1. Given quadratic trinomials $f_{1}(x)=x^{2}-2 x+a, f_{2}(x)=x^{2}+b x-2, f_{3}(x)=4 x^{2}+(b-6) x+3 a-2$ and $f_{4}(x)=4 x^{2}+(3 b-2) x-6+a$. Let the differences of their roots be $A, B, C$ and $D$ respectively. It is known that $|C| \neq|D|$. Find the ratio $\frac{A^{2}-B^{2}}{C^{2}-D^{2}}$. The values of $A, B, C, D, a, b$ are not given.
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

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

From this, it follows that $C^{2}-D^{2}=\frac{1}{16}\left(\left(b^{2}-48 a-12 b+68\right)-\left(9 b^{2}-16 a-12 b+100\right)\right)=\frac{1}{2}\left(-b^{2}-4 a-4\right)$, $A^{2}-B^{2}=-b^{2}-4 a-4$. Therefore, the desired ratio is 2.