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Given that, $ \left( x + \frac{1}{x} \right)^{2} – 3 \left( x + \frac{1}{x} \right) + 2 = 0 $

Let, $ x + \frac{1}{x} = k $

So, $ k^{2} – 3k + 2 = 0 $

$ \Rightarrow k^{2} – 2k – k + 2 = 0 $

$ \Rightarrow k (k-2) -1 (k-1) = 0 $

$ \Rightarrow (k-1)(k-2) = 0 $

$ \Rightarrow k = 1 ; k = 2 $

$ \Rightarrow x + \frac{1}{x} = 1 ; x + \frac{1}{x} = 2 $

We know that, $\text{AM} \geq \text{GM}$

$ \Rightarrow \dfrac{ x + \frac{1}{x} }{2} \geq \sqrt{ x \cdot \frac{1}{x}} $

$ \Rightarrow \boxed{ x + \frac{1}{x} \geq 2} $

So,

- $ x + \frac{1}{x} = 1 \quad (\text{Not possible}) $
- $ x + \frac{1}{x} = 2 \quad (\text{ Possible}) $

$\therefore$ The number distinct real roots of the equation $ = 1.$

Correct Answer$:1$