Given quadratic equation is
$$
x^2-2 x+4=0
$$
whose roots are $\alpha$ and $\beta$.
$$
\therefore \quad \alpha+\beta=2 \text { and } \alpha \beta=4
$$
Now,
$$
\begin{aligned}
& \alpha^9+\beta^9=\left(\alpha^3\right)^3+\left(\beta^3\right)^3 \\
& =\left(\alpha^3+\beta^3\right)\left(\alpha^6+\beta^6-\alpha^3 \beta^3\right) \\
= & (\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2\right)\left[\left(\alpha^2\right)^3+\left(\beta^2\right)^3-\alpha^3 \beta^3\right] \\
= & (\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right] \\
& \quad\left[\left(\alpha^2+\beta^2\right)\left(\alpha^4+\beta^4-\alpha^2 \beta^2\right)-\alpha^3 \beta^3\right] \\
= & (\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]\left[\left\{(\alpha+\beta)^2-2 \alpha \beta\right\}\right. \\
& \left.\left\{\left(\alpha^2+\beta^2\right)^2-3 \alpha^2 \beta^2\right\}-\alpha^3 \beta^3\right] \\
= & (\alpha+\beta)\left[(\alpha+\beta)^2-3 \alpha \beta\right]\left[\left\{(\alpha+\beta)^2-2 \alpha \beta\right\}\right. \\
& \left.\quad\left[\left\{(\alpha+\beta)^2-(2 \alpha \beta)\right]^2-3 \alpha^2 \beta^2\right\}-\alpha^3 \beta^3\right] \\
= & 2[4-12]\left[\{4-8\}\left\{(4-8)^2-48\right\}-64\right]
\end{aligned}
$$
[from Eq. (i)]
$$
\begin{aligned}
& =2(-8)\{(-4)(-32)(-64)\} \\
& =2(-8)(128-64) \\
& =2(-8)(64)=-2^{10}
\end{aligned}
$$