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Solving linear difference equation $f(n)=\frac{1}{6}f(n-1)+\frac{2}{3}f(n+1)$, $0<n<\infty$

with $f(0)=4, f(1)=3$

The text says to solve $f(n)=af(n-1)+bf(n+1)$ to try functions of the form $f(n)=\alpha^n$ and that such functions are solutions iff $\alpha^n=a\alpha^{n-1}+b\alpha^{n+1}$ or $\alpha=a+b\alpha^2$ and then using quadradic formula $\alpha=\frac{1+\sqrt{1-4ab}}{2b}$

So my thought was for this question I need to use $a=\frac{1}{6}$, $b=\frac{2}{3}$ which gives me $\alpha=\frac{1\pm\sqrt{1-4(\frac{1}{6})(\frac{2}{3})}}{2(\frac{2}{3})}$ but the text gives a solution as $\alpha=\frac{3\pm \sqrt{5}}{4}$ which I can't see how this is determined.

The general solution is then given as $f(n)=c_1(\frac{3+\sqrt{5}}{4})^n+c_2(\frac{3-\sqrt{5}}{4})^n$

AColoredReptile
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    Perhaps this will help https://math.stackexchange.com/questions/3292475/what-is-the-closed-form-of-the-f-with-f1-1-f2-7-and-fn-7fn-1-12f/3292559#3292559 – nonuser Oct 24 '21 at 21:19
  • @Aqua Yes that helped thanks. – AColoredReptile Oct 24 '21 at 21:46
  • Note that your solution formula can be compressed to $α=\dfrac{1\pm\sqrt{1-\frac49}}{\frac43}=\dfrac{1\pm\frac{\sqrt5}3}{\frac43}$, which simplifies in the next step to the reference solution. – Lutz Lehmann Oct 25 '21 at 05:49

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