Solve : `2|x+1|^(2)-|x+1|=3 , x in R `

Answer

To solve the equation \( 2|x+1|^2 - |x+1| = 3 \), we will consider two cases based on the definition of the absolute value function.

### Step 1: Define the cases for \( |x+1| \)

1. **Case 1:** \( x + 1 \geq 0 \) (i.e., \( x \geq -1 \))
- Here, \( |x+1| = x + 1 \).

2. **Case 2:** \( x + 1 < 0 \) (i.e., \( x < -1 \))
- Here, \( |x+1| = -(x + 1) = -x - 1 \).

### Step 2: Solve Case 1

For \( x \geq -1 \):
\[ 2|x+1|^2 - |x+1| = 3 \]
Substituting \( |x+1| = x + 1 \):
\[ 2(x + 1)^2 - (x + 1) = 3 \]
Expanding this:
\[ 2(x^2 + 2x + 1) - (x + 1) = 3 \]
\[ 2x^2 + 4x + 2 - x - 1 = 3 \]
Combining like terms:
\[ 2x^2 + 3x + 1 = 3 \]
Subtracting 3 from both sides:
\[ 2x^2 + 3x - 2 = 0 \]

### Step 3: Factor the quadratic equation

To factor \( 2x^2 + 3x - 2 = 0 \):
\[ (2x - 1)(x + 2) = 0 \]
Setting each factor to zero gives:
1. \( 2x - 1 = 0 \) → \( x = \frac{1}{2} \)
2. \( x + 2 = 0 \) → \( x = -2 \)

### Step 4: Check the domain for Case 1

Since we are in Case 1 where \( x \geq -1 \):
- \( x = \frac{1}{2} \) is valid.
- \( x = -2 \) is not valid (since \(-2 < -1\)).

### Step 5: Solve Case 2

For \( x < -1 \):
\[ 2|x+1|^2 - |x+1| = 3 \]
Substituting \( |x+1| = -x - 1 \):
\[ 2(-x - 1)^2 - (-x - 1) = 3 \]
Expanding this:
\[ 2(x^2 + 2x + 1) + (x + 1) = 3 \]
\[ 2x^2 + 4x + 2 + x + 1 = 3 \]
Combining like terms:
\[ 2x^2 + 5x + 3 = 3 \]
Subtracting 3 from both sides:
\[ 2x^2 + 5x = 0 \]

### Step 6: Factor the quadratic equation for Case 2

Factoring out \( x \):
\[ x(2x + 5) = 0 \]
Setting each factor to zero gives:
1. \( x = 0 \) (not valid since \( 0 \geq -1 \))
2. \( 2x + 5 = 0 \) → \( x = -\frac{5}{2} \)

### Step 7: Check the domain for Case 2

Since we are in Case 2 where \( x < -1 \):
- \( x = -\frac{5}{2} \) is valid (since \(-\frac{5}{2} < -1\)).

### Final Solutions

The valid solutions to the equation \( 2|x+1|^2 - |x+1| = 3 \) are:
- \( x = \frac{1}{2} \)
- \( x = -\frac{5}{2} \)