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    Question

    I. 9x2 + 45x + 26 = 0 II.

    7y2 – 59y − 36 = 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. 9x2 + 45x + 26 = 0 => 9x2 + 39x + 6x + 26 = 0 => 3x(3x + 13) + 2(3x + 13) = 0 => (3x + 2) (3x + 13) = 0 => x = -2/3, -13/3 II. 7y2 – 59y − 36 = 0 => 7y2 – 63y + 4y − 36 = 0 => 7y(y – 9) + 4(y - 9) = 0 => (7y + 4) (y – 9) = 0 => y = -4/7, 9 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 = -2/3, -13/3 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 = -4/7, 9 After comparing we can conclude that x < y.

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