4A. The product of one root of the quadratic equation $a x^{2}+b x+b=0$ with one root of the quadratic equation $a x^{2}+a x+b=0$ is 1.

Find the roots of the equations.
Solution. Let $y$ and $z$ be the roots of the equations

$$
a x^{2}+a x+b=0 \text { and } a x^{2}+b x+b=0
$$

respectively, for which $y z=1$. Then $z=\frac{1}{y}$ and $a \frac{1}{y^{2}}+b \frac{1}{y}+b=0$, i.e., $b y^{2}+b y+a=0$. If we add the equations $a y^{2}+a y+b=0$ and $b y^{2}+b y+a=0$, we get

$$
(a+b) y^{2}+(a+b) y+a+b=0
$$

i.e.,

$$
(a+b)\left(y^{2}+y+1\right)=0
$$

The expression $y^{2}+y+1$ is positive for every $y \in \mathbb{R}$. Therefore, $a+b=0$, i.e., $b=-a$. Thus, the equations are $a x^{2}+a x-a=0$ and $a x^{2}-a x-a=0$. Since these are quadratic equations, $a \neq 0$. Therefore, the given equations are

$$
x^{2}+x-1=0 \text { and } x^{2}-x-1=0 \text {. }
$$

Their solutions are $x_{1,2}=\frac{-1 \pm \sqrt{5}}{2}$ and $x_{1,2}=\frac{1 \pm \sqrt{5}}{2}$, respectively.