If $$x>1$$​, and $$x+\frac{1}{x}=\sqrt{29}$$​, what is the value of $$x-\frac{1}{x}$$​?

Correct option is C

Given:
x > 1 
​$$x + \frac{1}{x} = \sqrt{29}$$

Formula Used:

​$$\left( x - \frac{1}{x} \right)^2 = x^2 - 2 + \frac{1}{x^2}$$​​
Solution:
​$$x + \frac{1}{x} = \sqrt{29}$$​​

​$$\left( x + \frac{1}{x} \right)^2 = (\sqrt{29})^2$$​​
​$$x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 29$$​​

​$$x^2 + 2 + \frac{1}{x^2} = 29$$​​

​$$x^2 + \frac{1}{x^2} = 27$$​​

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

​$$\left( x - \frac{1}{x} \right)^2 = 27 - 2 = 25$$​​

​$$x - \frac{1}{x} = \pm \sqrt{25} = \pm 5$$​​

Given x > 1:

Since x > 1, $$\frac{1}{x}$$​ < 1 , so  $$x - \frac{1}{x} > 0$$​​
Therefore, $$x - \frac{1}{x} = 5$$​