Solve `|x^2+4x+3|=x+1`

Explanation

To solve the equation \( |x^2 + 4x + 3| = x + 1 \), we will follow these steps:

### Step 1: Rewrite the expression inside the absolute value
First, we can factor the quadratic expression inside the absolute value:

\[ x^2 + 4x + 3 = (x + 1)(x + 3) \]

### Step 2: Set up cases for the absolute value
The absolute value function splits our equation into two cases based on the sign of the expression inside the absolute value.

**Case 1:** \( x^2 + 4x + 3 \geq 0 \)

In this case, we have:

\[ x^2 + 4x + 3 = x + 1 \]

Rearranging gives us:

\[ x^2 + 4x + 3 - x - 1 = 0 \implies x^2 + 3x + 2 = 0 \]

Factoring this quadratic:

\[ (x + 1)(x + 2) = 0 \]

Thus, the solutions are:

\[ x = -1 \quad \text{and} \quad x = -2 \]

Next, we need to check if these solutions satisfy the condition \( x^2 + 4x + 3 \geq 0 \).

- For \( x = -1 \):

\[ (-1)^2 + 4(-1) + 3 = 1 - 4 + 3 = 0 \quad (\text{satisfies}) \]

- For \( x = -2 \):

\[ (-2)^2 + 4(-2) + 3 = 4 - 8 + 3 = -1 \quad (\text{does not satisfy}) \]

So, from Case 1, the only valid solution is \( x = -1 \).

### Step 3: Analyze Case 2
**Case 2:** \( x^2 + 4x + 3 < 0 \)

In this case, we have:

\[ -(x^2 + 4x + 3) = x + 1 \]

Rearranging gives us:

\[ -x^2 - 4x - 3 - x - 1 = 0 \implies -x^2 - 5x - 4 = 0 \implies x^2 + 5x + 4 = 0 \]

Factoring this quadratic:

\[ (x + 1)(x + 4) = 0 \]

Thus, the solutions are:

\[ x = -1 \quad \text{and} \quad x = -4 \]

Next, we need to check if these solutions satisfy the condition \( x^2 + 4x + 3 < 0 \).

- For \( x = -1 \):

\[ (-1)^2 + 4(-1) + 3 = 0 \quad (\text{does not satisfy}) \]

- For \( x = -4 \):

\[ (-4)^2 + 4(-4) + 3 = 16 - 16 + 3 = 3 \quad (\text{does not satisfy}) \]

So, from Case 2, there are no valid solutions.

### Conclusion
The only solution to the equation \( |x^2 + 4x + 3| = x + 1 \) is:

\[ \boxed{-1} \]