If $x^2-5x+1=0$ what is value of $\frac{x^{10}+1}{x^5}$

Question

If $x^2-5x+1=0$ what is value of $\frac{x^{10}+1}{x^5}$.

I tried with calculator but I don't think that,that was proper method ,if you got one pls post it

Answer 1

$\textbf{Hint:}$ $$x^2-5x+1=0\Rightarrow x+\frac{1}{x}=5$$ $$ x^2 + \frac{1}{x^2}=23$$ You can multiply $$ \left(x+\frac{1}{x}\right) \left(x^2 + \frac{1}{x^2}\right)$$ for $x^3 + \frac{1}{x^3}$.

Can you obtain $x^5+\frac{1}{x^5}$ progressing?

Answer 2

You can also use recursion. Let $a_n=x^n+\frac1{x^n}$. Then $a_0=2$ and $a_1=5$. We have $$a_{n+2}-5a_{n+1}+a_n=0.$$ So $a_2=5a_1-a_0=23$, $a_3=5a_2-a_1=110$, $a_4=5a_3-a_2=527$, and $a_5=5a_4-a_3=2525$. So, $$x^5+\frac{1}{x^5}=2525.$$