Question

$x = 2 \sqrt{3} then \: find \: x { } ^{2} + 1 \div x {?}^{2}$
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Answer 1

Answer: $\frac{145}{12}$

Step-by-step explanation:

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

⇒$(2\sqrt{3}) ^{2} +\frac{1}{(2\sqrt{3}) ^{2}}$

⇒$4*3+\frac{1}{4*3}$

⇒$12+\frac{1}{12}$

⇒$\frac{144+1}{12}$

⇒$\frac{145}{12}$

∴ $x^{2} +\frac{1}{x^{2} }$ = $\frac{145}{12}$

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Answer 2

Answer:

=> 145/12

Step-by-step explanation:

$= > \bf x = 2 \sqrt{3} \\ \\ = > \bf \frac{1}{x} = \frac{1}{2 \sqrt{3} } = \frac{ \sqrt{3} }{2 \sqrt{3} \times \sqrt{3} } = \frac{ \sqrt{3} }{6} \\ \\ = > \bf \: (x + \frac{1}{x} ) = 2 \sqrt{3} + { \frac{ \sqrt{3} }{6} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \: \: \: \: \: \: \: \: = \frac{12 \sqrt{3} + \sqrt{3} }{6} = \frac{13 \sqrt{3} }{6} \\ \\ \underline{ \bf \: \: formula \: \: to \: \: be \: \: used} \\ \\ = > \bf \: {x}^{2} + {y}^{2} = \it \: (x + y) {}^{2} - 2xy \\ \\ \bf \: thus \: \: \: similarly... \\ \\ \bf = > \: {x}^{2} + \frac{1}{ {x}^{2} } = (x + \frac{1}{x} ) {}^{2} - 2 \times x \times \frac{1}{x} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf = ( \frac{13 \sqrt{3} }{6} ) {}^{2} - 2 = \frac{13}{6} \times \frac{13}{6} \times 3 - 2 \\ \\ \bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{169}{12} - 2 = \frac{169 - 24}{12} = \frac{145}{12} = 12 \frac{1}{12}$