Multiple Choice Question (MCQ)

If a^2−4a−1=0,a≠0, then the value of a^2+3a+\dfrac{1}{a^2}-\dfrac{3}{a} is

  1. ×

    24

  2. ×

    26

  3. ×

    28

  4. 30

Collected in the board: Nike function

Steven Zheng posted 3 hours ago

Answer

  1. \because a^2−4a−1=0

    \therefore a -\dfrac{1}{a} =4
    (1)

    Square both side, then

    a^2+\dfrac{1}{a^2} = 18
    (2)

    a^2+3a+\dfrac{1}{a^2}-\dfrac{3}{a}

    =a^2+\dfrac{1}{a^2}+3(a -\dfrac{1}{a})

    =18+3\cdot 4 =30

    therefore

    D is the right choice

Steven Zheng posted 3 hours ago

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