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  1. #1
    Junior Member mathbuoy's Avatar
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    Arrow find the value of...

    Given that x^2-x-1=0, find x^8 + \frac{1}{x^8}.


    *show working and explanation PLS
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  2. #2
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    Quote Originally Posted by mathbuoy View Post
    Given that x^2-x-1=0, find x^8 + \frac{1}{x^8}.


    *show working and explanation PLS
    Hi
    x^2-x-1 = 0
    x^2 = x+1

    Multiplying par x
    x^3 = x^2+x = x+1+x = 2x+1

    Multiplying par x
    x^4 = 2x^2+x = 2(x+1)+x = 3x+2

    You can go on like this or use the Fibonacci sequence

    If x^n = a_n x+b_n

    Then x^{n+1} = a_n x^2+b_n x = a_n (x+1) + b_n x = (a_n + b_n) x + a_n
    Therefore a_{n+1} = a_n + b_n and b_{n+1} = a_n
    (an) follows the Fibonacci rule

    x^{16} = 987x+610

    x^8 = 21x+13

    x^8 + \frac{1}{x^8} = \frac{987x+611}{21x+13} = 47
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  3. #3
    Moo
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    Hello,

    Note that x \neq 0

    So you can divide x^2-x-1=0 by x on both sides, which will give :

    x-1-\frac 1x=0

    so x-\frac 1x=1


    Now, consider \left(x-\frac 1x\right)^2
    This is equal to x^2+\frac{1}{x^2}-2=1
    Hence x^2+\frac{1}{x^2}=3

    Square it again :
    9=\left(x^2+\frac{1}{x^2}\right)^2=x^4+\frac{1}{x^  4}+2
    Hence x^4+\frac{1}{x^4}=7

    Square it again :
    49=\left(x^4+\frac{1}{x^4}\right)^2=\boxed{x^8+\fr  ac{1}{x^8}}+2

    hence x^8+\frac{1}{x^8}=47
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  4. #4
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    I was pretty sure there was a more simple answer than mine !
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