Solution of the quadratic equation `(3 |x| -3) ^(2) = |x| +7, ` which belongs to the domain of the function `y = sqrt((x-4)x )` is :

Options

A. `pm 1/9 , pm 2`
B. `1/9, 8`
C. `-2, -1/9`
D. `-1/9 , 8`

Answer

To solve the quadratic equation \( (3 |x| - 3)^2 = |x| + 7 \) and find the solutions that belong to the domain of the function \( y = \sqrt{(x-4)x} \), we will follow these steps:

### Step 1: Find the domain of the function \( y = \sqrt{(x-4)x} \)

The expression inside the square root must be non-negative:
\[ (x-4)x \geq 0 \]
This can be factored as:
\[ x(x-4) \geq 0 \]
The critical points are \( x = 0 \) and \( x = 4 \). We will analyze the sign of the expression in the intervals determined by these points:

- For \( x < 0 \): Both \( x \) and \( x-4 \) are negative, so the product is positive.
- For \( 0 \leq x < 4 \): \( x \) is non-negative and \( x-4 \) is negative, so the product is negative.
- For \( x \geq 4 \): Both \( x \) and \( x-4 \) are non-negative, so the product is non-negative.

Thus, the domain of \( y \) is:
\[ (-\infty, 0] \cup [4, \infty) \]

### Step 2: Solve the quadratic equation \( (3 |x| - 3)^2 = |x| + 7 \)

First, we expand the left-hand side:
\[ (3 |x| - 3)^2 = 9 |x|^2 - 18 |x| + 9 \]
Setting the equation:
\[ 9 |x|^2 - 18 |x| + 9 = |x| + 7 \]
Rearranging gives:
\[ 9 |x|^2 - 19 |x| + 2 = 0 \]

### Step 3: Use the quadratic formula to find \( |x| \)

Using the quadratic formula \( |x| = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 9 \), \( b = -19 \), and \( c = 2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-19)^2 - 4 \cdot 9 \cdot 2 = 361 - 72 = 289 \]
Now, applying the quadratic formula:
\[ |x| = \frac{19 \pm \sqrt{289}}{2 \cdot 9} = \frac{19 \pm 17}{18} \]
Calculating the two possible values:
1. \( |x| = \frac{36}{18} = 2 \)
2. \( |x| = \frac{2}{18} = \frac{1}{9} \)

### Step 4: Determine the values of \( x \)

From \( |x| = 2 \):
- \( x = 2 \) or \( x = -2 \)

From \( |x| = \frac{1}{9} \):
- \( x = \frac{1}{9} \) or \( x = -\frac{1}{9} \)

### Step 5: Check which solutions belong to the domain

The solutions we have are:
- \( x = 2 \)
- \( x = -2 \)
- \( x = \frac{1}{9} \)
- \( x = -\frac{1}{9} \)

Now we check these against the domain \( (-\infty, 0] \cup [4, \infty) \):
- \( x = 2 \) is not in the domain.
- \( x = -2 \) is in the domain.
- \( x = \frac{1}{9} \) is not in the domain.
- \( x = -\frac{1}{9} \) is in the domain.

### Final Solutions

The solutions of the quadratic equation that belong to the domain of the function are:
\[ x = -2 \quad \text{and} \quad x = -\frac{1}{9} \]