The number of negative roots of `9^(x +2) - 6 (3^(x +1)) + 1 = 0` is

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the equation \(9^{(x + 2)} - 6(3^{(x + 1)}) + 1 = 0\) and find the number of negative roots, we can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting \(9^{(x + 2)}\) in terms of base 3:
\[ 9^{(x + 2)} = (3^2)^{(x + 2)} = 3^{(2(x + 2))} = 3^{(2x + 4)} \]
Thus, the equation becomes:
\[ 3^{(2x + 4)} - 6(3^{(x + 1)}) + 1 = 0 \]

### Step 2: Substitute \(y = 3^{(x + 1)}\)
Let \(y = 3^{(x + 1)}\). Then:
\[ 3^{(2x + 4)} = 3^{(2(x + 1) + 2)} = 3^2 \cdot (3^{(x + 1)})^2 = 9y^2 \]
Substituting this into the equation gives:
\[ 9y^2 - 6y + 1 = 0 \]

### Step 3: Solve the quadratic equation
Now we can solve the quadratic equation \(9y^2 - 6y + 1 = 0\) using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 9\), \(b = -6\), and \(c = 1\). Plugging in these values:
\[ y = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 9 \cdot 1}}{2 \cdot 9} \]
\[ y = \frac{6 \pm \sqrt{36 - 36}}{18} \]
\[ y = \frac{6 \pm 0}{18} \]
\[ y = \frac{6}{18} = \frac{1}{3} \]

### Step 4: Find \(x\) from \(y\)
Recall that \(y = 3^{(x + 1)}\):
\[ 3^{(x + 1)} = \frac{1}{3} \]
This can be rewritten as:
\[ 3^{(x + 1)} = 3^{-1} \]
By comparing the exponents, we have:
\[ x + 1 = -1 \implies x = -2 \]

### Step 5: Determine the number of negative roots
Since we found \(x = -2\) and this is the only solution, we conclude that there is **1 negative root**.

### Final Answer
The number of negative roots of the equation is **1**.

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