If x=2+√3 then
[tex] {x}^{2} + \frac{1}{ {x}^{2} } – 4(x + \frac{1}{x} )[/tex]


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1 thought on “If x=2+√3 then <br />[tex] {x}^{2} + \frac{1}{ {x}^{2} } – 4(x + \frac{1}{x} )[/tex]<br /><br /><br />​”

1. Hey mate!

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   Given :

   x = 2 +[tex] \sqrt{3}[/tex]

   To find :

   [tex]x + \frac{1}{x}[/tex]

   Solution :

   [tex]\begin{gathered}x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 – \sqrt{3} }{2 – \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 – \sqrt{3} }{(2) {}^{2} – ( \sqrt{3} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{2 – \sqrt{3} }{4 – 3} \\ \\ \frac{1}{x} = 2 – \sqrt{3} \end{gathered}[/tex]

   Now,

   [tex]\begin{gathered}x + \frac{1}{x} \\ \\ = 2 + \sqrt{3} + 2 – \sqrt{3} \\ \\ = 2 + 2 \\ \\ = 4\end{gathered}[/tex]

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   Thanks for the question !

   ☺️☺️☺️

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