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    Question

    I. 6p² + 17p + 12 = 0 II. 12q² - 25q +

    7 = 0 In each of these questions two equations numbered I and II are given. You have to solve both the equations and give answer:
    A P=Q or relationship cannot be established between p and q. Correct Answer Incorrect Answer
    B P>Q Correct Answer Incorrect Answer
    C P Correct Answer Incorrect Answer
    D P>=Q Correct Answer Incorrect Answer
    E P<=Q Correct Answer Incorrect Answer

    Solution

    6p² +17p + 12 = 0 6p² + 9p + 8p + 12 = 0 тЯ╣ 3 p (2p + 3) + 4 (2p + 3) = 0 тЯ╣ (3p + 4) (2p + 3) = 0 тЯ╣ p = - 4/3, - 3/2 II. 12q² - 25q + 7 = 0 тЯ╣ 12 q² - 21q - 4q + 7 =0 тЯ╣ 3q(4q - 7) - 1(4q -7)= 0 тЯ╣ (3q - 1) (4q - 7)= 0 тЯ╣ q =1/3 , 7/4 Hence p < q.

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