`F = 2x^(2)-3x-2`. Choose correct option.

Options

A. `x = - 1//2` is position of stable equilibrium
B. x = 2 is position of stable equilibrium
C. `x = - 1//2` is position of unstabl equilibrium
D. x = 2 is position of neutral equilibrium

Answer

To solve the problem given by the force equation \( F = 2x^2 - 3x - 2 \) and to find the points of equilibrium, we will follow these steps:

### Step 1: Set the Force Equation to Zero
To find the points of equilibrium, we need to set the force equation \( F \) to zero:
\[ 2x^2 - 3x - 2 = 0 \]

### Step 2: Factor the Quadratic Equation
Next, we will factor the quadratic equation. We can rearrange the equation:
\[ 2x^2 - 4x + x - 2 = 0 \]
Now, we can group the terms:
\[ (2x^2 - 4x) + (x - 2) = 0 \]
Factoring out common terms:
\[ 2x(x - 2) + 1(x - 2) = 0 \]
Now, we can factor by grouping:
\[ (2x + 1)(x - 2) = 0 \]

### Step 3: Solve for \( x \)
Setting each factor to zero gives us the possible values of \( x \):
1. \( 2x + 1 = 0 \) leads to \( x = -\frac{1}{2} \)
2. \( x - 2 = 0 \) leads to \( x = 2 \)

### Step 4: Determine Stability of Equilibrium Points
To determine the stability of these equilibrium points, we need to find the derivative of the force function:
\[ \frac{dF}{dx} = 4x - 3 \]
Now, we will evaluate the derivative at the equilibrium points:
1. For \( x = -\frac{1}{2} \):
\[ \frac{dF}{dx} = 4\left(-\frac{1}{2}\right) - 3 = -2 - 3 = -5 \quad (\text{negative, stable equilibrium}) \]
2. For \( x = 2 \):
\[ \frac{dF}{dx} = 4(2) - 3 = 8 - 3 = 5 \quad (\text{positive, unstable equilibrium}) \]

### Conclusion
The points of equilibrium are \( x = -\frac{1}{2} \) (stable) and \( x = 2 \) (unstable). Therefore, the correct option for stable equilibrium is:
\[ \text{Option A: } x = -\frac{1}{2} \]