If `F=2x^(2)-3x-2`, then select the correct statement

Options

A. `x=-(1)/(2)` is the position of stable equilibrium
B. `x=2` is the position of stable equilibrium
C. `x=-(1)/(2)` is the position of ubstable equilibrium
D. `x=2` is the position of neutral equilibrium

Answer

To solve the problem, we need to analyze the force function given by \( F = 2x^2 - 3x - 2 \) and determine the conditions for equilibrium, as well as whether these points correspond to stable or unstable equilibrium.

### Step 1: Set the Force to Zero
At equilibrium, the net force acting on the body is zero. Therefore, we set the force function equal to zero:

\[ 2x^2 - 3x - 2 = 0 \]

### Step 2: Solve the Quadratic Equation
To solve the quadratic equation, we can factor it. We can rewrite it as:

\[ 2x^2 - 4x + x - 2 = 0 \]

Now, we can factor by grouping:

\[ (2x^2 - 4x) + (x - 2) = 0 \]

Factoring gives us:

\[ 2x(x - 2) + 1(x - 2) = 0 \]

This can be factored further:

\[ (2x + 1)(x - 2) = 0 \]

### Step 3: Find the Roots
Setting each factor equal to zero gives us the equilibrium points:

1. \( 2x + 1 = 0 \) → \( x = -\frac{1}{2} \)
2. \( x - 2 = 0 \) → \( x = 2 \)

So, the equilibrium points are \( x = -\frac{1}{2} \) and \( x = 2 \).

### Step 4: Determine Stability of Equilibrium Points
To determine whether these points are stable or unstable equilibria, we need to analyze the derivative of the force function:

\[ \frac{dF}{dx} = 4x - 3 \]

### Step 5: Evaluate the Derivative at Each Equilibrium Point
1. **At \( x = 2 \)**:
\[ \frac{dF}{dx} = 4(2) - 3 = 8 - 3 = 5 \]
Since \( \frac{dF}{dx} > 0 \), this indicates that if the system is displaced slightly, the force will act in the direction of the displacement, leading to an increase in displacement. Thus, \( x = 2 \) is an **unstable equilibrium**.

2. **At \( x = -\frac{1}{2} \)**:
\[ \frac{dF}{dx} = 4\left(-\frac{1}{2}\right) - 3 = -2 - 3 = -5 \]
Since \( \frac{dF}{dx} < 0 \), this indicates that if the system is displaced slightly, the force will act to restore the system back to the equilibrium position. Thus, \( x = -\frac{1}{2} \) is a **stable equilibrium**.

### Conclusion
The correct statement is that \( x = -\frac{1}{2} \) corresponds to a stable equilibrium.

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