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

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

• Find $x$ on base of equality and substitute. Nov 15 '18 at 11:12
• You mean root of that equation , that's irrational dude,send me that lengthy calculation you would make to get that to 10th power Nov 15 '18 at 11:14
• I need mathematical method not calculative Nov 15 '18 at 11:16
• For a light hint, there is an algebraic method here which exploits the symmetry in the coefficients. You should perhaps try to find it. And there are other ways of proceeding too, which will get you there in the end. Now you know there is an algebraic method, maybe you will have the confidence to try to find it? Nov 15 '18 at 11:21
• $x+\frac1x=5$. Square that to find $x^2+\frac1{x^2}$ and so on up to $x^5+\frac1{x^5}$ Nov 15 '18 at 11:24

$$\textbf{Hint:}$$ $$x^2-5x+1=0\Rightarrow x+\frac{1}{x}=5$$ $$x^2 + \frac{1}{x^2}=23$$ 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?
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, $$x^5+\frac{1}{x^5}=2525.$$