3. Variant 1.

Find the ratio $\frac{16 b^{2}}{a c}$, given that one of the roots of the equation $a x^{2}+b x+c=0$ is 4 times the other.
Answer: 100.

Solution.

By Vieta's theorem, $-\frac{b}{a}=x_{1}+x_{2}=5 x_{2}$ and $\frac{c}{a}=x_{1} \cdot x_{2}=4 x_{2}^{2}$. Express $x_{2}=-\frac{b}{5 a}$ from the first equation and substitute it into the second: $\frac{c}{a}=\frac{4 b^{2}}{25 a^{2}}$. Then find $\frac{b^{2}}{a c}=\frac{25}{4}$.