Question

if x=3+2√2
then x+1÷x​

Answer 1

Answer:

${ \bold{ \underline{ Given } : - }} \\ \\ x = 3 + 2 \sqrt{2} \\ \\ { \bold { \underline{ To \: \: find } : - }} \\ \\ x + \frac{1}{x} \\ \\ { \bold{ \underline{ SOLUTION} : - }} \\ \\ \: \: \: . \: \: first \: \: we \: \: have \: \: to \: \: find \: \: \frac{1}{x} \: \: \: - \\ \\ = > \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \\ = > \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ = > \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{ { (3)}^{2} - {( 2\sqrt{2}) }^{2} } \\ \\ = > \frac{1}{x} = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ = > \frac{1}{x} = 3 - 2\sqrt{2} \\ \\ \: \: so \: \: that \: - \\ \\ = > x + \frac{1}{x} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ = > x + \frac{1}{x} = 6 \\ \\ \\ { \bold{ \underline{used \: \: formula} : - }} \\ \\ \: \: (1) \: \: (a + b)(a - b) = {a}^{2} - {b}^{2}$

Answer 2

$\huge {\bold {\pink{Solution }}}$

$<b ><font color ="purple" ><font size ="5" >$

$Given :$

$x = 3 + 2 \sqrt{2 }$

$To\:find :$

$x+ \frac{ 1}{x} = \: ?$

$So,$

$= > x+\frac{ 1}{x }\\ \\ => 3+ 2 \sqrt{2}+\frac{1 }{3 + 2 \sqrt{2} } \\ \\ = > \frac{2 + 2 \sqrt{2} }{3 + 2 \sqrt{ 2 } } \\ \\ = > 3+ 2 \sqrt{2}+ \frac{1 }{3 + 2 \sqrt{2} } ×\frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{ 2 } } \\ \\ => 3+ 2 \sqrt{2}+ 3-2 \sqrt{2} \\ \\ => 6$

$Hence, \:x+\frac{ 1}{x }= 6$

$</b ></font ></font >$

$\huge {\bold {\pink {Hope \:it \:helps}}}$