The solution of the equation `|x+1|^(2)-2|x+2|-26=0` is

Options

A. `pm7`
B. `-7, sqrt(29)`
C. `pm sqrt(29)`
D. `-7, 29`

Answer

To solve the equation \( |x+1|^2 - 2|x+2| - 26 = 0 \), we will analyze the absolute values and break the problem into cases based on the values of \( x \).

### Step 1: Identify the critical points
The absolute values \( |x+1| \) and \( |x+2| \) change at the points \( x = -1 \) and \( x = -2 \). We will consider three cases based on these critical points:
1. \( x < -2 \)
2. \( -2 \leq x < -1 \)
3. \( x \geq -1 \)

### Step 2: Case 1: \( x < -2 \)
In this case, both \( x + 1 \) and \( x + 2 \) are negative. Therefore:
\[ |x+1| = -(x+1) = -x - 1 \]
\[ |x+2| = -(x+2) = -x - 2 \]
Substituting these into the equation:
\[ (-x - 1)^2 - 2(-x - 2) - 26 = 0 \]
Calculating this gives:
\[ (x^2 + 2x + 1) + 2x + 4 - 26 = 0 \]
\[ x^2 + 4x - 21 = 0 \]
Now, we can factor this quadratic:
\[ (x + 7)(x - 3) = 0 \]
Thus, the solutions are:
\[ x = -7 \quad \text{or} \quad x = 3 \]
Since we are in the case \( x < -2 \), we accept \( x = -7 \).

### Step 3: Case 2: \( -2 \leq x < -1 \)
In this case, \( x + 1 \) is negative and \( x + 2 \) is non-negative. Therefore:
\[ |x+1| = -(x+1) = -x - 1 \]
\[ |x+2| = x + 2 \]
Substituting these into the equation:
\[ (-x - 1)^2 - 2(x + 2) - 26 = 0 \]
Calculating this gives:
\[ (x^2 + 2x + 1) - 2x - 4 - 26 = 0 \]
\[ x^2 - 29 = 0 \]
This simplifies to:
\[ x^2 = 29 \]
Thus, the solutions are:
\[ x = \sqrt{29} \quad \text{or} \quad x = -\sqrt{29} \]
Since we are in the case \( -2 \leq x < -1 \), we accept \( x = -\sqrt{29} \).

### Step 4: Case 3: \( x \geq -1 \)
In this case, both \( x + 1 \) and \( x + 2 \) are non-negative. Therefore:
\[ |x+1| = x + 1 \]
\[ |x+2| = x + 2 \]
Substituting these into the equation:
\[ (x + 1)^2 - 2(x + 2) - 26 = 0 \]
Calculating this gives:
\[ (x^2 + 2x + 1) - 2x - 4 - 26 = 0 \]
\[ x^2 - 29 = 0 \]
This simplifies to:
\[ x^2 = 29 \]
Thus, the solutions are:
\[ x = \sqrt{29} \quad \text{or} \quad x = -\sqrt{29} \]
Since we are in the case \( x \geq -1 \), we accept \( x = \sqrt{29} \).

### Final Solutions
The solutions to the original equation are:
\[ x = -7, \quad x = -\sqrt{29}, \quad x = \sqrt{29} \]