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

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

From this, it follows that $C^{2}-D^{2}=\frac{1}{4}\left(\left(b^{2}-12 b-24 a+28\right)-\left(9 b^{2}-12 b+8 a+28\right)\right)=-2\left(b^{2}+4 a\right)$, $A^{2}-B^{2}=-\left(b^{2}+4 a\right)$. Therefore, the desired ratio is 2.