1. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $f(3 x+2)-2 f(2 x-1)$ also has exactly one root. Find the root of the trinomial $f(x)$.
Answer: -7.

Solution. Since dividing all the coefficients of the quadratic polynomial $f(x)$ by $a$ does not change its roots or the roots of the polynomial $g(x)=f(3 x+2)-2 f(2 x-1)$, we can assume that $a=1$. A quadratic polynomial has exactly one root if and only if its discriminant is zero. Therefore, $b^{2}=4 c$ and, hence, $f(x)=x^{2}+b x+\frac{b^{2}}{4}$. Then

\[
\begin{aligned}
g(x) & =f(3 x+2)-2 f(2 x-1)=(3 x+3)^{2}+b(3 x+2)+\frac{b^{2}}{2}-\left((2 x-1)^{2}+b(2 x-1)+\frac{b^{2}}{4}\right)= \\
& =x^{2}+(20-b) x+\left(2+4 b-\frac{b^{2}}{4}\right)
\end{aligned}
\]

Therefore,

\[
0=(20-b)^{2}-4\left(2+4 b-\frac{b^{2}}{4}\right)=2 b^{2}-56 b+392=2(b-14)^{2}
\]

Thus, the discriminant of the polynomial $g(x)$ is 0 only when $b=14$, which means $f(x)=$ $x^{2}+14 x+49$ and its only root is -7.