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    Question

    I. x2 – 13x + 36 = 0 II.

    3y2 – 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 ≥ y Correct Answer Incorrect Answer
    C if x < y Correct Answer Incorrect Answer
    D if x ≤ 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 – 13x + 36 = 0 => x2 – 9x – 4x + 36 = 0 => x (x – 9) – 4 (x – 9) = 0 => x = 9, 4 II. 3y2 – 29y + 18 = 0 => 3y2 – 27y – 2y + 18 = 0 => 3y (y – 9) – 2(y – 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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