How many real numbers x satisfy the equation `3^(2x+2)-3^(x+3)-3^(x)+3 = 0`?

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation \( 3^{2x+2} - 3^{x+3} - 3^{x} + 3 = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the original equation:
\[ 3^{2x+2} - 3^{x+3} - 3^{x} + 3 = 0 \]
We can rewrite \( 3^{2x+2} \) as \( 3^{2x} \cdot 3^2 = 9 \cdot 3^{2x} \) and \( 3^{x+3} \) as \( 3^x \cdot 3^3 = 27 \cdot 3^x \). Thus, the equation becomes:
\[ 9 \cdot 3^{2x} - 27 \cdot 3^{x} - 3^{x} + 3 = 0 \]

### Step 2: Substitute \( t = 3^x \)
Let \( t = 3^x \). Then \( 3^{2x} = t^2 \). Substituting these into the equation gives:
\[ 9t^2 - 27t - t + 3 = 0 \]
This simplifies to:
\[ 9t^2 - 28t + 3 = 0 \]

### Step 3: Solve the quadratic equation
Now we can solve the quadratic equation \( 9t^2 - 28t + 3 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 9 \), \( b = -28 \), and \( c = 3 \). Plugging in these values:
\[ t = \frac{28 \pm \sqrt{(-28)^2 - 4 \cdot 9 \cdot 3}}{2 \cdot 9} \]
Calculating the discriminant:
\[ (-28)^2 = 784, \quad 4 \cdot 9 \cdot 3 = 108 \quad \Rightarrow \quad 784 - 108 = 676 \]
Now substituting back:
\[ t = \frac{28 \pm \sqrt{676}}{18} = \frac{28 \pm 26}{18} \]

### Step 4: Find the values of \( t \)
Calculating the two possible values for \( t \):
1. \( t = \frac{28 + 26}{18} = \frac{54}{18} = 3 \)
2. \( t = \frac{28 - 26}{18} = \frac{2}{18} = \frac{1}{9} \)

### Step 5: Back substitute to find \( x \)
Now we convert back to \( x \):
1. For \( t = 3 \):
\[ 3^x = 3 \quad \Rightarrow \quad x = 1 \]
2. For \( t = \frac{1}{9} \):
\[ 3^x = \frac{1}{9} \quad \Rightarrow \quad 3^x = 3^{-2} \quad \Rightarrow \quad x = -2 \]

### Conclusion
Thus, the two values of \( x \) that satisfy the equation are \( x = 1 \) and \( x = -2 \). Therefore, the number of real numbers \( x \) that satisfy the equation is:
\[ \boxed{2} \]