I'm recalling this question from memory, so I may be messing it up a bit.
Let $a/3+b/2+c=0$. Show that $ax^2+bx+c=0$ has at least one root in $[0,1]$ using the Mean Value Theorem.
Let $f(x)=ax^2+bc+c$. Then $f(0)=c$ and $f(1)=a+b+c$. Also $f'(x)=2ax+b$. So there exists $f(\xi)=[f(1)-f(0)]/1=a+b-c$. Then $a+b-c=2a\xi+b \Rightarrow (a-c)/2a=\xi$.
I'm not sure if this is right or where to go from here.