#### Answer

As long as a is greater than $\sqrt{12}$ or less than $-\sqrt{12}$, there will be two equilibria.

#### Work Step by Step

To find the equilibria, we take the derivative of the potential energy function and set it equal to zero. Thus, we find:
$0 = U_0(\frac{3x^2}{x_0^3}+\frac{2ax}{x_0^2}+\frac{4}{x_0})$
We only care about the determinant of the quadratic formula, so we find:
$\sqrt{(2a)^2 - 4(3)(4)}$
Here, we see that as long as a is greater than $\sqrt{12}$ or less than $-\sqrt{12}$, there will be two equilibria.
We take the second derivative of the original function to find:
$\frac{dU^2}{d^2x}=U_0(\frac{6x}{x_0^3}+\frac{2a}{x_0^2})$
Whether an equlibria is stable or unstable depends on the value of $x_0$. However, we know that if $\frac{dU^2}{d^2x}$ is positive, the eqilibrium will be stable, and if $\frac{dU^2}{d^2x}$ is negative, the equilibrium will be unstable.