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    Question

    I. 3x² - 22 x + 40 = 0    II. 4y² + 22y  +

    24 = 0     In the following questions, two equations numbered I and II are given. 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 > y Correct Answer Incorrect Answer
    D If x < y Correct Answer Incorrect Answer
    E if x = y or relationship cannot be established. Correct Answer Incorrect Answer

    Solution

    I. 3x² - 22 x + 40 = 0     3x² - 12x - 10x + 40 = 0     3x (x – 4) – 10 (x – 4) (3x – 10 ) (x – 4) x = 10/3, 4 II. 4y² + 22y  + 24 = 0    4y² + 16 y + 6 y + 24 = 0     4y (y + 4) + 6( y + 4) (4y + 6) (y + 4) x =  - 4 , - 3/2 Hence, x > y.

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