Brazilian Math Olympiad

Let $Q(x)=x^3-12x$. Then $Q^{\prime }(x)=3x^2-12$ has roots $-2$ and $2$, and $Q$ has a local minimum at $(2,-16)$ and a local maximum at $(-2,16)$. So $Q(x)=-t$ has three (not necessarily distinct) real roots for $-16\le t\le 16$ and one real root for $t<-16$ and $t>16$, which means that $\Delta (t)=0$ for $t<-16$ or $t>16$. So from now on we consider only $t\in [-16,16]$.

Let $u\le v\le w$ be the roots. Since $P_t^{\prime }(x)=0⟺x=-2$ or $x=2$ we have $u\le -2\le v\le 2\le w$. In particular, $-2\le v\le 2$, and $v$ can assume any value in this interval: if $t=-16$, $v=-2$ and if $t=16$, $v=2$.

But we know that $u+v+w=0$ and $uv+vw+uw=-12$, so $u+w=-v$ and $uw=-12-v(u+w)=v^2-12$, so $(\Delta (t){)}^2=(w-u{)}^2=(w+u{)}^2-4uw=48-3v^2$, which lies in the range $[48-3\cdot 2^2,48]=[36,48]$. So $36\le (\Delta (t){)}^2\le 48⟺6\le \Delta (t)\le 4\sqrt{3}$.

So the range of $\Delta (t)$ is $\{0\}\cup [6,4\sqrt{3}]$.