Find the product of the roots of the equation `9x^2-18|x|+5=0`

Answer

To find the product of the roots of the equation \( 9x^2 - 18|x| + 5 = 0 \), we will follow these steps:

### Step 1: Substitute \( |x| \) with a variable
Let \( t = |x| \). Then the equation becomes:
\[ 9x^2 - 18t + 5 = 0 \]
Since \( x^2 = t^2 \), we can rewrite the equation as:
\[ 9t^2 - 18t + 5 = 0 \]

### Step 2: Use the quadratic formula to find the roots
The roots of a quadratic equation \( at^2 + bt + c = 0 \) can be found using the formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 9 \), \( b = -18 \), and \( c = 5 \). Plugging in these values:
\[ t = \frac{18 \pm \sqrt{(-18)^2 - 4 \cdot 9 \cdot 5}}{2 \cdot 9} \]
Calculating the discriminant:
\[ (-18)^2 - 4 \cdot 9 \cdot 5 = 324 - 180 = 144 \]
Now substituting back into the formula:
\[ t = \frac{18 \pm \sqrt{144}}{18} = \frac{18 \pm 12}{18} \]
This gives us two values for \( t \):
\[ t_1 = \frac{30}{18} = \frac{5}{3}, \quad t_2 = \frac{6}{18} = \frac{1}{3} \]

### Step 3: Find the corresponding values of \( x \)
Since \( t = |x| \), we have:
1. For \( t_1 = \frac{5}{3} \): \( x = \frac{5}{3} \) or \( x = -\frac{5}{3} \)
2. For \( t_2 = \frac{1}{3} \): \( x = \frac{1}{3} \) or \( x = -\frac{1}{3} \)

### Step 4: Calculate the product of the roots
The roots of the original equation are:
\[ x_1 = \frac{5}{3}, \quad x_2 = -\frac{5}{3}, \quad x_3 = \frac{1}{3}, \quad x_4 = -\frac{1}{3} \]
The product of the roots can be calculated as:
\[ P = x_1 \cdot x_2 \cdot x_3 \cdot x_4 = \left(\frac{5}{3}\right) \cdot \left(-\frac{5}{3}\right) \cdot \left(\frac{1}{3}\right) \cdot \left(-\frac{1}{3}\right) \]
Calculating this step-by-step:
1. \( \left(\frac{5}{3}\right) \cdot \left(-\frac{5}{3}\right) = -\frac{25}{9} \)
2. \( \left(\frac{1}{3}\right) \cdot \left(-\frac{1}{3}\right) = -\frac{1}{9} \)
3. Now multiply these two results:
\[ P = -\frac{25}{9} \cdot -\frac{1}{9} = \frac{25}{81} \]

### Final Answer
The product of the roots of the equation \( 9x^2 - 18|x| + 5 = 0 \) is:
\[ \frac{25}{81} \]