# CAT 2020 Quant Question [Slot 1] with Solution 25

Question

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

Option: 1
Solution:

Let $x + \frac{1}{x} = a$

The given equation becomes, ${a^2} - 3a + 2 = 0$ $a = 2$ or 1 i.e $x + \frac{1}{x} = 2$ or $x + \frac{1}{x} = 1$

since $x$ is real, $x + \frac{1}{x} \ne 1;\therefore x + \frac{1}{x} = 2$

$\therefore$ The number of solutions = 1

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