ACGNmath wrote: Do there exist three different quadratic trinomials $f(x), g(x), h(x)$ such that the roots of the equation $f(x)=g(x)$ are $1$ and $4$, the roots of the equation $g(x)=h(x)$ are $2$ and $5$, and the roots of the equation $h(x)=f(x)$ are $3$ and $6$? Proposed by A. Golovanov Just write : $f-g=a(x-1)(x-4)$ $g-h=b(x-2)(x-5)$ $h-f=c(x-3)(x-6)$ And so $a(x-1)(x-4)+b(x-2)(x-5)+c(x-3)(x-6)=0$ $\forall x$ Which implies $a=b=c=0$ So no.

$f(x)=g(x)+a(x-1)(x-4)=h(x)+b(x-2)(x-5)+a(x-1)(x-4)$ $f(3)=h(3),f(6)=h(6) \to a=b=0 \to $ contradiction

No . use the roots and build system of equation by the coefficient of f,g and h .you will get that coefficients of x^2 are zero