Question

if x = √6- √2/ √6+ √2 find x + 1/x​

Answer 1

Given:-

Value of x = $\frac{\sqrt{6} -\sqrt{2} }{\sqrt{6} +\sqrt{2} }$

To find:-

Value of $x+\frac{1}{x}$

Solution:-

$x+\frac{1}{x}$

= $\frac{\sqrt{6} -\sqrt{2} }{\sqrt{6} +\sqrt{2} } + \frac{1}{\frac{\sqrt{6} -\sqrt{2} }{\sqrt{6} +\sqrt{2} }}$

= $\frac{\sqrt{6} -\sqrt{2} }{\sqrt{6} +\sqrt{2} }+\frac{\sqrt{6} +\sqrt{2} }{\sqrt{6}-\sqrt{2} }$

= $\frac{(\sqrt{6}-\sqrt{2} )^{2}+(\sqrt{6}+\sqrt{2} )^{2} }{(\sqrt{6}+{\sqrt{2} } )(\sqrt{6}-\sqrt{2} )}$

= $\frac{6+2-2(\sqrt{6} )(\sqrt{2} )+6+2+2(\sqrt{6} )(\sqrt{2} )}{6-2}$

= $\frac{16-2\sqrt{12} +2\sqrt{12} }{4}$

= $\frac{16}{4}$

= 4

Hence, the solution of the question is 4