Here is the official theorem I'll use: 
Since the Fibonacci sequence is defined as $F_n=F_{n-1}+F_{n-2}$, we solve the equation $x^2-x-1=0$ to find that $r_1 = \frac{1+\sqrt 5}{2}$ and $r_2 = \frac{1-\sqrt 5}{2}$
So we have $F_n = c_1\left(\frac{1+\sqrt 5}{2}\right)^n + c_2\left(\frac{1-\sqrt 5}{2}\right)^n$
We know that $F_0 = F_1 = 1$. So we can solve the following system to find the values of $c_1$ and $c_2$:
$1 = c_1 + c_2$
$1 = c_1\left(\frac{1+\sqrt 5}{2}\right) + c_2\left(\frac{1-\sqrt 5}{2}\right)$
Solving this system does not give $c_1 = 1/\sqrt5, c_2 = -1/\sqrt 5$ , even though that is apparently the right answer, i.e. the closed form of the Fibonacci sequence is apparently $$\frac1{\sqrt 5}\left(\frac{1+\sqrt 5}{2}\right) -\frac1{\sqrt 5}\left(\frac{1-\sqrt 5}{2}\right)$$
Where did I go wrong? Why doesn't solving the system of equations give me $c_1 = 1/\sqrt5, c_2 = -1/\sqrt 5$?
Let's see...
$$f_n = \left\{ \begin{array}{ll} 0 & \text{ for } n = 0 \\ 1 & \text{ for } n = 1 \\ f_{n-1} + f_{n-2} & \text{ for } n>1 \end{array} \right.$$
Now, the recursion can be written as $$f_n - f_{n-1} - f_{n-2} = 0,$$ so characteristic equation is $$x^2-x-1=0.$$ Now, the roots of the equation are $$X_{1,2} = \frac{1 \pm \sqrt{5}}2,$$ so general solution is $$f_n = C_1\cdot\left(\frac{1 + \sqrt{5}}2\right)^n + C_2\left(\frac{1 - \sqrt{5}}2\right)^n$$ From the $f_1$ and $f_2$ we get \begin{eqnarray} 0 &=& C_1 + C_2 \\ 1 &=& C_1\left(\frac{1 + \sqrt{5}}2\right) + C_2\left(\frac{1 - \sqrt{5}}2\right) \end{eqnarray} From the first equation we get $$C_2 = -C_1,$$ so \begin{equation} 1 = C_1\left(\frac{1 + \sqrt{5}}2\right) -C_1\left(\frac{1 - \sqrt{5}}2\right) \end{equation} Now, we have $$C_1\left[\frac{1 + \sqrt{5}}2 - \frac{1 - \sqrt{5}}2\right] = 1$$ or $$C_1\cdot\sqrt{5} =1$$ So, $$C_1 = \frac{1}{\sqrt{5}}.$$ Now, $$C_2 = -\frac{1}{\sqrt{5}}.$$ The particular solution for the equation is therefore $$f_n = \frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}2\right)^n - \left(\frac{1-\sqrt{5}}2\right)^n\right]$$