1. Given quadratic trinomials $f_{1}(x)=x^{2}-a x-3, f_{2}(x)=x^{2}+2 x-b, f_{3}(x)=3 x^{2}+(2-2 a) x-6-b$ and $f_{4}(x)=3 x^{2}+(4-a) x-3-2 b$. Let the differences of their roots be $A, B, C$ and $D$ respectively. It is known that $|C| \neq|D|$. Find the ratio $\frac{A^{2}-B^{2}}{C^{2}-D^{2}}$. The values of $A, B, C, D, a, b$ are not given.
Answer: 3.

Solution. Let $\alpha x^{2}+\beta x+\gamma$ be a quadratic trinomial with a positive discriminant $T$. Then its roots are determined by the formula $x_{1,2}=\frac{-b \pm \sqrt{T}}{2 a}$, so $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=\frac{\sqrt{T}}{|a|}$. Applying this formula four times, we get

$$
A=\sqrt{a^{2}+12}, B=\sqrt{4+4 b}, C=\frac{1}{3} \sqrt{(2-2 a)^{2}+12(6+b)}, D=\frac{1}{3} \sqrt{(4-a)^{2}+12(3+2 b)}
$$

From this, it follows that $C^{2}-D^{2}=\frac{1}{9}\left(\left(4 a^{2}-8 a+12 b+76\right)-\left(a^{2}-8 a+24 b+52\right)\right)=\frac{1}{3}\left(a^{2}-4 b+8\right), A^{2}-$ $B^{2}=a^{2}-4 b+8$. Therefore, the desired ratio is 3.