1. The quadratic trinomial $f(x)=a x^{2}+b x+c$ has exactly one root, the quadratic trinomial $2 f(2 x-3)-f(3 x+1)$ also has exactly one root. Find the root of the trinomial $f(x)$.
Answer: -11.

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 $g(x)=2 f(2 x-3)-f(3 x+1)$, 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,

\[
\begin{aligned}
g(x) & =2 f(2 x-3)-f(3 x+1)=2(2 x-3)^{2}+2 b(2 x-3)+\frac{b^{2}}{2}-\left((3 x+1)^{2}+b(3 x+1)+\frac{b^{2}}{4}\right)= \\
& =-x^{2}+(b-30) x+\left(17-7 b+\frac{b^{2}}{4}\right)
\end{aligned}
\]

Therefore,

\[
0=(b-30)^{2}+4\left(17-7 b+\frac{b^{2}}{4}\right)=2 b^{2}-88 b+968=2(b-22)^{2}
\]

Thus, the discriminant of the polynomial $g(x)$ is 0 only when $b=22$, which means $f(x)=$ $x^{2}+22 x+121$ and its only root is -11.