Question

$if \: x = 7 + 4 \sqrt{3} \: the \: solve \: x + \frac{1}{x } =$
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Answer 1

$\LARGE{\rm{\underline{\underline{Solution:-}}}}$

Given:-

$x = 7 + 4 \sqrt{3}$

To Find:-

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

Value of ${\frac{1}{x}}$ = ${\frac{1}{ 7 + 4 \sqrt{3}}}$

Rationalizing Factor = $7 - 4 \sqrt{3}$

$={\frac{1}{ 7 + 4 \sqrt{3}}}$$\times{\frac{7 - 4 \sqrt{3}}{7 - 4 \sqrt{3}}}$

$={\frac{7 - 4 \sqrt{3}}{(7)^{2} - (\sqrt{48})^{2}}}$

$={\frac{7 - 4 \sqrt{3}}{49 - 48}}$

$={\frac{7 - 4 \sqrt{3}}{1}}$

$= 7 - 4 \sqrt{3}$

Now as we know:-

$x = 7 + 4 \sqrt{3}$

${\frac{1}{x}}$$= 7 - 4 \sqrt{3}$

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

$= 7 + 4 \sqrt{3}$$+ 7 - 4 \sqrt{3}$

$= 7 + 7 + 4\sqrt{3} - 4\sqrt{3}$

$= 14$

Therefore,

$x + {\frac{1}{x}}$$= 14$

Answer 2

$\large \sf \underline{ \underline{ \red{given : }}} \:$

$\tt{x = 7 + 4 \sqrt{3} }$

$\\ \\\large \sf \underline{ \underline{ \red{to \: find : }}} \\ \\ \tt{x + \frac{1}{x} }$

$\\ \\ \large \sf \underline{ \underline{ \red{solution : }}}$

We know ,

• x = 7 + 4√3

We will find 1/x.

$\\ \tt{ \frac{1}{x} = \frac{1}{7 + 4 \sqrt{3} } } \\ \\ \\ \boxed{\sf{rationalising \: factor \: = 7 - 4 \sqrt{3} }} \\ \\ \\ \tt{ = > \frac{1}{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } } \\ \\ \\ \tt{ = > \frac{7 - 4 \sqrt{3} }{ {7}^{2} - ( {4 \sqrt{3} )}^{2} } } \\ \\ \\ \tt{ = > \frac{7 - 4 \sqrt{3} }{49 - 48} } \\ \\ \\ \tt{ \implies \frac{1}{x} = 7 - 4 \sqrt{3} }$

So,

$\tt{x + \frac{1}{x} = 7 + \cancel{4 \sqrt{3} } + 7 - \cancel{ 4 \sqrt{3} }} \\ \\ \\ \ \boxed{\tt{ \orange{x + \frac{1}{x} = 14 }}}$