If $$x=\frac{\sqrt5+1}{\sqrt5-1}$$​ and $$y=\frac{1}{x}$$​, then what will be the value of $$x^2+y^2$$​?

Correct option is C

​Given:​

$$x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}, \quad y = \frac{1}{x}$$

​Formula Used:

​$$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2$$​​

Solution:

​$$x = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} \cdot \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{(\sqrt{5} + 1)^2}{5 - 1} = \frac{5 + 2\sqrt{5} + 1}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}$$​​

​$$\frac{1}{x} =\frac{2}{3 + \sqrt{5}} \cdot \frac{3 - \sqrt{5}}{3 - \sqrt{5}} = \frac{2(3 - \sqrt{5})}{9 - 5} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$$​​

Now:

​$$x + \frac{1}{x} = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = \frac{6}{2} = 3$$​​

​$$x^2 + \frac{1}{x^2} = 3^2 - 2 = 9 - 2 = 7$$​