Let the roots of $ax^2+bx+c=0$ be $r$ and $s$. The equation with roots $ar+b$ and $as+b$ is:
$\textbf{(A)}\ x^2-bx-ac=0\qquad \textbf{(B)}\ x^2-bx+ac=0 \qquad\\ \textbf{(C)}\ x^2+3bx+ca+2b^2=0 \qquad \textbf{(D)}\ x^2+3bx-ca+2b^2=0 \qquad\\ \textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0$
From [Vieta's Formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas), $r+s=\frac{-b}{a}$ and $rs=\frac{c}{a}$ in the original quadratic.
The sum of the roots in the new quadratic is
\[ar + b + as + b\]
\[a(r+s) + 2b\]
\[b\]
The product of the roots in the new quadratic is
\[(ar + b)(as + b)\]
\[a^2rs + arb + asb + b^2\]
\[a^2rs + ab(r+s) + b^2\]
\[ac - b^2 + b^2\]
\[ac\]
Thus, the new quadratic is $x^2-bx+ac$.  The answer is $\boxed{\textbf{(B)}}$.