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

Let $y = x + \dfrac{1}{x}$

When we have repeated expressions in an equation, substitution transforms a complex equation into a simpler one.

Our equation becomes:

$y^2 - 3y + 2 = 0$

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This is a simple quadratic equation. We can factor it:

$y^2 - 3y + 2 = (y - 1)(y - 2) = 0$

Factoring check: $(y-1)(y-2) = y^2 - 2y - y + 2 = y^2 - 3y + 2$

Therefore: $y = 1$ or $y = 2$

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Now we need to solve for $x$ in each case.

Case 1: When $y = 2$

$x + \dfrac{1}{x} = 2$

Both sides by $x$ (note: $x \neq 0$ since we have $\dfrac{1}{x}$ in the original equation):

$x^2 + 1 = 2x$

$x^2 - 2x + 1 = 0$

$(x - 1)^2 = 0$

Therefore: $x = 1$

Case 2: When $y = 1$

$x + \dfrac{1}{x} = 1$

Both sides by $x$:

$x^2 + 1 = x$

$x^2 - x + 1 = 0$

Using the quadratic formula:

$x = \dfrac{1 \pm \sqrt{1 - 4}}{2} = \dfrac{1 \pm \sqrt{-3}}{2}$

Since the discriminant is negative ($-3 < 0$), this case gives us complex roots, not real roots.

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From our analysis:

• Case 1 gives us: $x = 1$ (real root)

• Case 2 gives us: complex roots (not real)

Even though $x = 1$ came from $(x-1)^2 = 0$, it's still counted as one distinct root.

Therefore, the number of distinct real roots is 1.