If x² - 3x + 1 = 0 then what is the value of $$x² + x + \frac{1}{x} + \frac{1}{x²}$$​?

Correct option is D

Given:

x² - 3x + 1 = 0

Formula Used:

$$(a+b)^2 = a^2 + b^2 +2ab$$​​

Solution:

x² - 3x + 1 = 0

Divide by x both side:

​$$\frac{x^2}{x} - \frac{3x}{x} + \frac{1}{x} = 0$$

​$$x - 3 + \frac{1}{x} = 0$$

​​$$x + \frac{1}{x} = 3$$  ......(1).

Squaring both side:

​$$\left(x + \frac{1}{x}\right)^2 = 3^2$$

$$x^2 + \frac{1}{x^2} + 2 = 9$$

​​​$$x^2 + \frac{1}{x^2} = 9 - 2$$

$$x^2 + \frac{1}{x^2} = 7$$ ....(2)

Add equation (1) and (2);
​$$x^2 + \frac{1}{x^2}+ x + \frac{1}{x} = 3 + 7$$

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

Thus, the correct answer is (d).