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      Question

      I. 3x2 - 16x - 12 = 0 II.

      2y2 + 11y + 9 = 0 In the following questions, two equations numbered I and II are given.┬а You have to solve both the equations and give answer.
      A If x > y Correct Answer Incorrect Answer
      B If x < y Correct Answer Incorrect Answer
      C If x тЙе y Correct Answer Incorrect Answer
      D If x тЙд y Correct Answer Incorrect Answer
      E If x = y or Relationship between x & y cannot be established Correct Answer Incorrect Answer

      Solution

      I. 3x2┬а- 16x - 12 = 0 => 3x2┬а- 18x + 2x - 12 = 0 => 3x(x - 6) + 2(x - 6) = 0 => (x - 6) (3x + 2) = 0 => x = 6, -2/3 II. 2y2┬а+ 11y + 9 = 0 => 2y2┬а+ 9y + 2y + 9 = 0 => y (2y + 9) + 1(2y + 9) = 0 => (y + 1) (2y + 9) = 0 => y = -1, -9/2 Hence, x > y

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