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      Question

      I. x2 โ€“ 13x + 36 = 0 II.

      3y2 โ€“ 29y + 18 = 0 In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer.
      A if x > y Correct Answer Incorrect Answer
      B if x โ‰ฅ y Correct Answer Incorrect Answer
      C if x < y Correct Answer Incorrect Answer
      D if x โ‰ค y Correct Answer Incorrect Answer
      E if x = y or the relation between x and y cannot be established Correct Answer Incorrect Answer

      Solution

      I. x2 โ€“ 13x + 36 = 0 => x2 โ€“ 9x โ€“ 4x + 36 = 0 => x (x โ€“ 9) โ€“ 4 (x โ€“ 9) = 0 => x = 9, 4 II. 3y2 โ€“ 29y + 18 = 0 => 3y2 โ€“ 27y โ€“ 2y + 18 = 0 => 3y (y โ€“ 9) โ€“ 2(y โ€“ 9) = 0 => y = 9, 2/3 Hence, the relation between x and y cannot be established. Alternate Method: if signs of quadratic equation is - ve and +ve respectively then the roots of equation will be +ve and +ve. So, roots of first equation = x = 9, 4 So, roots of second equation = y = 9, 2/3 After comparing we can conclude that no relationship can be establish in x and y.

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