Question

If
${x}^{2} + \frac{1}{ {x}^{2} } = 18 \:$
find the values of
$x + \frac{1}{x} \: \: and \: \: x - \frac{1}{x}$



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Answer 1

Solution :-

x² + 1/x² = 18

Adding 2 on both sides

⇒ x² + 1/x² + 2 = 18 + 2

⇒ x² + (1/x)² + 2(x)(1/) = 20

⇒ (x + 1/x)² = 20

⇒ x + 1/x = √20 = ± 2√5

x² + 1/x² = 18

Subtracting 2 on both sides

⇒ x² + 1/x² - 2 = 18 - 2

⇒ x² + (1/x)² - 2(x)(1/x) = 16

⇒ (x - 1/x)² = 16

⇒ x - 1/x = √16 = ± 4

Answer 2

Answer:

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