Find the root of the answer (2)

Relevant wiki: Vieta's Formula Problem Solving - Basic

$\begin{aligned} x^2 - 4x + 13 & = 0 \end{aligned}$

$\begin{aligned} \implies x^2 & = 4x - 13 & \small \color{#3D99F6}{\implies \alpha^2 = 4\alpha - 13 \text{, since } \alpha \text{ is a root.}} \\ \implies x^3 & = 4x^2 - 13x \\ & = 4(4x - 13) - 13x \\ & = 3x - 52 & \small \color{#3D99F6}{\implies \beta^3 = 3\beta - 52} \\ \implies x^4 & = 3x^2 - 52x \\ & = 3(4x - 13) - 52x \\ & = -40x - 39 & \small \color{#3D99F6}{\implies \alpha^4 = -40\alpha - 39} \end{aligned}$

Then, we have:

$\begin{aligned} X & = \alpha^4 + \beta^3 + \alpha^2 - 39\beta + 2264 \\ & = -40\alpha - 39 + 3\beta - 52 + 4\alpha - 13 - 39\beta + 2264 \\ & = - 36 \alpha -36 \beta + 2160 \\ & = -36(\color{#3D99F6}{\alpha+\beta}) + 2160 \quad \quad \small \color{#3D99F6}{\text{By Vieta's formula: }\alpha + \beta = 4} \\ & = -36(\color{#3D99F6}{4}) + 2160 \\ & = -144 + 2160 \\ & = \boxed{2016} \end{aligned}$

$\Rightarrow \alpha^4+\alpha^2+\beta^3-39\beta+2264$

$=\underbrace{\color{#BA33D6}{\alpha^2}}_{\color{#20A900}{4\alpha}-13}(\color{#BA33D6}{\alpha^2}+1)+\underbrace{\color{#3D99F6}{\beta^3}}_{4\beta^2-13\beta}-39\beta+2264$

$=(16-4\beta-13)(16-4\beta-12)+4\beta^2 -52\beta+2264$

$=(3-4\beta)(4-4\beta)+4\beta^2-52\beta+2264$

$=12-12\beta-16\beta-52\beta+16\beta^2+4\beta^2+2264$

$=\color{#EC7300}{20\beta^2-80\beta}+2276$

$\implies 2276-260=\boxed{2016}$

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•Using Vieta's Formula .

$\alpha+\beta=4$
$\implies \color{#20A900}{4\alpha}=16-4\beta$

•Since $\alpha$ is root of $x^2-4x+13=0$ ,

$\implies \color{#BA33D6}{\alpha^2}=4\alpha-13$

Also $\beta$ is root of $x^2-4x+13=0$ ,

$\implies \color{#3D99F6}{\beta^3}=4\beta^2-13\beta$ .

And,

Multiplying by $20$ both sides.

$\implies \color{#EC7300}{20\beta^2-80\beta}+260=0$

We have:
$\alpha^2 - 4\alpha + 13 =0$ ;.................... $(1)$
$\beta^2-4\beta + 13 =0$ ;............................ $(2)$
Now,
$\alpha^2 = 4\alpha - 13 \implies \alpha^4 = 16\alpha^2 + 169 - 104\alpha$
$\beta^2 = 4\beta - 13 \implies \beta^3 = 4\beta^2 - 13\beta$
Therefore,
$\alpha^4 + \beta^3 + \alpha^2 - 39\alpha + 2264 = 17\alpha^2-104\alpha + 4 \beta^2-52\beta+2433$
[Multiplying $(1)$ by 17 and $(2)$ by 4 and then subtracting both from the equation];
= $-36(\alpha+\beta) + 2160$
= $-36*(4)+ 2160$ (Since, $\color{#3D99F6} {\alpha + \beta=4}$ , by Vieta's Formula.)
= $\boxed{2016}$