Algebra

x^2 + 8x + 4 will have the same roots as 2(x^2 + 8x + 4).

2(x^2 + 8x + 4) = 2x^2 + 16x + 8

2 + 8 = 10

=^._.^=

The roots of  \(x^2 + 8x + 4\) are the same as
the roots of \(Ax^2 + 16x + B\) .
What is \(A + B\)?

\(\begin{array}{|rcll|} \hline -(x_1+x_2) = 8 &=& \dfrac{16}{A} \\\\ 8A &=& 16 \\ \mathbf{A} &=& \mathbf{2} \\ \hline x_1*x_2 =4 &=& \dfrac{B}{A} \\\\ B &=& 4A \\ B &=& 4*2 \\ \mathbf{B} &=& \mathbf{8} \\ \hline A+ B &=& 2+8 \\ \mathbf{A+B} &=& \mathbf{10} \\ \hline \end{array}\)

By Vieta's, we have $r_1 + r_2 = - \frac{8}{1} = -8$ and $r_1 r_2 = \frac{4}{1} = 4.$

We also have $r_1 + r_2 = - \frac{16}{A}$ and $r_1 r_2 = \frac{B}{A}.$

Thus, we have $- \frac{16}{A} = - \frac{8}{1} \Rightarrow \frac{2}{A} = 1 \Rightarrow A = 2.$

We also have $4 = \frac{B}{A} \Rightarrow 4 = \frac{B}{2} \Rightarrow B = 8.$

$2+8 = \boxed{10}.$

I could've noted the fact that $a_{n-1} = \frac{b_{n-1}}{2},$ but visualizations are somewhat unreliable.