Solve : `|x^(2) + 3x| + x^(2) - 2 ge 0`

Options

A. `(-oo, -2//3]`
B. `(-oo, -2//3] cup [1, oo)`
C. `(-oo, -2//3] cup [1//2, oo)`
D. None of these

Answer

To solve the inequality \( |x^2 + 3x| + x^2 - 2 \geq 0 \), we will consider two cases based on the definition of the absolute value.

### Step 1: Identify the cases for the absolute value
The expression inside the absolute value, \( x^2 + 3x \), can be positive or negative. We need to find the points where it equals zero:
\[ x^2 + 3x = 0 \]
Factoring gives:
\[ x(x + 3) = 0 \]
Thus, the critical points are \( x = 0 \) and \( x = -3 \).

### Step 2: Case 1: \( x^2 + 3x \geq 0 \)
In this case, the absolute value can be removed:
\[ |x^2 + 3x| = x^2 + 3x \]
Substituting into the inequality:
\[ x^2 + 3x + x^2 - 2 \geq 0 \]
This simplifies to:
\[ 2x^2 + 3x - 2 \geq 0 \]

### Step 3: Factor the quadratic
We will factor \( 2x^2 + 3x - 2 \):
\[ 2x^2 + 4x - x - 2 \geq 0 \]
Grouping gives:
\[ (2x^2 + 4x) + (-x - 2) \geq 0 \]
Factoring out common terms:
\[ 2x(x + 2) - 1(x + 2) \geq 0 \]
Factoring further:
\[ (2x - 1)(x + 2) \geq 0 \]

### Step 4: Find the critical points
The critical points from the factors are:
1. \( 2x - 1 = 0 \) ⇒ \( x = \frac{1}{2} \)
2. \( x + 2 = 0 \) ⇒ \( x = -2 \)

### Step 5: Test intervals
We will test the intervals determined by the critical points \( -2 \) and \( \frac{1}{2} \):
- Interval 1: \( (-\infty, -2) \)
- Interval 2: \( (-2, \frac{1}{2}) \)
- Interval 3: \( (\frac{1}{2}, \infty) \)

1. **For \( x < -2 \)** (e.g., \( x = -3 \)):
\[ (2(-3) - 1)(-3 + 2) = (-7)(-1) = 7 \geq 0 \quad \text{(True)} \]

2. **For \( -2 < x < \frac{1}{2} \)** (e.g., \( x = 0 \)):
\[ (2(0) - 1)(0 + 2) = (-1)(2) = -2 \geq 0 \quad \text{(False)} \]

3. **For \( x > \frac{1}{2} \)** (e.g., \( x = 1 \)):
\[ (2(1) - 1)(1 + 2) = (1)(3) = 3 \geq 0 \quad \text{(True)} \]

### Step 6: Combine results for Case 1
The solution for this case is:
\[ x \in (-\infty, -2] \cup \left[\frac{1}{2}, \infty\right) \]

### Step 7: Case 2: \( x^2 + 3x < 0 \)
In this case, we have:
\[ |x^2 + 3x| = -(x^2 + 3x) \]
Substituting into the inequality:
\[ -(x^2 + 3x) + x^2 - 2 \geq 0 \]
This simplifies to:
\[ -3x - 2 \geq 0 \]
Rearranging gives:
\[ -3x \geq 2 \quad \Rightarrow \quad x \leq -\frac{2}{3} \]

### Step 8: Combine results for Case 2
The solution for this case is:
\[ x \in \left(-\infty, -\frac{2}{3}\right] \]

### Final Solution
Combining the results from both cases, we have:
\[ x \in (-\infty, -2] \cup \left(-\infty, -\frac{2}{3}\right] \cup \left[\frac{1}{2}, \infty\right) \]
This simplifies to:
\[ x \in (-\infty, -\frac{2}{3}] \cup \left[\frac{1}{2}, \infty\right) \]