If `2-3x-2x^(2) ge 0`, then

Options

A. `x le -2`
B. `-2 le x le(1)/(2)`
C. `x ge -2`
D. `x le (1)/(2)`

Answer

To solve the inequality \(2 - 3x - 2x^2 \geq 0\), we will follow these steps:

### Step 1: Rewrite the Inequality
Start with the given inequality:
\[ 2 - 3x - 2x^2 \geq 0 \]

### Step 2: Multiply by -1
To simplify the expression, we can multiply the entire inequality by -1. Remember that multiplying by a negative number reverses the inequality sign:
\[ - (2 - 3x - 2x^2) \leq 0 \]
This simplifies to:
\[ -2 + 3x + 2x^2 \leq 0 \]
or rearranging gives:
\[ 2x^2 + 3x - 2 \leq 0 \]

### Step 3: Factor the Quadratic Expression
Next, we will factor the quadratic expression \(2x^2 + 3x - 2\). We can split the middle term:
\[ 2x^2 + 4x - x - 2 \leq 0 \]
Grouping the terms:
\[ (2x^2 + 4x) + (-x - 2) \leq 0 \]
Factoring out common terms:
\[ 2x(x + 2) - 1(x + 2) \leq 0 \]
Now we can factor out \((x + 2)\):
\[ (2x - 1)(x + 2) \leq 0 \]

### Step 4: Find Critical Points
Set each factor to zero to find the critical points:
1. \(2x - 1 = 0 \Rightarrow x = \frac{1}{2}\)
2. \(x + 2 = 0 \Rightarrow x = -2\)

### Step 5: Test Intervals
Now we will test the intervals determined by the critical points \(-2\) and \(\frac{1}{2}\):
- Choose a test point in each interval:
- For \(x < -2\) (e.g., \(x = -3\)):
\((2(-3) - 1)(-3 + 2) = (-7)(-1) = 7 > 0\)
- For \(-2 < x < \frac{1}{2}\) (e.g., \(x = 0\)):
\((2(0) - 1)(0 + 2) = (-1)(2) = -2 < 0\)
- For \(x > \frac{1}{2}\) (e.g., \(x = 1\)):
\((2(1) - 1)(1 + 2) = (1)(3) = 3 > 0\)

### Step 6: Determine the Solution Set
The inequality \((2x - 1)(x + 2) \leq 0\) holds true in the interval \([-2, \frac{1}{2}]\). Since the inequality includes equal to zero, we include the endpoints.

### Final Solution
Thus, the solution to the inequality \(2 - 3x - 2x^2 \geq 0\) is:
\[ x \in [-2, \frac{1}{2}] \]