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    Question

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

    are given.You have to solve both the equations and give answer    I. 7x² - 19x + 10 = 0                                         II. 8y² + 2y – 3 = 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

    7x² - 19x + 10 = 0 7x² - 14x – 5x + 10 = 0 7x (x - 2) – 5 (x - 2) = 0 x = 2, 5/7 II. 8y² + 2y – 3 = 0 8y² + 6y – 4y – 3 = 0 2y(4y + 3) – 1 (4y + 3) = 0  y = −3/4, 1/2 Hence, x > 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 = 2, 5/7 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 = -3/4, ½ After comparing roots of quadratic equation we can conclude that x> y.

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