Question

If
$x = (2 + \sqrt[]{3)}$
find the value of
$(x {}^{2} + \frac{1}{x {}^{2} } )$
​

Answer 1

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$\bold \color{red}{Given,\:\:x = (2 + \sqrt[]{3}) }$

$∴ \: \: \bold \color{blue}x {}^{2} + \frac{1}{x {}^{2} }$

$= \: \: \bold \color{blue}(x) {}^{2} +( \frac{1}{x } )^{2}$

$= \bold \color{blue} (x + \frac{1}{x} )^{2} - 2 \times x \times \frac{1}{x}$

$= \bold \color{blue} (x + \frac{1}{x} )^{2} - 2$

$= \bold \color{blue} (2 + \sqrt{3} + \frac{1}{2 + \sqrt{3} } )^{2} - 2$

$= \bold \color{blue} (2 + \sqrt{3} + \frac{1}{2 + \sqrt{3} } )^{2} - 2$

$= \bold \color{blue} (2 + \sqrt{3} + \frac{(2 - \sqrt{3}) }{(2 + \sqrt{3})(2 - \sqrt{3}) } )^{2} - 2$

$= \bold \color{blue} (2 + \sqrt{3} + \frac{(2 - \sqrt{3}) }{ {2}^{2} - {( \sqrt{3} )}^{2} } )^{2} - 2$

$= \bold \color{blue} (2 + \sqrt{3} + \frac{(2 - \sqrt{3}) }{4 - 3})^{2} - 2$

$= \bold \color{blue} (2 + \sqrt{3} + \frac{(2 - \sqrt{3}) }{1})^{2} - 2$

$= \bold \color{blue} (2 + \sqrt{3} + {2 - \sqrt{3} }{})^{2} - 2$

$= \bold \color{blue} 4^{2} - 2$

$= \bold \color{blue} 16 - 2$

$= \boxed{\bold \color{green}14}$

∴ The value of $(x {}^{2} + \frac{1}{x {}^{2} } )$ is 14 .

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★ Identity used —

$\boxed { \tt{{x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy}}$