Show whether there exists a cubic polynomial with real coefficients and distinct real roots $(r_1, r_2, r_3)$ such that its first derivative has roots $r_1,r_2$ and second derivative has root $r_3$.
By fiddling around with this, I've figured out that $r_3=\frac{r_1+r_2}{2}$ because the point where the derivative of a quadratic is $0$ (the root of its derivative) is always the midpoint of its own real roots (assuming they exist (which they do for the purposes of this problem)). Then the cubic's roots must be symmetric. But I've gotten somewhat lost from there about what conclusions I can draw to help me along. I only came up with this in my sleep last night, so I don't necessarily expect that there exists a nice, neat solution, but anything would be helpful. Thanks!
Let the roots be $a,b,c$ for simplicity.
We have three equations
\begin{align} x^3-(a+b+c)x^2+(ab+bc+ca)\,x-abc&=0 \tag{1}\label{1} ,\\ x^2-\tfrac23\,(a+b+c)\,x+\tfrac13\,(ab+bc+ca)&=0 \tag{2}\label{2} ,\\ x-\tfrac13\,(a+b+c)&=0 \tag{3}\label{3} . \end{align}
From the last one we can find the last root, say, $c$:
\begin{align} c-\tfrac13\,(a+b+c)&=0 \tag{4}\label{4} , \end{align}
\begin{align} c&=\tfrac12\,(a+b) \tag{5}\label{5} . \end{align}
Substitution of \eqref{5} into \eqref{2} gives
\begin{align} x^2-(a+b)\,x+\tfrac16\,(a^2+b^2)+\tfrac23\,ab \tag{6}\label{6} . \end{align}
In order to enforse $a,b$ to be the roots of \eqref{6}, we need to provide
\begin{align} \tfrac16\,(a^2+b^2)+\tfrac23\,ab &=ab \tag{7}\label{7} \\ \text{or, }\quad (a-b)^2&=0 \tag{8}\label{8} . \end{align}
This implies
\begin{align} b&=a .\quad c=a . \end{align}
So, the answer is: a cubic polynomial with real coefficients and distinct real roots with given properties does not exist.