To solve the equation \( |x^2 + 4x + 3| + 2x + 5 = 0 \), we will break it down into two cases based on the definition of absolute value.
### Step 1: Analyze the expression inside the absolute value
The expression inside the absolute value is \( x^2 + 4x + 3 \). We can factor this quadratic expression:
\[
x^2 + 4x + 3 = (x + 3)(x + 1)
\]
The roots of this quadratic are \( x = -3 \) and \( x = -1 \). This means the expression changes sign at these points.
### Step 2: Determine the intervals
We will analyze the sign of \( x^2 + 4x + 3 \) in the intervals defined by the roots:
1. \( (-\infty, -3) \)
2. \( (-3, -1) \)
3. \( (-1, \infty) \)
### Step 3: Case 1: \( x^2 + 4x + 3 \geq 0 \)
This occurs in the intervals \( (-\infty, -3] \) and \( [-1, \infty) \).
**Subcase 1: \( x \in (-\infty, -3] \)**
In this interval, we can drop the absolute value:
\[
x^2 + 4x + 3 + 2x + 5 = 0
\]
Simplifying gives:
\[
x^2 + 6x + 8 = 0
\]
Factoring:
\[
(x + 4)(x + 2) = 0
\]
Thus, the solutions are:
\[
x = -4 \quad \text{and} \quad x = -2
\]
Now we check if these solutions lie in the interval \( (-\infty, -3] \):
- \( x = -4 \) is valid.
- \( x = -2 \) is not valid.
**Subcase 2: \( x \in [-1, \infty) \)**
In this interval, we again drop the absolute value:
\[
x^2 + 4x + 3 + 2x + 5 = 0
\]
This simplifies to:
\[
x^2 + 6x + 8 = 0
\]
We already found the roots \( x = -4 \) and \( x = -2 \). Neither of these solutions is valid in the interval \( [-1, \infty) \).
### Step 4: Case 2: \( x^2 + 4x + 3 < 0 \)
This occurs in the interval \( (-3, -1) \).
Here, we need to consider:
\[
-(x^2 + 4x + 3) + 2x + 5 = 0
\]
This simplifies to:
\[
-x^2 - 4x - 3 + 2x + 5 = 0
\]
Combining like terms gives:
\[
-x^2 - 2x + 2 = 0
\]
Multiplying through by -1:
\[
x^2 + 2x - 2 = 0
\]
Using the quadratic formula:
\[
x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-2)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 8}}{2} = \frac{-2 \pm \sqrt{12}}{2} = \frac{-2 \pm 2\sqrt{3}}{2} = -1 \pm \sqrt{3}
\]
Thus, the solutions are:
\[
x = -1 + \sqrt{3} \quad \text{and} \quad x = -1 - \sqrt{3}
\]
Now we check if these solutions lie in the interval \( (-3, -1) \):
- \( x = -1 - \sqrt{3} \) is valid (approximately -2.73).
- \( x = -1 + \sqrt{3} \) is not valid (approximately 0.73).
### Final Solution
The valid solutions to the original equation are:
\[
x = -4 \quad \text{and} \quad x = -1 - \sqrt{3}
\]

Solve `|x^2+4x+3|+2x+5=0.`

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