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    Question

    I. 12x2 + 22x + 8 = 0 II.

    4y2 - y − 3 = 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. 12x2 + 22x + 8 = 0 => 12x2 + 16x + 6x + 8 = 0 => 4x(3x + 4) + 2(3x + 4) = 0 => (3x + 4) (4x + 2) = 0 => x = -4/3, -1/2 II. 4y2 - y − 3 = 0 => 4y2 - 4y + 3y - 3 = 0 => 4y(y – 1) + 3(y – 1) = 0 => (y – 1) (4y + 3) = 0 => y = 1, y = -3/4 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 = -4/3, -1/2 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 larger root) So, roots of second equation = y = 1, -3/4 After comparing we can conclude that relationship cannot be established between x and y.

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