CAT 2020 Slot 1 QA Question 6: The number of distinct real roots of the equation $(x + \dfrac{1}{x})^2 - 3(x + \dfrac{1}{x}) + 2 = 0$ equals

Question

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

Accepted Answer

Let $t = x + \dfrac{1}{x}$. Then the given equation becomes a quadratic in $t$.

$t^2 - 3t + 2 = 0$
$\Rightarrow (t - 1)(t - 2) = 0$
$\Rightarrow t = 1$ or $t = 2$

For real $x$, $x + \dfrac{1}{x} \geq 2$ or $x + \dfrac{1}{x} \leq -2$. Hence, $t = 1$ is impossible.

For $t = 2$, $x + \dfrac{1}{x} =

The number of distinct real roots of the equation (x+1x)2−3(x+1x)+2=0(x + \dfrac{1}{x})^2 - 3(x + \dfrac{1}{x}) + 2 = 0(x+x1​)2−3(x+x1​)+2=0 equals