Question

If x=2+√3 then
${x}^{2} + \frac{1}{ {x}^{2} } - 4(x + \frac{1}{x} )$


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Answer 1

Hey mate!

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Given :

x = 2 +$\sqrt{3}$

To find :

$x + \frac{1}{x}$

Solution :

$\begin{gathered}x = 2 + \sqrt{3} \\ \\ \frac{1}{x} = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{(2) {}^{2} - ( \sqrt{3} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} \\ \\ \frac{1}{x} = 2 - \sqrt{3} \end{gathered}$

Now,

$\begin{gathered}x + \frac{1}{x} \\ \\ = 2 + \sqrt{3} + 2 - \sqrt{3} \\ \\ = 2 + 2 \\ \\ = 4\end{gathered}$

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Thanks for the question !

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