Let alpha, beta ; alphabeta, be the roots | vedclass

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Question

$ x^2-\sqrt{2 x}-\sqrt{3}=0\left\langle_\beta^\alphaight. $

$ \alpha^{n+2}-\sqrt{2} \alpha^{n+1}-\sqrt{3} \alpha^n=0 $

$ 	ext { and } \beta^{

Vedclass · Accepted Answer

$10 \sqrt{3} \mathrm{P}_9$

Vedclass · Answer

$ x^2-\sqrt{2 x}-\sqrt{3}=0\left\langle_\beta^\alphaight. $

$ \alpha^{n+2}-\sqrt{2} \alpha^{n+1}-\sqrt{3} \alpha^n=0 $

$ 	ext { and } \beta^{n+2}-\sqrt{2} \beta^{n+1}-\sqrt{3} \beta^n=0$

Subtracting

$ \left(\alpha^{n+2}-\beta^{n+2}ight)-\sqrt{2}\left(\alpha^{n+1}-\beta^{n+1}ight)-\sqrt{3}\left(\alpha^n-\beta^night)=0 $

$ \Rightarrow P_{n+2}-\sqrt{2} P_{n+1}-\sqrt{3} P_n=0$

Put $\mathrm{n}=10$

$ \mathrm{P}_{12}-\sqrt{2} \mathrm{P}_{11}-\sqrt{3} \mathrm{P}_{10}=0 $

$ \mathrm{n}=9 $

$ \mathrm{P}_{11}-\sqrt{2} \mathrm{P}_{10}-\sqrt{3} \mathrm{P}_9=0 $

$ 11\left(\sqrt{3} \cdot \mathrm{P}_{10}+\sqrt{2} \mathrm{P}_{11}-\mathrm{P}_{11}ight)-10\left(\sqrt{2} \mathrm{P}_{10}-\mathrm{P}_{11}ight) $

$ =0-10\left(-\sqrt{3} \mathrm{P}_9ight)=10 \sqrt{3} \mathrm{P}_9$

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

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