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    Question

    I. 2y2– 19y + 35 = 0 II.

    4x2– 16x + 15 = 0
    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 relationship cannot be established Correct Answer Incorrect Answer

    Solution

    I. 2y2 – 19y + 35 = 0 2y2 – 5y – 14 y + 35 = 0 y (2 y – 5) – 7(2 y – 5) = 0 (2 y – 5)(y – 7) = 0 y = 7, 5/2 II. 4x2 – 16x + 15 = 0 4x2 – 6x – 10 x + 15 = 0 2 x(2 x – 3) – 5(2 x – 3) = 0 (2 x – 3)(2x – 5 )=0 x = 5/2, 3/2 Hence, x ≤ y 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 = y = 7, 5/2 So, roots of second equation = x = 5/2, 3/2 After comparing we can conclude that x ≤ y.

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