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(A) Given $\alpha^2 + \beta^2 = 5$ and $3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)$.Let $S_n = \alpha^n + \beta^n$. We have $S_2 = 5$.Using the id

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(A) Given $\alpha^2 + \beta^2 = 5$ and $3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)$.Let $S_n = \alpha^n + \beta^n$. We have $S_2 = 5$.Using the identity $\alpha^5 + \beta^5 = (\alpha^2 + \beta^2)(\alpha^3 + \beta^3) - \alpha^2 \beta^2(\alpha + \beta)$ is not direct,so we use the recurrence relation for $x^2 - px + q = 0$ where $p = \alpha + \beta$ and $q = \alpha \beta$.Since $\alpha^2 + \beta^2 = 5$,we have $p^2 - 2q = 5$.The relation $3 S_5 = 11 S_3$ can be solved by testing values. If $\alpha \beta = 2$,then $\alpha^2 + \beta^2 = 5$ implies $\alpha, \beta$ are roots of $x^2 - px + 2 = 0$. Since $\alpha^2 + \beta^2 = p^2 - 2(2) = 5$,$p^2 = 9$,so $p = \pm 3$.If $p = 3$,the roots are $1$ and $2$. Checking the condition: $3(1^5 + 2^5) = 3(1 + 32) = 3(33) = 99$. Also $11(1^3 + 2^3) = 11(1 + 8) = 11(9) = 99$.Since $99 = 99$,the value $\alpha \beta = 2$ satisfies the given equations.

$A. \alpha + \beta + \gamma$	$(1) -\frac{43}{4}$
$B. \alpha^2 + \beta^2 + \gamma^2$	$(2) -\frac{7}{8}$
$C. \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$	$(3) 86$
$D. \frac{\alpha}{\beta \gamma} + \frac{\beta}{\gamma \alpha} + \frac{\gamma}{\alpha \beta}$	$(4) 0$
	$(5) 10$

If $\alpha$ and $\beta$ are two real numbers satisfying $\alpha^2 + \beta^2 = 5$ and $3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)$,then the value of $\alpha \beta$ is:

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