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(B) Given the equation $x^2 - \sqrt{2}x - \sqrt{3} = 0$,the roots $\alpha$ and $\beta$ satisfy:$\alpha^2 - \sqrt{2}\alpha - \sqrt{3} = 0 \Rightarrow \

Vedclass · Accepted Answer

$10\sqrt{3}P_9$

Vedclass · Answer

(B) Given the equation $x^2 - \sqrt{2}x - \sqrt{3} = 0$,the roots $\alpha$ and $\beta$ satisfy:$\alpha^2 - \sqrt{2}\alpha - \sqrt{3} = 0 \Rightarrow \alpha^{n+2} - \sqrt{2}\alpha^{n+1} - \sqrt{3}\alpha^n = 0$$\beta^2 - \sqrt{2}\beta - \sqrt{3} = 0 \Rightarrow \beta^{n+2} - \sqrt{2}\beta^{n+1} - \sqrt{3}\beta^n = 0$Subtracting these equations,we get:$P_{n+2} - \sqrt{2}P_{n+1} - \sqrt{3}P_n = 0 \Rightarrow P_{n+2} = \sqrt{2}P_{n+1} + \sqrt{3}P_n$For $n=10$,$P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}$For $n=9$,$P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9 \Rightarrow \sqrt{3}P_9 = P_{11} - \sqrt{2}P_{10}$Now,consider the expression $E = (11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}$Substitute $P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10}$:$E = 11\sqrt{3}P_{10} - 10\sqrt{2}P_{10} + 11\sqrt{2}P_{11} + 10P_{11} - 11(\sqrt{2}P_{11} + \sqrt{3}P_{10})$$E = 11\sqrt{3}P_{10} - 10\sqrt{2}P_{10} + 11\sqrt{2}P_{11} + 10P_{11} - 11\sqrt{2}P_{11} - 11\sqrt{3}P_{10}$$E = -10\sqrt{2}P_{10} + 10P_{11} = 10(P_{11} - \sqrt{2}P_{10})$Since $P_{11} - \sqrt{2}P_{10} = \sqrt{3}P_9$,we have $E = 10\sqrt{3}P_9$.

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

Let $\alpha, \beta$ with $\alpha > \beta$ be the roots of the equation $x^2 - \sqrt{2}x - \sqrt{3} = 0$. Let $P_n = \alpha^n - \beta^n$ for $n \in \mathbb{N}$. Then $(11\sqrt{3} - 10\sqrt{2})P_{10} + (11\sqrt{2} + 10)P_{11} - 11P_{12}$ is equal to:

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