Question

Find: please help me!!!

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\bf{Given:-}\begin{cases}\sf{x=3+\sqrt{8}}\end{cases}$

• We have to find the value of

$\sf\implies x{}^{2}+\frac{1}{x{}^{2}}$

$\bf\underline{\underline{According\:To\: Question:-}}$

• First find the value of $\sf \frac{1}{x}$

$\sf\implies x=3+\sqrt{8}\\ \\ \sf\implies \frac{1}{x}=\frac{1}{3+\sqrt{8}}\times (3-\sqrt{8}){3-\sqrt{8}}\\ \\ \sf\implies \frac{1}{x}=\frac{3-\sqrt{8}}{(3){}^{2}-(\sqrt{8}){}^{2}}\\ \\ \sf\implies \frac{1}{x}=\frac{3-\sqrt{8}}{9-8}\\ \\ \sf\implies \frac{1}{x}=\frac{3-\sqrt{8}}{1}\\ \\ \sf\implies \frac{1}{x}=3-\sqrt{8}$

$\bf The\:value\:of\:\frac{1}{x}\\ \\ \sf\longmapsto 3-\sqrt{8}$

$\tt\implies Value\:of\: x+\frac{1}{x}\\ \\ \sf\implies 3+\cancel{\sqrt{8}}+3-\cancel{\sqrt{8}}\\ \\ \sf\implies x+\frac{1}{x}=6$

$\bf\implies [x+\frac{1}{x}]{}^{2}=x{}^{2}+\frac{1}{x{}^{2}}+2\times \cancel{x}\times \frac{1}{\cancel{x}}\\ \\ \bf\implies [x+\frac{1}{x}]{}^{2}=x{}^{2}+\frac{1}{x{}^{2}}+2$

$\sf\implies [x+\frac{1}{x}]{}^{2}=(6){}^{2}\\ \\ \sf\implies x{}^{2}+\frac{1}{x{}^{2}}+2=36\\ \\ \sf\implies x{}^{2}+\frac{1}{x{}^{2}}=36-2\\ \\ \sf\implies x{}^{2}+\frac{1}{x{}^{2}}=34$

$\boxed{\boxed{\red{\sf{x{}^{2}+\frac{1}{x{}^{2}}=34}}}}$

#BAL

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