A classical mechanics problem by Kirti Sharma

Knowing that $v = \frac{dx}{dt}$ and that $a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt}$ , let's first determine $\frac{dx}{dt}$ at $x = 2$ :

$v^2 = (\frac{dx}{dt})^2 = 2\alpha - 8\beta \Rightarrow \frac{dx}{dt} = \sqrt{2\alpha - 8\beta} = 0$ (i)

Now for acceleration at $x=2$ :

$\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = \frac{\alpha - 3\beta x^2}{2\sqrt{\alpha x - \beta x^3}} \cdot \sqrt{\alpha x - \beta x^3} = \frac{\alpha - 3\beta x^2}{2} \Rightarrow -8 = \frac{\alpha - 12\beta}{2}$ (ii)

Taking (i) and (ii) as a system of equations in $\alpha, \beta$ , we now obtain:

$\alpha = 8, \beta = 2 \Rightarrow \alpha + \beta = \boxed{10}.$