Root(s) of the equatio `9x^(2) - 18|x|+5 = 0` belonging to the domain of definition of the function `f(x) = log(x^(2) - x - 2)`, is (are)

Options

A. `(-5)/(3),(-1)/(3)`
B. `5/3, 1/3`
C. `(-5)/(3)`
D. `(-1)/(3)`

Answer

To solve the equation \(9x^2 - 18|x| + 5 = 0\) and find the roots that belong to the domain of the function \(f(x) = \log(x^2 - x - 2)\), we will follow these steps:

### Step 1: Rewrite the equation
The given equation is:
\[ 9x^2 - 18|x| + 5 = 0 \]
We can rewrite this by considering the cases for \(|x|\).

### Step 2: Consider cases for \(|x|\)
1. **Case 1**: \(x \geq 0\) (then \(|x| = x\))
\[ 9x^2 - 18x + 5 = 0 \]
2. **Case 2**: \(x < 0\) (then \(|x| = -x\))
\[ 9x^2 + 18x + 5 = 0 \]

### Step 3: Solve Case 1
For \(9x^2 - 18x + 5 = 0\), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 9\), \(b = -18\), and \(c = 5\):
\[ x = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 9 \cdot 5}}{2 \cdot 9} \]
\[ x = \frac{18 \pm \sqrt{324 - 180}}{18} \]
\[ x = \frac{18 \pm \sqrt{144}}{18} \]
\[ x = \frac{18 \pm 12}{18} \]
Calculating the roots:
\[ x_1 = \frac{30}{18} = \frac{5}{3}, \quad x_2 = \frac{6}{18} = \frac{1}{3} \]

### Step 4: Solve Case 2
For \(9x^2 + 18x + 5 = 0\):
\[ x = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 9 \cdot 5}}{2 \cdot 9} \]
\[ x = \frac{-18 \pm \sqrt{324 - 180}}{18} \]
\[ x = \frac{-18 \pm \sqrt{144}}{18} \]
Calculating the roots:
\[ x_3 = \frac{-18 + 12}{18} = \frac{-6}{18} = -\frac{1}{3}, \quad x_4 = \frac{-18 - 12}{18} = \frac{-30}{18} = -\frac{5}{3} \]

### Step 5: List all roots
The roots of the equation are:
\[ x = \frac{5}{3}, \quad x = \frac{1}{3}, \quad x = -\frac{1}{3}, \quad x = -\frac{5}{3} \]

### Step 6: Determine the domain of \(f(x)\)
The function \(f(x) = \log(x^2 - x - 2)\) is defined when:
\[ x^2 - x - 2 > 0 \]
Factoring gives:
\[ (x - 2)(x + 1) > 0 \]
The critical points are \(x = 2\) and \(x = -1\). Testing intervals:
- For \(x < -1\), both factors are negative, so the product is positive.
- For \(-1 < x < 2\), the product is negative.
- For \(x > 2\), both factors are positive, so the product is positive.

Thus, the domain of \(f(x)\) is:
\[ (-\infty, -1) \cup (2, \infty) \]

### Step 7: Check which roots belong to the domain
- \(x = \frac{5}{3}\) (not in domain)
- \(x = \frac{1}{3}\) (not in domain)
- \(x = -\frac{1}{3}\) (not in domain)
- \(x = -\frac{5}{3}\) (in domain)

### Conclusion
The only root that belongs to the domain of \(f(x)\) is:
\[ \boxed{-\frac{5}{3}} \]