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

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

$$
A=\sqrt{a^{2}-8}, B=\sqrt{9-4 b}, C=\frac{1}{3} \sqrt{(3-2 a)^{2}-12(4+b)}, D=\frac{1}{3} \sqrt{(6-a)^{2}-12(2+b)}
$$

From this, it follows that $C^{2}-D^{2}=\frac{1}{9}\left(\left(4 a^{2}-12 a-12 b-39\right)-\left(a^{2}-12 a-24 b+12\right)\right)=\frac{1}{3}\left(a^{2}+4 b-17\right)$, $A^{2}-B^{2}=a^{2}+4 b-17$. Therefore, the desired ratio is $\frac{1}{3}$.