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    Question

    I. 27x6 - 152x3 + 125 = 0

    II. 216y6 - 91y3 + 8 = 0 In the following questions, two equations numbered I and II are given. You have to solve both the equations and find out the correct 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 cannot be established Correct Answer Incorrect Answer

    Solution

    I. 27x6 - 152x3 + 125 = 0 Let x³ = a 27a2 - 152a + 125 = 0 27a2 - 125a - 27a + 125 = 0 a(27a - 125) - 1(27a - 125)=0 a=1,125/27 x³=1,125/27 x=1,5/3 II. (216y)6 - 91y3 + 8 = 0 Let y3 = b 216b2 - 91b + 8 = 0 216b2 - 27b - 64b + 8 = 0 27b(8b - 1) - 8(8b - 1) = 0 b = 1/8,8/27 y3 = 1/8,8/27 y = 1/2,2/3 Hence, x > y

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