2. Each of the equations $a x^{2}-b x+c=0$ and $c x^{2}-a x+b=0$ has two distinct real roots. The sum of the roots of the first equation is non-negative, and the product of the roots of the first equation is 9 times the sum of the roots of the second equation. Find the ratio of the sum of the roots of the first equation to the product of the roots of the second equation.
# Solution.

From the condition, it follows that the coefficients $a, c \neq 0$.

By Vieta's theorem, from the condition it follows that $\frac{c}{a}=9 \frac{a}{c}$. Hence, $c^{2}=9 a^{2}$, which means $\left[\begin{array}{l}c=3 a, \\ c=-3 a .\end{array}\right.$

1 case.

$c=3 a$.

We get the equations $a x^{2}-b x+3 a=0$ and $3 a x^{2}-a x+b=0$.

By the condition $\frac{b}{a} \geq 0, b^{2}-12 a^{2}>0, a^{2}-12 a b>0$.

That is, $\frac{b}{a} \geq 0, \frac{b^{2}}{a^{2}}>12,1-12 \frac{b}{a}>0$. Making the substitution $\frac{b}{a}=t$, we get the system of inequalities:

$\left\{\begin{array}{l}t \geq 0, \\ t^{2}>12, \\ t0, a^{2}+12 a b>0$.

These inequalities are satisfied for any $a$ and $b$ such that $\frac{b}{a} \geq 0$.

The sum of the roots of the first equation is $\frac{b}{a}$. The product of the roots of the second equation is $-\frac{b}{3 a}$. The ratio is -3.

Answer: -3.

## Recommendations for checking.

Only the value $3-2$ points found.

Values $\pm 3-3$ points found.