If $$x^2-3x+1=0$$​, then the value of $$x^2+x+(\frac{1}{x})+(\frac{1}{x^2})$$​ is:

Correct option is B

​Given:​

$$x^2 - 3x + 1 = 0$$

​​$$x^2 + x + \frac{1}{x} + \frac{1}{x^2}$$ = ?

Formula Used:
If x $$+ \frac{1}{x} =$$ a, then

​$$x^2 + \frac{1}{x^2} = a^2 - 2$$​

Solution:
From the equation:

​$$x^2 + 1 = 3x$$​​

​$$\frac{x^2 + 1}{x} = 3$$​​

​$$x + \frac{1}{x} = 3$$​​

So,

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

Now add both:

​$$x^2 + x + \frac{1}{x} + \frac{1}{x^2}= x + \frac{1}{x} +x^2 + \frac{1}{x^2} = 3 + 7 = 10$$​