Let , ; , be the roots of the equation x^2-√2x-√3=0.

Question

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 $(11\sqrt{3}-10\sqrt{2}){P}_{10}+(11\sqrt{2}+10){P}_{11}-11{P}_{12}$ is equal to

Accepted Answer

$\begin{gathered} {x}^{2}-\sqrt{2}x-\sqrt{3}=0 \ {P}_{n}={\alpha }^{n}-{\beta }^{n} \end{gathered}$

$\alpha $ and $\beta $ are the roots of the equation

Using Newton's theorem

$\begin{gathered} {P}_{n+2}-\sqrt{2}{P}_{n+1}-\sqrt{3}{P}_{n}=0 \ n=10 \ {P}_{12}-\sqrt{2}{P}_{11}-\sqrt{3}{P}_{10}=0 \end{gathered}$

${P}_{12}=\sqrt{2}{P}_{11}+\sqrt{3}{P}_{10}$

Put $n=9$

$\begin{gathered} {P}_{11}-\sqrt{2}{P}_{10}-\sqrt{3}{P}_{9}=0 \ {P}_{11}=\sqrt{2}{P}_{10}+\sqrt{3}{P}_{9} \ (11\sqrt{3}-10\sqrt{2}){P}_{10}+(11\sqrt{2}+10){P}_{11}-11{P}_{12} \end{gathered}$

Put the value of ${P}_{12}$ \& ${P}_{12}$ in above equation.

$\begin{gathered} = (11\sqrt{3}-10\sqrt{2}){P}_{10}+(11\sqrt{2}+10)(\sqrt{2}{P}_{10}+\sqrt{3}{P}_{9})-11(\sqrt{2}{P}_{11}+\sqrt{3}{P}_{10}) \ = 11\sqrt{3}{P}_{10}-10\sqrt{2}{P}_{10}+22{P}_{10}+10\sqrt{2}{P}_{10}+11\sqrt{6}{P}_{9}+10\sqrt{3}{P}_{9}-11\sqrt{2}{P}_{11}-11\sqrt{3}{P}_{10} \ = 22{P}_{10}+11\sqrt{6}{P}_{9}+10\sqrt{3}{P}_{9}-11\sqrt{2}(\sqrt{2}{P}_{10}+\sqrt{3}{P}_{9}) \ = 22{P}_{10}+11\sqrt{6}{P}_{9}+10\sqrt{3}{P}_{9}-22{P}_{10}-11\sqrt{6}{P}_{9} \ = 10\sqrt{3}{P}_{9} \end{gathered}$

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