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 \( (3 |x| - 3)^2 = |x| + 7 \) that belongs to the domain of the function \( y = \sqrt{(x-4)x} \), we will follow these steps:

### Step 1: Determine the Domain of the Function
The function \( y = \sqrt{(x-4)x} \) is defined when the expression inside the square root is non-negative:
\[ (x-4)x \geq 0 \]
This inequality can be solved by finding the critical points:
- The critical points are \( x = 0 \) and \( x = 4 \).

Now, we analyze the intervals:
1. For \( x < 0 \): \( (x-4)x \) is positive (both factors are negative).
2. For \( 0 < x < 4 \): \( (x-4)x \) is negative (one factor is positive, the other is negative).
3. For \( x > 4 \): \( (x-4)x \) is positive (both factors are positive).

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

### Step 2: Substitute \( |x| \) with \( t \)
Let \( t = |x| \). The equation becomes:
\[ (3t - 3)^2 = t + 7 \]

### Step 3: Expand and Rearrange the Equation
Expanding the left side:
\[ (3t - 3)^2 = 9t^2 - 18t + 9 \]
Setting the equation:
\[ 9t^2 - 18t + 9 = t + 7 \]
Rearranging gives:
\[ 9t^2 - 19t + 2 = 0 \]

### Step 4: Factor the Quadratic Equation
To factor \( 9t^2 - 19t + 2 = 0 \), we can look for two numbers that multiply to \( 9 \times 2 = 18 \) and add to \(-19\). The factors are \(-18\) and \(-1\):
\[ 9t^2 - 18t - t + 2 = 0 \]
Grouping gives:
\[ (9t^2 - 18t) + (-t + 2) = 0 \]
Factoring out common terms:
\[ 9t(t - 2) - 1(t - 2) = 0 \]
Factoring further:
\[ (t - 2)(9t - 1) = 0 \]

### Step 5: Solve for \( t \)
Setting each factor to zero gives:
1. \( t - 2 = 0 \) → \( t = 2 \)
2. \( 9t - 1 = 0 \) → \( t = \frac{1}{9} \)

### Step 6: Convert Back to \( |x| \)
Since \( t = |x| \):
1. \( |x| = 2 \) → \( x = 2 \) or \( x = -2 \)
2. \( |x| = \frac{1}{9} \) → \( x = \frac{1}{9} \) or \( x = -\frac{1}{9} \)

### Step 7: Check Which Solutions Belong to the Domain
Now, we check which of these solutions belong to the domain \( (-\infty, 0] \cup [4, \infty) \):
- \( x = 2 \) (not in the domain)
- \( x = -2 \) (in the domain)
- \( x = \frac{1}{9} \) (not in the domain)
- \( x = -\frac{1}{9} \) (in the domain)

### Final Solutions
The solutions of the quadratic equation that belong to the domain of the function are:
\[ \boxed{-2 \text{ and } -\frac{1}{9}} \]