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    Question

    I. 2y² - 3y – 14 = 0 II. 3x² - 7x + 4 =

    0 In the following questions, two equations numbered I and II are given. You have to solve both the equations and find out the correct option. Give answer if ;
    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 or the relationship cannot be established Correct Answer Incorrect Answer
    E if x < y Correct Answer Incorrect Answer

    Solution

    I. 2y² - 3y – 14 = 0 2y² - 7y  + 4 y – 14 = 0 y (2 y – 7) + 2( 2 y – 7) = 0 (y + 2) (2 y – 7) = 0 ∴ y = -2, 7/2 II. 3x² - 7x + 4 = 0 3x² - 4x - 3 x  + 4 = 0 x (3 x – 4) – 1 (3 x – 4) = 0 ∴ x = 1, 4/3 Hence, relationship cannot be established between x and y Alternate Method: if signs of quadratic equation is - ve and -ve respectively then the roots of equation will be -ve and +ve. (note: -ve sign will come in smaller root) So, roots of first equation = y = -2, 7/2 if signs of quadratic equation is - ve and +ve respectively then the roots of equation will be +ve and +ve. So, roots of second equation = x = 1, 4/3 After comparing we can conclude that relationship cannot be established.

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