Solve each of the following equatins :
`9x^(2)+6x+1=0`

Answer

To solve the quadratic equation \(9x^2 + 6x + 1 = 0\), we can follow these steps:

### Step 1: Identify the coefficients
The given equation is in the standard form \(ax^2 + bx + c = 0\), where:
- \(a = 9\)
- \(b = 6\)
- \(c = 1\)

### Step 2: Rewrite the equation
We can rewrite \(9x^2\) as \((3x)^2\), \(6x\) as \(2 \cdot 3x \cdot 1\), and \(1\) as \(1^2\). Thus, we can express the equation as:
\[ (3x)^2 + 2 \cdot (3x) \cdot 1 + 1^2 = 0 \]

### Step 3: Apply the perfect square identity
We recognize that the left-hand side can be factored using the identity \(a^2 + 2ab + b^2 = (a + b)^2\). Here, \(a = 3x\) and \(b = 1\):
\[ (3x + 1)^2 = 0 \]

### Step 4: Solve for \(x\)
Now, we set the squared term equal to zero:
\[ 3x + 1 = 0 \]
Subtracting 1 from both sides gives:
\[ 3x = -1 \]
Dividing both sides by 3 results in:
\[ x = -\frac{1}{3} \]

### Final Solution
Thus, the solution to the equation \(9x^2 + 6x + 1 = 0\) is:
\[ x = -\frac{1}{3} \]

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