Question

$x = 1 - \sqrt{2 \: } \: \: find \: the \: \: value \: ofx }^{2} \: + \frac{1}{x {}^{2} }$
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Answer 1

★Given:

$x = 1 - \sqrt{2}$

★To Find:

$\rm the\:value\:of \: x^2 +\frac{1}{x^2}$

★Solution :

$x = 1 - \sqrt{2}\\ \\ \\ \frac{1}{x} = \frac{1}{1-\sqrt{2}} \\ \\ \\ \implies (by\: rationalising) \frac{1}{1 -\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}} \\ \\ \\ \bold {We\:know\:that,}\\ \\ \\ {\pink{\boxed{(a+b)(a-b) = a^2-b^2}}} \\ \\ \\ \implies \frac{1+\sqrt{2}}{(1)^2 - (\sqrt{2})^2}\\ \\ \\ \implies \frac{1+\sqrt{2}}{1-2}\\ \\ \\ \implies \frac{1+\sqrt{2}}{-1} \\ \\ \\ \implies -(1+\sqrt{2})\\ \\ \\ \implies -1-\sqrt{2}$

$x +\frac{1}{x} = \\ \\ \\ \implies (1-\sqrt{2}) + (-1- \sqrt{2}) \\ \\ \\ \implies -\sqrt{2}-\sqrt{2} \\ \\ \\ \implies -2\sqrt{2}$

$\bold {We \:know\:that}\\ \\ \\ {\purple{\boxed{(a+b)^2 = a^2+b^2+2ab}}}\\ \\ \\ (x+\frac{1}{x})^2 = x^2+\frac{1}{x^2} + 2(x \times \frac{1}{x}\\ \\ \\ by\: putting \:values \\ \\ \\ (-2\sqrt{2})^2 = x^2+\frac{1}{x^2} + 2\\ \\ \\ 8 = x^2+\frac{1}{x^2} + 2\\ \\ \\ x^2+\frac{1}{x^2} = 8-2 \\ \\ \\ {\blue{\boxed{x^2+\frac{1}{x^2} = 6}}}$

Answer 2

$\red{\color{white}{\fcolorbox{cyan}{black}{Answer:-}}}$

$\boxed{\sf{x = 1 - \sqrt{2} }} \\ \\⇝\sf \frac{1}{ x } = \frac{1}{1 - \sqrt{2} } \\ \\ \sf \: Rationalising \: the \: denominator :- \\\\⇝\sf \frac{1}{x} = \frac{1}{1 - \sqrt{2} } \times \frac{1 + \sqrt{2} }{1 + \sqrt{2} } \\\\ ⇝\sf \frac{1}{x} = \frac{1 + \sqrt{2} }{ {(1)}^{2} - { (\sqrt{2}) }^{2} } \\\\⇝\sf \frac{1}{x} = \frac{1 + \sqrt{2} }{1 - 2} \\\\ ⇝ \sf\frac{1}{x} = \frac{1 + \sqrt{2} }{ - 1} \\\\ \sf{\therefore}~~</p><p> \boxed{\sf{\frac{1}{x} = - 1 -\sqrt{ 2}}}$

$\sf Finding~value~ of ~x + \frac{1}{x}\\\\⇝\sf x + \frac{1}{x} = 1 - \sqrt{2} + ( - 1 - \sqrt{2} )\\\\ ⇝\sf x + \frac{1}{x} =1 - \sqrt{2} - 1 - \sqrt{2} \\\\ \sf {\therefore}~ ~ \boxed{\sf{ x + \frac{1}{x} = - 2 \sqrt{2}}} \\\\\sf Finding~value~ of ~{(x + \frac{1}{x})}^{2}\\\\⇝\sf {(x + \frac{1}{x} ) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \times x \times \frac{1}{x} \\\\⇝\sf {(x + \frac{1}{x} ) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\\\ ⇝ \sf {(- 2 \sqrt{2} ) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\\\⇝\sf 8 = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\\\⇝\sf 8 - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\\\ </p><h3>{\therefore}~~ \boxed{\sf{\red{{x}^{2} + \frac{1}{ {x}^{2} } = 6}}}$