Question

x=2+ root 3 find x square + 1/x square...?​

Answer 1

Answer:

Here is your solution

Given :-

x=2+√3

Now

\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} \\ \\ \\

x + \frac{1}{x} = 2 - \sqrt{3} + 2 + \sqrt{3} \\ x + \frac{1}{x} = 4 \\ Both \: sides \: squaring. \: \\ (x + \frac{1}{x} ) {}^{2} = 4 {}^{2} \\ x {}^{2} + \frac{1}{x {}^{2} } + 2 = 16 \\ x {}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ x {}^{2} + \frac{1}{x {}^{2} } = 14

Answer 2

Answer:

$x = 2 + \sqrt{3}$

${x}^{2} + \frac{1}{ {x}^{2} } =$

$\frac{2 + \sqrt{3} }{1} + \frac{1}{2 + \sqrt{3} }$

$\boxed{\green{rationalise \: the \: denominator: - }}$

$\frac{1}{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$= \frac{2 - \sqrt{3} }{ {2}^{2} - { \sqrt{3} }^{2} }$

$= \frac{2 - \sqrt{3} }{4 - 3}$

$= 2 - \sqrt{3}$

$\huge\red{ - - - - - - --- - }$

$2 + \sqrt{3} + 2 - \sqrt{3}$

$= > 2 + 2$

$\boxed{ = = > 4}$

$\huge\red{ - - - - - - --- - }$

$\huge \underline { \underline \blue{note : - }}$

$\boxed{ \purple{ {a}^{2} + {b}^{2} = (a + b)(a - b)}}$

$if \: we \: have \: to \: add \: two \: rational \\ number \: then \: we \: should \\ rationalise \: the \: denominator.$

$\huge\red{ - - - - - - --- }$

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