We are given the equation:

\[
x \sqrt{x} = x + \sqrt{x}
\]

To solve this equation, let's proceed step by step.

### Step 1: Isolate terms involving \( \sqrt{x} \)

Subtract \( \sqrt{x} \) from both sides:

\[
x \sqrt{x} - \sqrt{x} = x
\]

### Step 2: Factor out \( \sqrt{x} \)

Notice that \( \sqrt{x} \) is common on the left-hand side:

\[
\sqrt{x} (x - 1) = x
\]

### Step 3: Consider two cases

#### Case 1: \( \sqrt{x} = 0 \)

This happens when \( x = 0 \), so \( x = 0 \) is one solution.

#### Case 2: \( x - 1 = 0 \)

This implies \( x = 1 \). Substituting this into the equation, we get:

\[
\sqrt{1} (1 - 1) = 1 \quad \text{(False, discard this)}
\]

### Step 4: Verify \( x = 0 \)

Substitute \( x = 0 \) back into the original equation:

\[
0 \sqrt{0} = 0 + \sqrt{0}
\]

Both sides are equal, so \( x = 0 \) is indeed a solution.

### Final Answer:
\[
x = 0
\]

Would you like more details on a specific step, or have any other questions?

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Here are some related questions:
1. What happens if you apply different methods to solve similar radical equations?
2. How would the equation change if it were \( x^2 \sqrt{x} = x + \sqrt{x} \)?
3. How do you solve equations involving multiple radical terms?
4. What happens if you square both sides of radical equations?
5. How can the factoring method be applied to more complex equations?

**Tip:** When solving equations with radicals, consider squaring both sides to eliminate the square roots, but always check for extraneous solutions afterward!

This detailed solution explains how to solve the radical equation x√x = x + √x using factoring and properties of square roots. Follow the step-by-step breakdown to isolate the square root term and solve for x. Ideal for students in Grades 9-11, covering algebra and radical equations.

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