6) = 0, ∴ y = – 2, + 6, Now, While comparing the root values of x and y, x, y, –2= –2, –2< +6, –3< –2, –3< +6, Here, x ≤ y, Hence, option (B) is correct.

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27., , I. 25x2 – 90x + 72 = 0, ⇒ (5x – 6) (5y – 12) = 0, 6 12, ⇒x= ,, 5 5, II. y2 + 26y + 168 = 0, ⇒ (y + 12)(y + 14) = 0, ⇒ y = – 12, – 14, Hence, x > Y, Hence, option A is correct., , 28., , I. 3x2 – 8x – 16 = 0, ⇒ (3x + 4)(x – 4) = 0, 4, ⇒x=– ,4, 3, II. 3y2 – 19y + 28 = 0, ⇒ (3y – 7)(y – 4) = 0, 7, ⇒y= ,4, 3, Hence, relationship between x and y cannot be determined., Hence, option E is correct., , 29., , I. 12x2 – 4x – 5 = 0, ⇒ 12x2 – 10x + 6x – 5 = 0, ⇒ 6x (2x + 1) – 5(2x + 1) = 0, ⇒ (6x – 5)(2x + 1) = 0, 5 –1, ∴ x = or, 6, 2, II. 8y2 – 4y – 4 = 0, ⇒ 8y2 – 8y + 4y – 4 = 0, ⇒ 8y (y – 1) + 4(y – 1) = 0, ⇒ (8y + 4) (y – 1) = 0, –1, ∴y=, or y = 1, 2, So, here we can’t say anything., Hence, option E is correct.

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34., , I. x2 – 4√7x + 21 = 0, ⇒ x2 – √7x – 3√7x + 21 = 0, ⇒ x (x – √7) – 3√7 (x – √7) = 0, ⇒ (x – √7)(x – 3√7) = 0, ⇒ x = √7, 3√7, II. 2y2 – 8√5y – 50 = 0, ⇒ 2y2 – 8√5y – 50 = 0, Taking 2 as a common term, we get, ⇒ y2 – 4√5y – 25 = 0, ⇒ y2 + √5y – 5√5y – 25 = 0, ⇒ y( y + √5) – 5√5 (y + √5) = 0, ⇒ (y + √5) (y – 5√5) = 0, ⇒ y = – √5, 5√5, While comparing the root values of x and y, we find that root values of y lies between the x's root, values., Therefore, relationship between x and y can't be determined., Hence, option E is correct., , 35., , I. 3x2 – 8x – 16 = 0, ⇒ (3x + 4)(x – 4) = 0, 4, ⇒x=– ,4, 3, II. 3y2 – 19y + 28 = 0, ⇒ (3y – 7)(y – 4) = 0, 7, ⇒y= ,4, 3, Hence, relationship between x and y cannot be determined., Hence, option E is correct.

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38., , We have, 2x + 3y = 77................(i), 3x + 5y = 124...............(ii), Multiplying eq. (i) by 3 and eq. (ii) by 2, we get, 6x + 9y = 231 ..............(iii), 6x + 10y = 246 ..............(iv), Subtracting eq. (iv) from (iii) we get, y = 17, putting value of y = 17 in (i), we get, ⇒ 2x + 3 × 17 = 77, ⇒ 2x = 51, ∴ x = 13, Here, x < y ., Hence, option (D) is correct., , 39., , I. x2 – 4√2x – 4√5x + 16√10 = 0, ⇒ x (x – 4√2) – 4√5 (x – 4√2) = 0, ⇒ (x – 4√5) (x – 4√2) = 0, ∴ x = 4√2, 4√5, II. y2 – 5 (√3 + 2√2) y + 50√6 = 0, ⇒ y2 – 5√3y – 10√2y + 50√6 = 0, ⇒ y (y – 5√3) – 10√2 (y – 5√3) = 0, ⇒ (y – 10√2) (y – 5√3) = 0, ∴ y = 10√2, 5√3, on comparing the root value of x and y, x, 4√2, 4√2, 4√5, 4√5, , <, <, <, >, , y, 10√2, 5√3, 10√2, 5√3, , Here, Either x = y or relation can't be established., Hence, option E is correct.

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40., , I. x2 – 4√3x + 9 = 0, x2 – 3√3x – √3x + 9 = 0, x(x – 3√3) – √3 (x – 3√3) = 0, (x – 3√3)(x – √3) = 0, x = √3, 3√3, II. y2 – √3y – 18 = 0, y2 + 2√3y – 3√3y – 18 = 0, y(y + 2√3) – 3√3 (y + 2√3) = 0, (y + 2√2)(y – 3√3) = 0, y = –2√3, 3√3, For x = 3√3, and y = 3√3 x = y, For x = 3√3, and y = –2√3 x > y, For x = √3, and y = –2√3 x > y, For x = √3, and y = 3√3 x < y, Therefore, relationship can’t be established, Hence, option E is correct., , 41., , I. x2 – 9x + 20 = 0, x2 – 5x – 4x + 20 = 0, x(x – 5) – 4 (x – 5) = 0, (x – 5)(x – 4) = 0, x = 5, 4, II. 2y2 – 15y + 28 = 0, 2y2 – 8y – 7y + 28 = 0, 2y(y – 4) – 7(y – 4) = 0, (y – 4)(2y – 7) = 0, y = 4,, , 7, 2, , 7, For x = 5, and y = 4, x > y, 2, For x = 4, and y = 4 x = y, For x = 4, and y = 3.5 x > y, Therefore, x ≥ y, Hence, option C is correct.

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42., , I. x2 – x – 20 = 0, x2 – 5x + 4x – 20 = 0, x(x – 5) + 4(x – 5) = 0, (x – 5)(x + 4) = 0, x = 5, –4, II. y2 + y – 30 = 0, y2 + 6y – 5y – 30 = 0, y(y + 6) – 5(y + 6) = 0, (y – 5)(y + 6) = 0, y = –6, 5, For x = 5, and y = 5, –6, x ≥ y, For x = –4, and y = –6, x > y, For x = –4, and y = 5, x < y, Therefore, relationship can’t be established, Hence, option E is correct., , 43., , I. x2 – 9x + 18 = 0, x2 – 6x –3x + 18 = 0, x(x – 6) – 3(x – 6) = 0, (x – 6)(x – 3) = 0, x = 3, 6, II : y2 – 9√2y + 36 = 0, y2 – 3√2y – 6√2y + 36 = 0, y(y – 3√2) – 6√2 (y – 3√2) = 0, (y – 6√2)(y – 3√2) = 0, y = 6√2, 3√2, For x = 3 and y = 6√2 or 3√2 x < y, For x = 6 and y = 6√2, x<y, For x = 6 and y = 3√2, x>y, Therefore, relationship can’t be established, Hence, option E is correct.

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44., , I: x3 – 9 = 0, x = 3, –3, II: 2y2 + 13y + 21 = 0, 2y2 + 7y + 6y + 21 = 0, y(2y + 7) + 3(2y + 7) = 0, (y + 3)(2y + 7) = 0, 7, y = –3, – = –3.5, 2, For x = –3 and y = –3, x = y, For x = –3 and y = –3.5 x > y, For x = 3, and y = –3 or –3.5 x > y, Therefore, x ≥ y, Hence, option C is correct., , 45., , I. 5x2 + 11x – 12 = 0, ⇒ 5x2 + 15x – 4x – 12 = 0, ⇒ 5x (x + 3) – 4(x + 3) = 0, ⇒ (5x – 4) (x + 3) = 0, 4, ⇒x= ,–3, 5, II. 4y2 – 13y – 12 = 0, ⇒ 4y2 – 16y + 3y – 12 = 0, ⇒ 4y(y – 4) + 3 (y – 4) = 0, ⇒ (4y + 3) (y – 4) = 0, 3, ⇒y=– ,4, 4, While comparing the root values of x and y, we find that one root value of y lies between the root, values of x., Therefore, relationship between x and y can't be determined., Hence, option E is correct.

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46., , I. x2 + 16x + 63 = 0, x2 + 9x + 7x + 63 = 0, x(x + 9) + 7(x + 9) = 0, (x + 7)(x + 9) = 0, x = – 7, – 9, II. y2 + 13y + 42 = 0, y2 + 7y + 6y + 42 = 0, y (y + 7) + 6(y + 7) = 0, (y + 7)(y + 6) = 0, y = – 7, – 6, For x = – 7, and y = – 7,, x=y, For x = – 7, or – 9 and y = – 6 x < y, For x = – 9 and y = – 6, x<y, Therefore, x ≤ y, Hence, option B is correct., , 47., , I. 2x2 + 3x – 20 = 0, 2x2 + 8x – 5x – 20 = 0, 2x(x + 4) – 5(x + 4) = 0, (2x – 5)(x + 4) = 0, x = 2.5, – 4, II. 2y2 + 15y + 28 = 0, 2y2 + 8y + 7y + 28 = 0, 2y(y + 4) + 7(y + 4) = 0, (2y + 7)(y + 4) = 0, y = – 3.5, – 4, For x = – 4 and y = – 4, x = y, For x = 2.5, and y = – 3.5 or – 4, x>y, For x = – 4, and y = – 3.5, x<y, Therefore, relationship can’t be established, Hence, option E is correct.

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48., , I. x2 – 13.5x + 38 = 0, x2 – 9.5x – 4x + 38 = 0, x(x – 9.5) – 4(x – 9.5) = 0, (x – 9.5)(x – 4) = 0, x = 9.5, 4, II. y2 – 1.5y – 10 = 0, y2 – 4y + 2.5y – 10 = 0, y(y – 4) + 2.5(y – 4) = 0, (y – 4)(y + 2.5) = 0, y = 4, – 2.5, For x = 9.5 x > y, For x = 4, and y = 4, x = y, Therefore, x ≥ y, Hence, option C is correct., , 49., , I. x2 + 11x + 30 = 0, x2 + 5x + 6x + 30 = 0, x(x + 5) + 6(x + 5) = 0, (x + 6)(x + 5) = 0, x = – 6, – 5, II. y2 + y – 20 = 0, y2 + 5y – 4y – 20 = 0, y(y + 5) – 4(y + 5) = 0, (y – 4)(y + 5) = 0, y = – 5, 4, For x = – 5 and y = – 5, x = y, For x = – 6, y = – 5 or 4 , x < y, Therefore, x ≤ y, Hence, option B is correct.

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50., , I. 4x2 – 216 = 0, x2 = 54, x = ±3 × 61/2, II. 5y3 – 810√6 = 0, y3 = 162√6 = 3√6 × 3√6 × 3 × √6, y = 3√6, Hence, option B is correct.