Given the force F = 2x^2 - 3x - 2. Choose the | vedclass

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(A) For equilibrium,the net force $F$ must be zero: $2x^2 - 3x - 2 = 0$.Factoring the quadratic equation: $2x^2 - 4x + x - 2 = 0 \Rightarrow 2x(x - 2)

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$x = -1/2$ is a position of stable equilibrium.

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(A) For equilibrium,the net force $F$ must be zero: $2x^2 - 3x - 2 = 0$.Factoring the quadratic equation: $2x^2 - 4x + x - 2 = 0 \Rightarrow 2x(x - 2) + 1(x - 2) = 0 \Rightarrow (x - 2)(2x + 1) = 0$.The equilibrium positions are $x = 2$ and $x = -1/2$.The potential energy $U$ is related to force by $F = -dU/dx$,so $dU/dx = -F = -(2x^2 - 3x - 2)$.To determine stability,we check the second derivative $d^2U/dx^2 = -dF/dx = -(4x - 3)$.For $x = 2$: $d^2U/dx^2 = -(4(2) - 3) = -5 For $x = -1/2$: $d^2U/dx^2 = -(4(-1/2) - 3) = -(-2 - 3) = 5 > 0$. Since the second derivative is positive,$x = -1/2$ is a position of stable equilibrium.

Given the force $F = 2x^2 - 3x - 2$. Choose the correct option regarding the equilibrium positions.

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