If value of x+1/x is 3+√8 …

It is given that, 
x = 3 + {tex}\sqrt{8}{/tex}
Now, {tex}\frac{1}{x}{/tex} = {tex}\frac{1}{3+\sqrt{8}}{/tex} = {tex}\frac{1}{3+\sqrt{8}} \times \frac{3-\sqrt{8}}{3-\sqrt{8}}{/tex} ( rationalising the expression) 
= {tex}\frac{3-\sqrt{8}}{(3)^{2}-(\sqrt{8})^{2}}{/tex}
= {tex}\frac{3-\sqrt{8}}{9-8}{/tex} = 3 - {tex}\sqrt{8}{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{1}{x}{/tex} = 3 - {tex}\sqrt{8}{/tex}
We know that, (x + {tex}\frac{1}{x}{/tex})² = x² + {tex}\frac{1}{x^2}{/tex} + 2 {tex}\times{/tex} x {tex}\times{/tex} {tex}\frac{1}{x}{/tex}
{tex}\Rightarrow{/tex} (3 + {tex}\sqrt{8}{/tex} + 3 - {tex}\sqrt{8}{/tex})² = x² + {tex}\frac{1}{x^2}{/tex} + 2
{tex}\Rightarrow{/tex} (6)² = x² + {tex}\frac{1}{x^2}{/tex} + 2
{tex}\Rightarrow{/tex} 36 = x² + {tex}\frac{1}{x^2}{/tex} + 2
{tex}\Rightarrow{/tex} x² + {tex}\frac{1}{x^2}{/tex} = 36 - 2 = 34
Hence, x² + {tex}\frac{1}{x^2}{/tex} = 34