What is the solution set `root(5)(x^2+4x)=2?`

Options

A. `{4}`
B. `{8}`
C. `{-4,8}`
D. `{-8,4}`

Answer

To solve the equation \( \sqrt[5]{x^2 + 4x} = 2 \), we will follow these steps:

### Step 1: Eliminate the root
To eliminate the fifth root, we will raise both sides of the equation to the power of 5.

\[ \left(\sqrt[5]{x^2 + 4x}\right)^5 = 2^5 \]

### Step 2: Simplify the equation
This simplifies to:

\[ x^2 + 4x = 32 \]

### Step 3: Rearrange the equation
Next, we will rearrange the equation to set it to zero:

\[ x^2 + 4x - 32 = 0 \]

### Step 4: Factor the quadratic equation
Now we will factor the quadratic equation. We need two numbers that multiply to \(-32\) and add to \(4\). The numbers \(8\) and \(-4\) work:

\[ (x + 8)(x - 4) = 0 \]

### Step 5: Solve for \(x\)
Now we will set each factor to zero:

1. \(x + 8 = 0 \Rightarrow x = -8\)
2. \(x - 4 = 0 \Rightarrow x = 4\)

### Step 6: Write the solution set
Thus, the solution set is:

\[ \{ -8, 4 \} \]