Solve the linear homogeneous recurrence relation with constant coefficients $$9a_{n} = 6a_{n-1}-a_{n-2}, a_{0}=6, a_{1}=5$$
So
$$x^n = (6x^{n-1}-x^{n-2})\div9$$
thus
$$[x^2 = (6x-1)\div9] \equiv [x^2 - \frac{2}{3}x + \frac{1}{9} = 0], x=\frac{1}{3}$$
also
$$a_{2}=\frac{8}{3}, a_{3}=\frac{31}{27}$$
How do I plug in that x/root to solve for the given recurrence relation?
I tried
$$9a_{n} = 6(\frac{1}{3})^{n-1} - (\frac{1}{3})^{n-2}$$
which for n=1 gives 5... which is correct, for a_1 though, not 9a_1? So that's not right.

Here are the steps
$$
a_{0}=6
$$
$$
a_{1}=5
$$
$$
9a_{n} = 6a_{n-1}-a_{n-2}
$$
Lets rewrite this recurrence as
$$
a_{n} = \frac69a_{n-1}-\frac19a_{n-2}
$$
$$
a_{n} = \frac23a_{n-1}-\frac1{3^2}a_{n-2}
$$
Now let's multiply by $3^n$
$$
3^0a_{0}=3^06=6
$$
$$
3^1a_{1}=3^15=15
$$
$$
3^na_{n} = 2\cdot 3^{n-1}a_{n-1}-3^{n-2}a_{n-2}
$$
Let $S_n=3^na_n$, then
$$
S_0=6
$$
$$
S_{1}=15
$$
$$
S_n= 2S_{n-1}-S_{n-2}
$$
It follows that
$$
S_n= S_1n-S_0(n-1)=9n+6=3(3n+2)
$$
Therefore 

$$
a_n=\frac{3(3n+2)}{3^n}=3^{1-n}(3n+2)
$$

