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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 $S_2 = 5$ and $3S_5 =

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$2$

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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 $S_2 = 5$ and $3S_5 = 11S_3$.Using the identity $\alpha^5 + \beta^5 = (\alpha^2 + \beta^2)(\alpha^3 + \beta^3) - \alpha^2 \beta^2(\alpha + \beta)$,this is not direct. Instead,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 $p^2 - 2q = 5$.Also,$S_n = p S_{n-1} - q S_{n-2}$.For $n=3$: $S_3 = p S_2 - q S_1 = 5p - qp = p(5-q)$.For $n=5$: $S_5 = p S_4 - q S_3$.$S_4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2 = 25 - 2q^2$.So,$S_5 = p(25 - 2q^2) - q(p(5-q)) = p(25 - 2q^2 - 5q + q^2) = p(25 - 5q - q^2)$.Substitute into $3S_5 = 11S_3$:$3p(25 - 5q - q^2) = 11p(5 - q)$.Assuming $p 
eq 0$,$3(25 - 5q - q^2) = 11(5 - q) \Rightarrow 75 - 15q - 3q^2 = 55 - 11q$.$3q^2 + 4q - 20 = 0 \Rightarrow (3q + 10)(q - 2) = 0$.Since $\alpha, \beta$ are real,$p^2 = 5 + 2q \ge 0$. If $q=2$,$p^2 = 9$ (real). If $q = -10/3$,$p^2 = 5 - 20/3 = -5/3$ (not real).Thus,$\alpha \beta = 2$.

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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