find the value of...

**mathbuoy** · December 6th 2008, 11:46 PM

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

*show working and explanation PLS

**running-gag** · December 7th 2008, 02:07 AM

Originally Posted by mathbuoy

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

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

$x^8 = 21x+13$

$x^8 + \frac{1}{x^8} = \frac{987x+611}{21x+13} = 47$

**Moo** · December 7th 2008, 02:18 AM

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

**running-gag** · December 7th 2008, 08:19 AM

I was pretty sure there was a more simple answer than mine !

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