Question

$x4 + 1x4 = 57 \\ \\ find \: x + 1 \div x =$
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Answer 1

you can do it by pithagaurousthoram

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

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$\bf\implies (x{}^{4}+\frac{1}{x{}^{4}})=47$

$</u></strong><strong><u>\</u></strong><strong><u>b</u></strong><strong><u>f</u></strong><strong><u>\huge \star$ TO finD

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$</u></strong><strong><u>\</u></strong><strong><u>b</u></strong><strong><u>f</u></strong><strong><u>\huge \star$ Solution:-

$\bf\implies (x{}^{4}+\frac{1}{x{}^{4}})=47\\ \implies [(x{}^{2}){}^{2}+(\frac{1}{x{}^{2}}){}^{2}]=47 \\ \implies [x{}^{2}+\frac{1}{x{}^{2}}]{}^{2}-2\times \cancel{x{}^{2}}\times \frac{1}{\cancel{x{}^{2}}}=47\\ \implies [(x+\frac{1}{x}){}^{2}-2\times \cancel {x}\times\frac{1}{\cancel{x}}]{}^{2}-2=47\\ \implies [(x+\frac{1}{x}){}^{2}-2]{}^{2}=47+2\\</p><p>\implies [(x+\frac{1}{x}){}^{2}-2]=\sqrt{49}\\ \implies [(x+\frac{1}{x}){}^{2}-2]=7\\</p><p>\implies (x+\frac{1}{x}){}^{2}=7+2\\</p><p>\implies (x+\frac{1}{x})=\sqrt{9}\\</p><p>\implies \boxed{\bf\red{x+\frac{1}{x}=3}}$

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$\bf\huge \star$ UseD ForMuLaS:-

$\rightarrow\bf\green{ A{}^{2}+B{}^{2}=(A+B){}^{2}-2A.B}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

$\mathcal{ \&#35;\mathcal{answer with quality }\: \: \&#38; \: \: \&#35;BAL }$