2. Non-zero numbers $a, b$ and $c$ are such that the doubled roots of the quadratic polynomial $x^{2}+a x+b$ are the roots of the polynomial $x^{2}+b x+c$. What can the ratio $a / c$ be?
Answer: $1 / 8$.

Solution. Let $x_{1}$ and $x_{2}$ be the roots of the first equation. Then, by Vieta's theorem, $a=-x_{1}-x_{2}$ and $b=x_{1} x_{2}$. Since $2 x_{1}$ and $2 x_{2}$ are the roots of the second equation, $b=-2 x_{1}-2 x_{2}=2 a$ and $c=4 x_{1} x_{2}=4 b$. Therefore, $\frac{a}{c}=$ $\frac{b / 2}{4 b}=\frac{1}{8}$.