​$$\text{If }x + \frac{ 1}{x} = 17 \text{ then }x^2 + \frac{ 1}{x^2} \text{ is:}$$

Correct option is D

​Given:

​$$x + \frac{1}{x} = 17$$
Formula Used:
​$$\left(x + \frac{1}{x}\right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = x^2 + 2 + \frac{1}{x^2}$$​
Solution:

​$$x + \frac{1}{x} = 17 \\ \ \\\left(x + \frac{1}{x}\right)^2 = 17^2 \\ \ \\x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 289 \\ \ \\x^2 + 2 + \frac{1}{x^2} = 289 \\ \ \\x^2 + \frac{1}{x^2} = 289- 2 \\ \ \\x^2 + \frac{1}{x^2} = 287$$