Question

$x = 9 - 4 \sqrt{5} find x {?}^{2} + 1 \div x {?}^{2}$

Answer 1

Hey friend,
Here is the answer you were looking for:
$x = 9 - 4 \sqrt{5} \\ \\ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5} } \times \frac{9 + 4 \sqrt{5} }{9 + 4 \sqrt{5} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{ {(9)}^{2} - {(4 \sqrt{5}) }^{2} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{81 - 80} \\ \\ \frac{1}{x} = 9 + 4 \sqrt{5} \\ \\ x + \frac{1}{x} = (9 - 4 \sqrt{5} ) + (9 + 4 \sqrt{5} ) \\ \\ = 9 - 4 \sqrt{5} + 9 + 4 \sqrt{5} \\ \\ = 18 \\ \\ {x}^{2} + {( \frac{1}{x} )}^{2} = {(x + \frac{1}{x}) }^{2} + 2 \times x \times \frac{1}{x} \\ \\ = {(x + \frac{1}{x} )}^{2} + 2 \\ \\ = {(18)}^{2} + 2 \\ \\ = 324 + 2 \\ \\ = 326 \\ \\ {x}^{2} + {( \frac{1}{x} )}^{2} = 326$

Hope this helps!!!

@Mahak24

Thanks...
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