📢 Too many exams? Don’t know which one suits you best? Book Your Free Expert 👉 call Now!

    ⚡ Month End Offer - Flat 52% Off On All Courses! Enroll Now ⚡
    00:00:00 AM Left

    Question

    I. x2 + 12√2 x + 22 = 0 II.

    y2 - 13√2 y – 28 = 0 In the following questions, two equations I and II are given. You have to solve both equations and give answer as,
    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 the relation cannot be established. Correct Answer Incorrect Answer

    Solution

    I. x2 + 12√2 x + 22 = 0 (x + 11√2) (x + √2) = 0 x = - 11√2, -√2  II. y2 - 13√2 y – 28 = 0 (y - 14√2) (y + √2) = 0 y = - √2, 14√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 = -11√2, -√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 smaller root) So, roots of second equation = y = - √2, 14√2 After comparing roots of quadratic equation we can conclude that x ≤ y.

    Practice Next
    More Quadratic equation Questions

    Relevant for Exams:

    ask-question