The solution of `|x^(2) + 3x|+x^(2)-2 ge 0` is `

Options

A. `(-oo,1)`
B. (0,1)
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 break it down into cases based on the expression inside the absolute value.

### Step 1: Identify Cases for the Absolute Value
The expression \( |x^2 + 3x| \) can be split into two cases:
1. Case 1: \( x^2 + 3x \geq 0 \)
2. Case 2: \( x^2 + 3x < 0 \)

### Step 2: Solve Case 1
In Case 1, since \( x^2 + 3x \geq 0 \), we can remove the absolute value:
\[ x^2 + 3x + x^2 - 2 \geq 0 \]
This simplifies to:
\[ 2x^2 + 3x - 2 \geq 0 \]

### Step 3: Factor the Quadratic
To solve \( 2x^2 + 3x - 2 \geq 0 \), we can factor it:
\[ 2x^2 + 4x - x - 2 \geq 0 \implies (2x - 1)(x + 2) \geq 0 \]

### Step 4: Find Critical Points
The critical points are found by setting the factors to zero:
1. \( 2x - 1 = 0 \) gives \( x = \frac{1}{2} \)
2. \( x + 2 = 0 \) gives \( x = -2 \)

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

Using a test point from each interval:
1. For \( x = -3 \): \( (2(-3) - 1)(-3 + 2) = (-7)(-1) > 0 \) (valid)
2. For \( x = 0 \): \( (2(0) - 1)(0 + 2) = (-1)(2) < 0 \) (invalid)
3. For \( x = 1 \): \( (2(1) - 1)(1 + 2) = (1)(3) > 0 \) (valid)

Thus, the solution for Case 1 is:
\[ x \in (-\infty, -2] \cup [\frac{1}{2}, \infty) \]

### Step 6: Solve Case 2
In Case 2, since \( x^2 + 3x < 0 \), we have:
\[ -(x^2 + 3x) + x^2 - 2 \geq 0 \]
This simplifies to:
\[ -3x - 2 \geq 0 \implies 3x + 2 \leq 0 \implies x \leq -\frac{2}{3} \]

### Step 7: Find the Range for Case 2
From Case 2, we also have the condition \( x^2 + 3x < 0 \) which factors to:
\[ x(x + 3) < 0 \]
The critical points are \( x = 0 \) and \( x = -3 \). Testing intervals:
1. For \( x = -2 \): \( (-2)(-2 + 3) > 0 \) (invalid)
2. For \( x = -4 \): \( (-4)(-4 + 3) > 0 \) (valid)
3. For \( x = -1 \): \( (-1)(-1 + 3) < 0 \) (valid)

Thus, the solution for Case 2 is:
\[ x \in (-3, 0) \]

### Step 8: Combine Solutions
Now we combine the solutions from both cases:
- From Case 1: \( (-\infty, -2] \cup [\frac{1}{2}, \infty) \)
- From Case 2: \( (-3, 0) \)

The combined solution is:
\[ x \in (-\infty, -2] \cup (-3, 0) \cup [\frac{1}{2}, \infty) \]

### Final Answer
Thus, the solution to the inequality \( |x^2 + 3x| + x^2 - 2 \geq 0 \) is:
\[ (-\infty, -2] \cup (-3, 0) \cup [\frac{1}{2}, \infty) \]