Problem 9.1. Find all real numbers $a$ for which the equation $x^{2}+a x+3 a^{2}-$ $7 a-19=0$ has real roots $x_{1}$ and $x_{2}$ such that

$$
\frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}=-\frac{2 a}{13}
$$

Peter Boyvalenkov
9.1. Using Vieta's formulas we get

$$
\frac{1}{x_{1}-2}+\frac{1}{x_{2}-2}=\frac{x_{1}+x_{2}-4}{\left(x_{1}-2\right)\left(x_{2}-2\right)}=-\frac{a+4}{3 a^{2}-5 a-15}
$$

Therefore $3 a^{2}-5 a-15 \neq 0$ and

$$
\frac{a+4}{3 a^{2}-5 a-15}=\frac{2 a}{13}
$$

Hence $6 a^{3}-10 a^{2}-43 a-52=0 \Longleftrightarrow(a-4)\left(6 a^{2}+14 a+13\right)=0$, i.e. $a=4$. In this case $x_{1,2}=-2 \pm \sqrt{15}$.