The number of solutions of the equation `9x^(2)-18|x|+5 = 0` belonging to the domain of definition of `log_(e) {(x+1)(x+2)}`, is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the equation \(9x^2 - 18|x| + 5 = 0\) and find the number of solutions that belong to the domain of the function \(\log_e{((x+1)(x+2))}\), we will follow these steps:

### Step 1: Rewrite the Equation
The given equation is:
\[ 9x^2 - 18|x| + 5 = 0 \]
We can express this as:
\[ 9(|x|^2) - 18|x| + 5 = 0 \]

### Step 2: Let \(y = |x|\)
Let \(y = |x|\). Then the equation becomes:
\[ 9y^2 - 18y + 5 = 0 \]

### Step 3: Factor the Quadratic Equation
To factor the quadratic equation, we can use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 9\), \(b = -18\), and \(c = 5\).

Calculating the discriminant:
\[ b^2 - 4ac = (-18)^2 - 4 \cdot 9 \cdot 5 = 324 - 180 = 144 \]

Now substituting into the quadratic formula:
\[ y = \frac{18 \pm \sqrt{144}}{2 \cdot 9} = \frac{18 \pm 12}{18} \]
This gives us:
\[ y = \frac{30}{18} = \frac{5}{3} \quad \text{and} \quad y = \frac{6}{18} = \frac{1}{3} \]

### Step 4: Find Values of \(x\)
Since \(y = |x|\), we have:
\[ |x| = \frac{5}{3} \quad \Rightarrow \quad x = \frac{5}{3} \quad \text{or} \quad x = -\frac{5}{3} \]
\[ |x| = \frac{1}{3} \quad \Rightarrow \quad x = \frac{1}{3} \quad \text{or} \quad x = -\frac{1}{3} \]

Thus, the possible values of \(x\) are:
\[ x = \frac{5}{3}, \quad x = -\frac{5}{3}, \quad x = \frac{1}{3}, \quad x = -\frac{1}{3} \]

### Step 5: Determine the Domain of the Logarithmic Function
The logarithmic function \(\log_e{((x+1)(x+2))}\) is defined when:
\[ (x+1)(x+2) > 0 \]
This inequality holds when:
1. \(x < -2\)
2. \(x > -1\)

### Step 6: Check Each Solution Against the Domain
Now we check which of the solutions fall within the domain:
- \(x = \frac{5}{3} \approx 1.67\) (valid since \(1.67 > -1\))
- \(x = -\frac{5}{3} \approx -1.67\) (not valid since \(-1.67 < -2\))
- \(x = \frac{1}{3} \approx 0.33\) (valid since \(0.33 > -1\))
- \(x = -\frac{1}{3} \approx -0.33\) (valid since \(-0.33 > -1\))

### Conclusion
The valid solutions that belong to the domain are:
- \(x = \frac{5}{3}\)
- \(x = \frac{1}{3}\)
- \(x = -\frac{1}{3}\)

Thus, the total number of solutions belonging to the domain is **3**.

### Final Answer
The number of solutions of the equation belonging to the domain is **3**.

---