Question

    In each of these questions, two equations (I) and (II)

    are given.You have to solve both the equations and give answer    I. 3x² –7x + 4 = 0                                     II. 2y² – 9y + 10 = 0
    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. 3x² –7x + 4 = 0 ⇒ 3x²– 4x - 3x +4 = 0    ⇒ (3x – 4) (x -1) = 0  x = 4/3 or 1  II. 2y² - 9y + 10 = 0  ⇒ 2y² - 4y - 5y + 10 = 0  ⇒ (2y - 5) (y -2) =0    ⇒ y = 5/2  or 2 Hence, y  > x 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 So, roots of second equation = y = 5/2, 2 After comparing roots of quadratice eqution we can conclude that y > x.

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