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

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 $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=$ $\frac{\sqrt{T}}{|a|} \cdot$ Applying this formula four times, we get

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

From this, it follows that $C^{2}-D^{2}=\frac{1}{16}\left(\left(b^{2}-6 b+48 a-23\right)-\left(9 b^{2}-6 b+16 a-95\right)\right)=\frac{1}{2}\left(9+4 a-b^{2}\right)$, $A^{2}-B^{2}=9+4 a-b^{2}$. Therefore, the desired ratio is $\frac{1}{2}$.