1. Given quadratic trinomials $f_{1}(x)=x^{2}-x+2 a, f_{2}(x)=x^{2}+2 b x+3, f_{3}(x)=4 x^{2}+(2 b-3) x+6 a+3$ and $f_{4}(x)=4 x^{2}+(6 b-1) x+9+2 a$. 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: $\frac{1}{2}$.

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

$$
A=\sqrt{1-8 a}, B=\sqrt{4 b^{2}-12}, C=\frac{1}{4} \sqrt{(2 b-3)^{2}-16(6 a+3)}, D=\frac{1}{4} \sqrt{(6 b-1)^{2}-16(9+2 a)}
$$

From this, it follows that $C^{2}-D^{2}=\frac{1}{16}\left(\left(4 b^{2}-12 b-96 a-39\right)-\left(36 b^{2}-12 b-32 a-143\right)\right)=\frac{1}{2}\left(13-8 a-4 b^{2}\right)$, $A^{2}-B^{2}=13-8 a-4 b^{2}$. Therefore, the desired ratio is $\frac{1}{2}$.