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The number of distinct real roots of the equation $\left ( x+\frac{1}{x}\right )^{2}-3\left ( x+\frac{1}{x} \right )+2= 0$ equals

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$:1$

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