Question

If x =2 root of 3 + 2 root of 2 ;find : 1/x,x+1/x,(x+1/x)^2

Answer 1

Answer:

$\large\bold\red{\frac{ \sqrt{3} - \sqrt{2} }{2}}\\\\\large\bold\red{\frac{5 \sqrt{3} + 3 \sqrt{2} }{2}}\\\\\large\bold\red{ \frac{93 + 30 \sqrt{6} }{4}}$

Step-by-step explanation:

Given,

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

Therefore,

We get,

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

Now,

$\frac{1}{x} = \frac{1}{2( \sqrt{3} + \sqrt{2} )} \\ \\ = > \frac{1}{x} = \frac{1}{2} \times \frac{1}{( \sqrt{3} + \sqrt{2} )} \times \frac{( \sqrt{3} - \sqrt{2}) }{( \sqrt{3} - \sqrt{2}) } \\ \\ = > \frac{1}{x} = \frac{( \sqrt{3} - \sqrt{2} )}{2( { (\sqrt{3}) }^{2} - { (\sqrt{2} )}^{2}) } \\ \\ = > \frac{1}{x} = \frac{( \sqrt{3} - \sqrt{2} ) }{2(3 - 2)} \\ \\ = > \large \bold{\frac{1}{x} = \frac{ \sqrt{3} - \sqrt{2} }{2} }$

Now,

$x + \frac{1}{x} \\ \\ = 2( \sqrt{3} + \sqrt{2} ) + \frac{( \sqrt{3} - \sqrt{2}) }{2} \\ \\ = \frac{4( \sqrt{3 } + \sqrt{2} ) + (\sqrt{3} - \sqrt{2}) }{2} \\ \\ = \frac{5 \sqrt{3} + 3 \sqrt{2} }{2} \\ \\ = > \large \bold{x + \frac{1}{x} = \frac{5 \sqrt{3} + 3 \sqrt{2} }{2} }$

Again,

${(x + \frac{1}{x}) }^{2} \\ \\ = { (\frac{5 \sqrt{3} + 3 \sqrt{2} }{2} )}^{2} \\ \\ = \frac{ {(5 \sqrt{3} + 3 \sqrt{2} )}^{2} }{ {2}^{2} } \\ \\ = \frac{ {(5 \sqrt{3}) }^{2} + {(3 \sqrt{2} )}^{2} + 2( 5\sqrt{3})(3 \sqrt{2} ) }{4} \\ \\ = \frac{75 + 18 + 30 \sqrt{6} }{4} \\ \\ = \frac{93 + 30 \sqrt{6} }{4} \\ \\ = > \large \bold{ {(x + \frac{1}{x} )}^{2} = \frac{93 + 30 \sqrt{6} }{4} }$