Find the number or real values of x satisfying the equation `9^(2log_(9)x)+4x+3=0`.

Answer

To solve the equation \( 9^{2\log_{9}x} + 4x + 3 = 0 \), we will follow these steps:

### Step 1: Simplify the Exponential Term
We start with the equation:
\[ 9^{2\log_{9}x} + 4x + 3 = 0 \]
Using the property of logarithms, we know that \( a^{\log_{a}b} = b \). Therefore, we can rewrite \( 9^{2\log_{9}x} \) as:
\[ (9^{\log_{9}x})^2 = x^2 \]
Thus, the equation simplifies to:
\[ x^2 + 4x + 3 = 0 \]

### Step 2: Factor the Quadratic Equation
Next, we need to factor the quadratic equation \( x^2 + 4x + 3 = 0 \). We can do this by finding two numbers that multiply to \( 3 \) (the constant term) and add up to \( 4 \) (the coefficient of \( x \)). The numbers \( 1 \) and \( 3 \) satisfy these conditions. Therefore, we can factor the equation as:
\[ (x + 1)(x + 3) = 0 \]

### Step 3: Solve for x
Setting each factor equal to zero gives us:
\[ x + 1 = 0 \quad \text{or} \quad x + 3 = 0 \]
This leads to:
\[ x = -1 \quad \text{or} \quad x = -3 \]

### Step 4: Check for Validity
Since \( x \) is inside a logarithm in the original equation, we need to ensure that \( x > 0 \) for \( \log_{9}x \) to be defined. Both solutions \( x = -1 \) and \( x = -3 \) are negative, which means they do not satisfy the condition \( x > 0 \).

### Conclusion
Thus, there are no real values of \( x \) that satisfy the original equation \( 9^{2\log_{9}x} + 4x + 3 = 0 \).