1. Given quadratic trinomials $f_{1}(x)=x^{2}+a x+3, f_{2}(x)=x^{2}+2 x-b, f_{3}(x)=x^{2}+2(a-1) x+b+6$ and $f_{4}(x)=x^{2}+(4-a) x-2 b-3$. 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: 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 $\left|x_{2}-x_{1}\right|=\left|\frac{-b+\sqrt{T}-(-b-\sqrt{T})}{2 a}\right|=$ $\frac{\sqrt{T}}{|a|} \cdot$ Applying this formula four times, we get

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

From this, it follows that $C^{2}-D^{2}=\left(\left(4 a^{2}-8 a-4 b-20\right)-\left(a^{2}-8 a+8 b+28\right)\right)=3\left(a^{2}-4 b-16\right)$, $A^{2}-B^{2}=a^{2}-4 b-16$. Therefore, the desired ratio is 3.