If f(x) = `x^(2) - 4` , for what real number values of x will f(f(x))= 0 ?

Options

A. 2.4
B. `pm 2.4`
C. 2 or 6
D. `pm 1.4 or pm 2.4`

Explanation

To solve the problem where we need to find the real number values of \( x \) such that \( f(f(x)) = 0 \) for the function \( f(x) = x^2 - 4 \), we will follow these steps:

### Step 1: Define the function
We start with the given function:
\[ f(x) = x^2 - 4 \]

### Step 2: Find \( f(f(x)) \)
Next, we need to compute \( f(f(x)) \):
\[ f(f(x)) = f(x^2 - 4) \]
Substituting \( x^2 - 4 \) into the function \( f \):
\[ f(f(x)) = (x^2 - 4)^2 - 4 \]

### Step 3: Expand the expression
Now, we will expand \( (x^2 - 4)^2 \):
\[ (x^2 - 4)^2 = x^4 - 8x^2 + 16 \]
So,
\[ f(f(x)) = x^4 - 8x^2 + 16 - 4 = x^4 - 8x^2 + 12 \]

### Step 4: Set the equation to zero
We want to find the values of \( x \) such that:
\[ f(f(x)) = 0 \]
This gives us the equation:
\[ x^4 - 8x^2 + 12 = 0 \]

### Step 5: Substitute to simplify
Let \( y = x^2 \). Then the equation becomes:
\[ y^2 - 8y + 12 = 0 \]

### Step 6: Solve the quadratic equation
Now, we will solve the quadratic equation using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -8, c = 12 \):
\[ y = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \]
\[ y = \frac{8 \pm \sqrt{64 - 48}}{2} \]
\[ y = \frac{8 \pm \sqrt{16}}{2} \]
\[ y = \frac{8 \pm 4}{2} \]
Calculating the two possible values:
1. \( y = \frac{12}{2} = 6 \)
2. \( y = \frac{4}{2} = 2 \)

### Step 7: Back substitute for \( x \)
Now we substitute back for \( x \):
1. From \( y = 6 \):
\[ x^2 = 6 \implies x = \pm \sqrt{6} \]
2. From \( y = 2 \):
\[ x^2 = 2 \implies x = \pm \sqrt{2} \]

### Step 8: Final answer
Thus, the real number values of \( x \) that satisfy \( f(f(x)) = 0 \) are:
\[ x = \pm \sqrt{6}, \pm \sqrt{2} \]