Question

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Answer 1

If
${x}^{2} + \frac{1}{ {x}^{2} } = 27$


Steps:a)
$( {x}^{2} + \frac{1}{ {x}^{2} } ) = 27 \\ = > {(x + \frac{1}{x}) }^{2} - 2 = 27 \\ = > ( {x + \frac{1}{x} })^{2} = 29 \\ = > (x + \frac{1}{x} ) = + \sqrt{29} \: or \: - \sqrt{29}$


b)
$( {x - \frac{1}{x} })^{2} + 2 = 27 \\ = > ( {x - \frac{1}{x} })^{2} = 25 \\ = > (x - \frac{1}{x} ) = + 5 \: or \: - 5$

Answer 2

$\sf{GIVEN:-}\\\sf{x^2 + \frac{1}{x^2} = 27}$-------------( 1 )

$\Large{\sf{NOW, \:\: we \:\:know}}$

a). $\sf{(x + 1/x)^2 = x^2 + 1/x^2 + 2x * 1/x }\\\\\sf{(x^2 + 1/x^2) + 2 }\\\\\sf{27 + 2}$
$\sf{\to\to\to\to\to\to\to\therefore\boxed{x^2 + 1/x^2 = 27}}$

$\sf{(x + 1/x)^2 = 29}\\\\\sf{x + 1/x = ±\sqrt{29}}$

$\Large{\sf{ \:\: we \:\:know}}$

b). $\sf{(x - 1/x)^2 = x^2 + 1/x^2 - 2x * 1/x }\\\\\sf{(x^2 + 1/x^2) - 2 }\\\sf{=27 - 2}$
$\sf{\to\to\to\to\to\to\to\therefore\boxed{x^2 + 1/x^2 = 27}}$

$\sf{(x - 1/x)^2 = 25}\\\\\sf{x - 1/x = ±5}$