1. Given quadratic trinomials $f_{1}(x)=x^{2}-2 a x+3, f_{2}(x)=x^{2}+x+b, f_{3}(x)=3 x^{2}+(1-4 a) x+6+b$ and $f_{4}(x)=3 x^{2}+(2-2 a) x+3+2 b$. Let the differences of their roots be $A, B, C$ and $D$ respectively. It is known 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 given.
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

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

From this, it follows that $C^{2}-D^{2}=\frac{1}{9}\left(\left(16 a^{2}-8 a-12 b-71\right)-\left(4 a^{2}-8 a-24 b-32\right)\right)=\frac{1}{3}\left(4 a^{2}+4 b-13\right)$, $A^{2}-B^{2}=4 a^{2}+4 b-13$. Therefore, the desired ratio is $\frac{1}{3}$.