Step 1: Substitute \( y = \frac{x+1}{x} \) into the equation. This results in the quadratic equation:
\(y^2 - 3y + 2 = 0\).
Step 2: Solve the quadratic equation \(y^2 - 3y + 2 = 0\) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = -3, c = 2 \).
\(y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}\)
\(y = \frac{3 \pm \sqrt{9 - 8}}{2}\)
\(y = \frac{3 \pm 1}{2}\)
The solutions for \(y\) are \(y = 2\) and \(y = 1\).
Step 3: Substitute back \( y = \frac{x+1}{x} \) to find \( x \):
Case 1: For \( y = 2\):
\( \frac{x+1}{x} = 2\)
\( x+1 = 2x\)
\( x = 1\).
Case 2: For \( y = 1\):
\( \frac{x+1}{x} = 1\)
\( x+1 = x\)
This simplifies to \( 1=0 \), which is an impossible statement. Therefore, there is no solution for \( x \) in this case.
The only distinct real root is \(x = 1\).
Consequently, the number of distinct real roots is 1.