Question

$x - \frac{1}{x} = 5 \\ x {}^{2} - \frac{1}{x { }^{2} } = what$
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Answer 1

$\huge{\mathfrak{\green{Ans}}} {\mathfrak{wer \: :}}$

★ $\Large{\underline{\sf{Given \: }}} \sf{:}$

$\sf{x - \frac{1}{x} = 5}$

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★ $\Large{\underline{\sf{To \: Find }}} \sf{:}$

We have to find the value of

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

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★ $\Large{\underline{\sf{Solution \: }}} \sf{:}$

$\sf{x - \frac{1}{x} = 5}$ ....... ①

(Squaring Both sides) in eq ①

$\sf{(x - \frac{1}{x})^2 = (5)^2}$

Using Identity :

$\Large{\star{\boxed{\sf{(a - b)^2 = a^2 + b^2 - 2ab}}}}$

____________[Put Values]

$\sf{→x^2 + (\frac{1}{x})^2 - 2(x)(\frac{1}{x}) = 25} \\ \\ \sf{→x^2 + \frac{1}{x^2} - 2(\cancel{x})(\frac{1}{\cancel{x}}) = 25} \\ \\ \sf{→x^2 + \frac{1}{x^2} - 2 = 25} \\ \\ \sf{→x^2 + \frac{1}{x^2} = 25 + 2} \\ \\ \sf{→x^2 + \frac{1}{x^2} = 27}$

$\Large{\star{\boxed{\sf{x^2 + \frac{1}{x^2} = 27}}}}$