The function $f(x) = x^3-4a^2x$ has primitive function $F(x)$ . Find the constant $a$ so that $F(x)$ has the minimum value $0$ and $F(2)=4$. 
The function $f(x) = x^3-4a^2x$ has primitive function $F(x)$ . Find the constant $a$ so that $F(x)$ has the minimum value $0$ and $F(2)=4$.

This is what I tried below.
I am not sure how to use the minimum value information for the $F(x)$ function.
I tried to find $c$ in terms of $a$ to help simplify the problem, but I don't think it helps.
I think I might have to set up some sort of system of equations with the two pieces of information.

 A: If $F(x)=\frac14x^4-2a^2x^2+c$, the minimum of $F$ is attained at a point $x$ such that $f(x)=0$, that is, it is attained at $\pm a$ or at $0$. Actually, it cannot be attained at $0$, since $F''(0)=f'(0)=-4a^2$, and therefore $F$ has a local maximum at $0$. On the other hand,$$F(\pm a)=c-\frac74a^4,$$and therefore $F(\pm a)=0\iff a=0\vee c=\frac74a^4$.
If $a=0$, then you just take $c=0$, so that $F(2)=4$. And if $c=\frac74a^4$, then$$F(x)=\frac14x^4-2a^2x^2+\frac74a^4,$$and then $F(2)=4-8a^2+\frac74a^4$. So, $F(2)=4\iff a=0\vee a=\pm4\sqrt{\frac27}$.
A: Following your work with $F(2) = 4 \implies c = 8a^2$. Thus: $F(x) = \dfrac{x^4}{4} -2a^2x^2+8a^2= \left(\dfrac{x^2}{2} - 2a^2\right)^2+8a^2 - 4a^4\ge 8a^2-4a^4$. This minimum value equals to $0$. So $8a^2 - 4a^4 = 0\implies 4a^2(2-a^2)=0\implies a = 0,\pm \sqrt{2}$.
A: The equation you have arrived at is a quadratic in $x^2$.
$$
\frac{1}{4} x^4 - 2a^2x^2 + 8a^2 = 0
$$
is equivalent to
$$
\left(\frac{1}{4}\right) u^2 - (2a^2) u + (8a^2) = 0
$$
where I have substituted $u=x^2$. If the minimum value of this quadratic is zero, it means its vertex (peak) touches the $x$-axis. In that case, it cannot intersect the $x$-axis at two points: it will have only one root. This implies that the discriminant of this quadratic is equal to 0.
\begin{align*}
(2a^2)^2 - 4\left(\frac{1}{4}\right)(8a^2) &= 0 \\
4a^4 &= 8a^2 \\
a &\in \{0, \sqrt{2}, \sqrt{- 2} \}
\end{align*}
