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  9. Q98

IBPS PO 2020 · Question 98 of 99

In the following question, two equations numbered I and II are given. You have to solve both the equations and give answer: I. 3x^2 - 11x + 10 = 0 II. 4y^2 + 24y + 35 = 0

Quadratic Equations

  1. Ax > yCorrect
  2. Bx >= y
  3. Cx < y
  4. Dx <= y
  5. Ex = y or the relation cannot be determined

Answer: A. x > y

Explanation

Equation I: 3x^2 - 11x + 10 = 0 => 3x^2 - 5x - 6x + 10 = 0 => x(3x - 5) - 2(3x - 5) = 0 => (x - 2)(3x - 5) = 0 => x = 2 or x = 5/3. Equation II: 4y^2 + 24y + 35 = 0 => 4y^2 + 14y + 10y + 35 = 0 => 2y(2y + 7) + 5(2y + 7) = 0 => (2y + 5)(2y + 7) = 0 => y = -5/2 or y = -7/2. Value of x / Value of y / Relation: 2, -5/2 -> x > y; 2, -7/2 -> x > y; 5/3, -5/2 -> x > y; 5/3, -7/2 -> x > y. Therefore, x > y.

Source: IBPS PO 10-Oct-2020 Prelims Memory Based Paper Shift 1 (Prepp) · memory-based

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