3. The coefficients of the quadratic trinomials: $f_{i}(x)=x^{2}+b_{i} x+c_{i}$, satisfy the equations $\frac{b_{i+1}}{b_{i}}=2, c_{i}=-32 \cdot b_{i}-1024(i=1,2, \cdots)$. It is known that the roots of the polynomial $f_{1}(x)$ are the numbers 32 and -31. a) Find the roots of the quadratic trinomial $f_{12}$; b) find the roots of the quadratic trinomial $f$.
Answer: $2016: 32$.

Solution: By Vieta's theorem: $b_{1}=-(32+(-31))=-1, c_{1}=-32 \cdot(-1)-1024=$ $32-1024=-992$. Therefore, $b_{i}=-2^{i-1}, c_{i}=-32 \cdot\left(-2^{i-1}\right)-1024=2^{i+4}-2^{10}$.

Solving the quadratic equation

$$
x^{2}-2^{i-1} x+2^{i+4}-2^{10}=0
$$

we find:

$D=\left(2^{i-1}\right)^{2}-4\left(2^{i+4}-2^{10}\right)=\left(2^{i-1}\right)^{2}-2 \cdot 2^{i-1} \cdot 2^{6}+2^{12}=\left(2^{i-1}-2^{6}\right)^{2}, \sqrt{D}=\left|2^{i-1}-2^{6}\right|$.
a) For $i>6 x_{1}=\frac{2^{i-1}+2^{i-1}-2^{6}}{2}=2^{i-1}-2^{5}, x_{2}=\frac{2^{i-1}-2^{i-1}+2^{6}}{2}=2^{5}$.

For $i \leq 6 x_{1}=\frac{2^{i-1}-2^{i-1}+2^{6}}{2}=2^{5}, x_{2}=\frac{2^{i-1}+2^{i-1}-2^{6}}{2}=2^{i-1}-2^{5}$.

b) For $i=12$ we obtain

$$
x_{1}=2016, x_{2}=32
$$

Evaluation criteria. Correct solution - 20 points; sign not considered when finding the root of the discriminant - 15 points; only part a) solved - 12 points; first terms of the sequence found correctly, but answer is incorrect - 6 points; only the correct answer - 3 points.