Problem 4. Given a quadratic trinomial $g(x)$, which has exactly one root. Find this root, if it is known that the polynomial $g(a x+b)+g(c x+d)$ $(a \neq c)$ also has exactly one root.
Solution. Let $x_{0}$ be the unique root of the polynomial $g(x)$. Then the function $g(x)$ maintains its sign at all points $x \neq x_{0}$ and only $g\left(x_{0}\right)=0$. Therefore, the root of the polynomial $f(x)=g(a x+b)+g(c x+d)$ is only such a point $x_{1}$ that $a x_{1}+b=c x_{1}+d=x_{0}$. We obtain the parametric equation

$$
(a-c) x_{1}=(d-b)
$$

Since $a \neq c$ by condition, then $x_{1}=(d-b) /(a-c)$,

$$
x_{0}=a(d-b) /(a-c)+b=(a d-b c) /(a-c)
$$

Answer: $x_{0}=(a d-b c) /(a-c)$.