4. Given three non-zero real numbers $a, b, c$ such that the equations: $a x^{2}+b x+c=0, b x^{2}+c x+a=0, c x^{2}+a x+b=0$ each have two roots. How many of the roots of these equations can be negative?
Answer: 2.

Sketch of the solution.

If the numbers $a, b, c$ are replaced by their opposites, then the "new" equations will have the same set of roots as the original ones. There are two possible cases: the numbers $a, b, c$ are of the same sign; among them, there are both positive and negative numbers.

First case. Due to the initial observation, we can assume that they are positive. And due to the cyclic symmetry of the coefficients of the equations, we can assume that $0<a \leq b, 0<a \leq c$. Multiplying these inequalities, we get $\mathrm{a}^{2} \leq \mathrm{bc}<4 \mathrm{bc}$. From this inequality, it follows that the discriminant of the equation $c x^{2}+a x+b=0$ is negative, and the equation has no roots. This case is impossible.

Second case. Due to the initial observation and the cyclic symmetry of the coefficients of the equations, we can assume that $0<a, 0<c, b<0$. From these inequalities, it follows that in the equations $b x^{2}+c x+a=0$, $c x^{2}+a x+b=0$, the product of the roots is negative, and thus exactly one root is negative. In the equation $a x^{2}+b x+c=0$, the product of the roots is positive and the sum is positive, meaning both roots are positive. In the end, we have two negative roots.