affanafaqbaig
affanafaqbaig affanafaqbaig
  • 16-12-2021
  • Mathematics
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thi sis the question

thi sis the question class=

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loneguy
loneguy loneguy
  • 16-12-2021

[tex]\text{Given that,}\\\\x^2 + \dfrac1{x^2} = 38\\\\\\(i)\\\\\left(x-\dfrac 1x \right)^2 = x^2 + \dfrac 1{x^2} - 2 \cdot x \cdot \dfrac 1x\\\\\\\implies \left(x-\dfrac 1x \right)^2 = 38 -2 = 36\\\\\implies x - \dfrac 1x = \pm\sqrt{36} = \pm 6 \\\\\\(ii)\\\\x^2 + \dfrac 1{x^2} = 38\\\\\\\implies \left(x^2 + \dfrac 1{x^2} \right)^2 = 38^2\\\\\\\implies x^4 + \dfrac 1{x^4} + 2\cdot x^2 \cdot \dfrac 1{x^2} = 1444\\\\\\\implies x^4 + \dfrac 1{x^4} + 2 = 1444\\\\\\[/tex]

[tex]\implies x^4 +\dfrac 1{x^4} = 1444-2 = 1442[/tex]

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