You're correct that the given equation is a rational equation, not a quadratic one. Let's revisit the solution with the correct approach.

The original equation is:

\[
\frac{x}{x + 1} + \frac{3}{x} = \frac{5}{2x}
\]

### Step 1: Eliminate the Denominators

To eliminate the fractions, multiply both sides by the least common denominator (LCD). The LCD of \(x\), \(x + 1\), and \(2x\) is \(2x(x + 1)\).

Multiply each term by the LCD:

\[
2x(x + 1) \cdot \frac{x}{x + 1} + 2x(x + 1) \cdot \frac{3}{x} = 2x(x + 1) \cdot \frac{5}{2x}
\]

This simplifies to:

\[
2x^2 + 6(x + 1) = 5(x + 1)
\]

### Step 2: Expand and Simplify

Distribute and expand:

\[
2x^2 + 6x + 6 = 5x + 5
\]

### Step 3: Move All Terms to One Side

Rearrange the equation to bring all terms to one side:

\[
2x^2 + 6x - 5x + 6 - 5 = 0
\]

Simplify further:

\[
2x^2 + x + 1 = 0
\]

### Step 4: Solve the Quadratic Equation

We can solve the quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \(a = 2\), \(b = 1\), and \(c = 1\). Plugging in these values:

\[
x = \frac{-1 \pm \sqrt{1^2 - 4(2)(1)}}{2(2)}
\]
\[
x = \frac{-1 \pm \sqrt{1 - 8}}{4}
\]
\[
x = \frac{-1 \pm \sqrt{-7}}{4}
\]

Since the discriminant is negative (\(\sqrt{-7}\)), the solutions are complex:

\[
x = \frac{-1 + \sqrt{7}i}{4} \quad \text{or} \quad x = \frac{-1 - \sqrt{7}i}{4}
\]

### Final Answer:

The solutions to the equation are complex:

\[
x = \frac{-1 + \sqrt{7}i}{4} \quad \text{or} \quad x = \frac{-1 - \sqrt{7}i}{4}
\]

If you're looking for solutions in the context of rational equations, these are the results you get. However, since the original problem involves a rational equation, it’s important to check the domain and verify that no solution would make any original denominator zero (which would make the original equation undefined).

Would you like further details or clarification on any specific part? Here are some related questions to explore:

1. How does the concept of extraneous solutions apply to rational equations?
2. What are some methods to solve rational equations effectively?
3. How can complex numbers be interpreted in real-world problems?
4. What are some common pitfalls when solving rational equations?
5. Why is it important to check the solutions back in the original equation?

**Tip:** When solving rational equations, always remember to check your solutions in the original equation to ensure they don’t make any denominator zero, which would invalidate them.

Learn how to solve the rational equation x / (x + 1) + 3 / x = 5 / (2x) with complex solutions. This problem involves applying the quadratic formula to solve for complex numbers, suitable for advanced high school students.

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