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

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


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


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

    Multiplying par x
    $\displaystyle x^3 = x^2+x = x+1+x = 2x+1$

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

    You can go on like this or use the Fibonacci sequence

    If $\displaystyle x^n = a_n x+b_n$

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

    $\displaystyle x^{16} = 987x+610$

    $\displaystyle x^8 = 21x+13$

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

    Note that $\displaystyle x \neq 0$

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

    $\displaystyle x-1-\frac 1x=0$

    so $\displaystyle x-\frac 1x=1$


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

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

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

    hence $\displaystyle x^8+\frac{1}{x^8}=47$
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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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