Question

plz solve this also no spam ❌​

Answer 1

Answer:

(x-1/x)^2=x^2+1/x^2-2

x-1/x=4 as 16=4^2

And x^2+1/x^2=18

hope it helps

Answer 2

GIVEN:

$\sf \mapsto {{x}^{2} + \dfrac{1}{ {x}^{2} } = 18}$

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TO FIND: ‎ ‎ ‎

$\sf{ \mapsto x - \dfrac{1}{x} }$ ‎ ‎ ‎

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SOLUTION:

${ \text{ we \: have \: to \: find} \: \sf{x - \frac{1}{x} }}$

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Squaring this term will give:

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

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$\sf{{x}^{2} + \dfrac{1}{{x }^{2} } } \: is \: given \: to \: be \: 18.$

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$\sf{{ \{ x - \frac{1}{x} \} }^{2} } = 18 - 2 \cancel{(x} ) \times ( \frac{1}{ \cancel{x} })$

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$\sf{{ \{ x - \frac{1}{x} \} }^{2} } = 18 - 2$

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$\sf{{ \{ x - \frac{1}{x} \} }^{2} } = 16$

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$\sf{{ \{ x - \frac{1}{x} \} }} = \sqrt{16}$

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$\sf{{ \{ x - \frac{1}{x} \} }} = 4$

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$\sf{{ Hence \: value \: of \: ( x - \frac{1}{x} )}} = 4.$