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    Question

    I. 2x2 + 13x + 21 = 0 II.

    3y2 + 34y + 63 = 0 In these questions, two equations numbered I and II are given. You have to solve both equations and mark the appropriate option. 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 between x and y can’t be established. Correct Answer Incorrect Answer

    Solution

    I. 2x2 + 13x + 21 = 0 => 2x2 + 6x + 7x + 21 = 0 => 2x(x + 3) + 7(x + 3) = 0 => (x + 3) (2x + 7) = 0 => x = -3, -7/2 II. 3y2 + 34y + 63 = 0 => 3y2 + 27y + 7y + 63 = 0 => 3y(y + 9) + 7(y + 9) = 0 => (y + 9) (3y + 7) = 0 => y = -9, -7/3 Hence, relation 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. So, roots of first equation = x = -3, -7/2 So, roots of second equation = y = -9, -7/3 After comparing we can conclude that relationship cannot be established between x and y.

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