If `5x^(2) + 4xy + y^(2) + 2x + 1= 0` then find the value of x, y

Answer

To solve the equation \( 5x^2 + 4xy + y^2 + 2x + 1 = 0 \), we will rearrange and factor it.

### Step-by-Step Solution:

1. **Rearrange the equation**:
The given equation is:
\[ 5x^2 + 4xy + y^2 + 2x + 1 = 0 \]

2. **Group terms**:
We can group the terms to help us factor:
\[ (5x^2 + 4xy + y^2) + (2x + 1) = 0 \]

3. **Complete the square**:
We notice that \( 5x^2 + 4xy + y^2 \) can be rearranged into a perfect square. We can express it as:
\[ (2x + y)^2 + (2x + 1) = 0 \]
Here, we recognize that \( (2x + y)^2 \) is a perfect square.

4. **Set each part to zero**:
For the sum of squares to equal zero, each square must independently be zero:
\[ (2x + y)^2 = 0 \quad \text{and} \quad (2x + 1) = 0 \]

5. **Solve the first equation**:
From \( (2x + y) = 0 \):
\[ y = -2x \]

6. **Solve the second equation**:
From \( (2x + 1) = 0 \):
\[ 2x = -1 \quad \Rightarrow \quad x = -\frac{1}{2} \]

7. **Substitute \( x \) back to find \( y \)**:
Substitute \( x = -\frac{1}{2} \) into \( y = -2x \):
\[ y = -2 \left(-\frac{1}{2}\right) = 1 \]

8. **Final values**:
Thus, the values of \( x \) and \( y \) are:
\[ x = -\frac{1}{2}, \quad y = 1 \]

### Summary of the Solution:
The solution to the equation \( 5x^2 + 4xy + y^2 + 2x + 1 = 0 \) gives us:
\[ x = -\frac{1}{2}, \quad y = 1 \]