Solve the following. [(x^(1/2)) - (4x^(1/4)) = 0]

To solve the equation:

\[
x^{\frac{1}{2}} - 4x^{\frac{1}{4}} = 0
\]

we can approach it by simplifying the terms with substitution. Here’s a step-by-step solution:

### Step 1: Substitute a New Variable
Let:
\[
u = x^{\frac{1}{4}}
\]
Then:
\[
u^2 = x^{\frac{1}{2}}
\]

This substitution changes the equation to:
\[
u^2 - 4u = 0
\]

### Step 2: Factor the Equation
The equation \( u^2 - 4u = 0 \) can be factored as:
\[
u(u - 4) = 0
\]

### Step 3: Solve for \( u \)
Set each factor equal to zero:
1. \( u = 0 \)
2. \( u - 4 = 0 \Rightarrow u = 4 \)

### Step 4: Substitute Back for \( x \)
Now, recall that \( u = x^{\frac{1}{4}} \). Substitute back to find \( x \) for each solution of \( u \).

1. If \( u = 0 \):
   \[
   x^{\frac{1}{4}} = 0 \Rightarrow x = 0
   \]

2. If \( u = 4 \):
   \[
   x^{\frac{1}{4}} = 4 \Rightarrow x = 4^4 = 256
   \]

### Final Answer
The solutions to the equation are:
\[
x = 0 \quad \text{and} \quad x = 256
\]

Would you like more details on any step?

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Here are five related questions for practice:
1. Solve \( x^{\frac{1}{3}} - 3x^{\frac{1}{6}} = 0 \).
2. Find all solutions to \( x^{\frac{2}{3}} - 5x^{\frac{1}{3}} = 0 \).
3. Solve the equation \( y^2 - 6y = 0 \) using factoring.
4. What are the real solutions to \( z^2 - 2z = 0 \)?
5. Solve \( x^{\frac{1}{5}} - 2x^{\frac{1}{10}} = 0 \).

**Tip:** When faced with fractional exponents, substitution can simplify the problem and reveal hidden factorizations.

Learn how to solve the equation x^(1/2) - 4x^(1/4) = 0 using substitution and factoring. This guide provides a step-by-step approach to handle equations with fractional exponents, making it ideal for high school algebra students in Grades 10-12.

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