Question

if x=√2+1 then prove that x^2+{1/x^2}=6

Answer 1

$x = \sqrt{2} + 1 \\ \\ \frac{1}{x} = \frac{1}{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1 }{ \sqrt{2} - 1 } \\ \\ \\ => \frac{1}{x} = \frac{ \sqrt{2} - 1}{ {( \sqrt{2} )}^{2} - {1}^{2} } \\ \\ \\ => \frac{1}{x} = \frac{ \sqrt{2} - 1}{2 - 1} \\ \\ \\ = > \frac{1}{x} = \sqrt{2} - 1 \\ \\ \\ \\ x + \frac{1}{x} \\ \\ = > \sqrt{2} + 1 + \sqrt{2} - 1 \\ \\ => 2\sqrt{2} \\ \\ \\ x + \frac{1}{x} = 2 \sqrt{2}$
.
Square on both sides,


${x }^{2} + \frac{1}{ {x}^{2} } + 2 = {(2 \sqrt{2} )}^{2} \\ \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 8 - 2 \\ \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 6$



Hence, proved.

Answer 2

x=√2-1 and 1/x =√2-1