Question

    I.  3y2 + 13y - 16 = 0 II.

    3x2 – 13x + 14 = 0 In each of these questions, two equations (I) and (II) are given.You have to solve both the equations and give answer   
    A X > Y Correct Answer Incorrect Answer
    B X < Y Correct Answer Incorrect Answer
    C X ≥Y Correct Answer Incorrect Answer
    D X ≤ Y Correct Answer Incorrect Answer
    E Relation cannot be established Correct Answer Incorrect Answer

    Solution

    I. 3y2 + 13y - 16 = 0 3y2 + 16y – 3y - 16 = 0 y (3y+16) – 1 (3y+16) = 0 (3y+16) (y-1) = 0 y = -16/3, 1 II. 3x2 – 13x + 14 = 0 3x2 – 6x - 7x + 14 = 0 3x (x-2) – 7 (x-2) = 0 (x-2) (3x-7) = 0 x = 2, 7/3 ؞ x > 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 larger root)   So, roots of first equation = y = -16/3, 1 If signs of quadratic equation are -ve and +ve then both roots of x will come with +ve signs So, roots of second equation = x = 2, 7/3 After comparing roots of quadratice equations we can conclude that x > y.

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