Condition for a Given Number to Be a Root

Since

$$\alpha,\beta$$ are roots of $$x^2-6x+4=0,$$ they satisfy $$\alpha^2=6\alpha-4$$ and $$\beta^2=6\beta-4.$$

Multiplying by $$\alpha^{n-2}$$ and $$\beta^{n-2}$$ respectively,

$$\alpha^n=6\alpha^{n-1}-4\alpha^{n-2}$$

$$\beta^n=6\beta^{n-1}-4\beta^{n-2}$$

Subtracting,

$$a_n=6a_{n-1}-4a_{n-2}$$

Therefore,

$$a_{10}=6a_9-4a_8$$

Hence,

$$a_{10}+4a_8=6a_9$$

So,

$$\frac{a_{10}+4a_8}{2a_9}=\frac{6a_9}{2a_9}=3$$

Thus, $$3$$ is a root of $$x^2-kx+12=0.$$

Substituting

$$x=3,$$

$$3^2-3k+12=0$$

$$9-3k+12=0$$

$$21=3k$$

$$k=7$$