Question

$if \: x = 2 + \sqrt{3} \: find \: the \: value \: of \: {x}^{2} + \frac{1}{ {x}^{2} }$

Answer 1

Hey mate!

Here's ur answer dear:-)
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$x = 2 + \sqrt{3} \\ \\ \frac{1}{x } = \frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{(2) {}^{2} - ( \sqrt{3} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{2 - \sqrt{3} }{4 - 3} = 2 - \sqrt{3}$

Now,

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

Squaring both sides..

$(x + \frac{1}{x} ) {}^{2} = (4) {}^{2} \\ \\ x {}^{2} + 2 \times x \times \frac{1}{x} +(\frac{1}{x} ) {}^{2} = 16 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 16 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ \\ x {}^{2} + \frac{1}{x {}^{2} } = 14$

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HOPE IT HELPS!

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Answer 2

Hola !!❤❤

Here is your answer in the given attachment ......

HOPE IT HELPS...✌✌

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