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    Question

    If (x+1/x)^2 = 3, then the value of ┬аx^206 + x^200 +

    x^90 + x^84 + x^42 + x^36 + x^12 + x^6 + 5 is
    B 1 Correct Answer Incorrect Answer
    C 5 Correct Answer Incorrect Answer
    D 4 Correct Answer Incorrect Answer

    Solution

    (x+1/x)^2= 3 тЖТ x+1/x = тИЪ3 By cubing both sides, (x+1/x)^3= (тИЪ3)^3 x^3+1/x^3 ┬а+ 3 ├Ч x ├Ч 1/x (x+1/x) = 3тИЪ3 x^3+1/x^3 ┬а+ 3 ├Ч тИЪ3├Ч = 3тИЪ3 x^3+1/x^3 ┬а= 3тИЪ3 - 3тИЪ3 x^3+1/x^3 ┬а= 0 (x^6+1)/x^3 ┬а= 0 тЖТ x^6+1 = 0 x^206 + x^200 + x^90 + x^84 + x^42 + x^36 + x^12 + x^6 + 5 x^200 (x^6 ┬а+1) + x^84 (x^6 ┬а+1) + x^36 (x^6 ┬а+1) + x^6 (x^6 ┬а+1) + 5 By putting the value (x^6 ┬а+1=0 )┬а x^200 ├Ч 0 + x^84 ├Ч 0 + x^36 ├Ч 0 + x^6 ├Ч 0 + 5 = 5

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