Question

if x is equal to 3 minus root 5 find the value of x minus one upon X whole square​

Answer 1

$\large{ \bold{ \red{ \underline{ \underline{Question...}}}}}$

$\large{ \sf{ \: if \: x = 3 - \sqrt{5} \: find \: (x - \frac{1}{x} ) {}^{2} }}$

$\large{ \bold{ \green{ \underline{ \underline{Answer..}}}}}$

$\large{ \sf{ = \frac{103 - 45 \sqrt{5} }{8} }}$

$\large{ \red{ \bold{ \underline{ \underline{Solution...}}}}}$

$\sf{ \blue{ \to \: given}} \\ \sf{ x = 3 - \sqrt{5}} \\ \sf{ \blue{ \to \: to \: find}} \\ \sf{(x - \frac{1}{x} ) {}^{2} }$

$\large{ \sf{ \to \: x = 3 - \sqrt{5} }} \\ \\ \large{ \sf{ = > \frac{1}{x} = \frac{1}{3 - \sqrt{5} } }} \\ \\ \large{ \sf{ = \frac{ 3 + \sqrt{5} }{(3 - \sqrt{5})(3 + \sqrt{5} ) }}} \\ \\ \large{ \sf{ = \frac{3 + \sqrt{5} }{ {3}^{2} - {( \sqrt{5} )}^{2} } }} \\ \\ \large{ \sf{ = \frac{3 + \sqrt{5} }{9 - 5} }} \\ \\ \large{ \sf{ \to \: \frac{1}{x} = \frac{3 + \sqrt{5} }{4} }}(rationalised \: denominator) \\ \\ \red{ \sf{ \large{then \: (x - \frac{1}{x} ) {}^{2} }}} \\ \\ \large{ \sf{ = (3 - \sqrt{5} - ( \frac{3 + \sqrt{5} }{4} ) ){}^{2} }} \\ \\ \large{ \sf{ =( \frac{4(3 - \sqrt{5}) - 3 - \sqrt{5} }{4}) {}^{2} }} \\ \\ \large{ \sf{ = ( \frac{12 - 4 \sqrt{5} - 3 - \sqrt{5} }{4}) {}^{2} }} \\ \\ \large{ \sf{ = (\frac{9 - 5\sqrt{5} }{4}) {}^{2} }} \\ \\ \large{ \sf{ = \frac{(9 - 5 \sqrt{5}) {}^{2} }{16} }} \\ \\ \large{ \sf{ = \frac{81 + 125 - 90 \sqrt{5} }{16} }} \\ \\ \large{ \sf{ = \frac{206 - 90 \sqrt{5} }{16} }}(take \: 2 \: common \: in \: deno \: and \: num) \\ \\ \large{ \sf{ \boxed{ \green{ = \frac{103 - 45 \sqrt{5} }{8} }}}}$

$\large{ \sf{ \orange{ required \: answer \: is \: \frac{103 - 45 \sqrt{5} }{8} }}}$