Question

If
$(x - \frac{1}{x} ) = 12$
find the value of (x2+x21​) and (x4+x41​)

Answer 1

$x - \frac{1}{x} = 12 \\ \\ squaring \: both \: sides \\ \\ (x - \frac{1}{x}) {}^{2} = (12) {}^{2} \\ \\ x {}^{2} + \frac{1}{x {}^{2} } - 2 = 144 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 146 \: \: \: .......ans \\ \\ and \: \\ \\ (x {}^{2} + \frac{1}{x {}^{2} } ) {}^{2} = (146) {}^{2} \\ \\ x {}^{4} + \frac{1}{x {}^{4} } = 2116- 2 \\ \\ x {}^{4} + \frac{1}{x {}^{4} } =2114$

Thanks

Answer 2

$\huge{\mathcal{Hi\: there!}}$

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Given :

$x - \frac{1}{x} = 12 \\ \\ \bold{on \: squaring \: both \: sides : } \\ \\ (x - \frac{1}{x} ) {}^{2} = (12) {}^{2} \\ \\ x{}^{2} + \frac{1}{x {}^{2} } - 2 = 144 \\ \\ x{}^{2} + \frac{1}{x {}^{2} } = 144 + 2 \\ \\ \boxed{ \bold{x{}^{2} + \frac{1}{x {}^{2} } = 146}}$


Now,

$\bold{again \: on \: squaring \: both \: sides : } \\ \\ (x {}^{2} + \frac{1}{x {}^{2} } ) {}^{2} = (146) {}^{2} \\ \\ x {}^{4} + \frac{1}{x {}^{4} } + 2 = 21316 \\ \\ x {}^{4} + \frac{1}{x {}^{4} } = 21316 - 2 \\ \\ \boxed{ \bold{x {}^{4} + \frac{1}{x {}^{4} } = 21314}}$

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Thanks for the question!

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