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    Question

    Solve both equations I & II and form a new equation III

    in variable ‘r’ (reduce to lowest possible factor) using roots of equation I and II as per instructions given below. I. 4 - (22/p) + (18/P2) = 0 II. (q-4)2 = (16/9) If roots of equation III are highest root of equation I and lowest root of equation II, then which of the following will be equation III. (a) 6r2- 68r + 72 = 0 (b) 6r2- 20r + 36 = 0 (c) 6r2- 33r + 62 = 0 (d) 6r2- 43r + 72= 0 (e) None of these
    A a Correct Answer Incorrect Answer
    B b Correct Answer Incorrect Answer
    C c Correct Answer Incorrect Answer
    D d Correct Answer Incorrect Answer
    E e Correct Answer Incorrect Answer

    Solution

    ATQ, From I. 4- (22/p)+(18/p)2 = 0   4p2-22p+18=0  4p2-4p-18p+18=0  4p(p-1)-18(p-1)=0  (4p-18)  (p-1)=0  p=1,9/2 From II.  (q-4)2=16/9  q-4 = ± 4/3  q=16/3, 8/3 New eq. = r^2-ar+b=0 a = sum of root, b= product of root So, highest root of equation I and lowest root of equation II is (9/2) and (8/3) respectively. Required equation = r2- (43/6)r + 12 = 0   6r2- 43r + 72 = 0 

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