Absolute Value Tangents

We can split $g(x)=|ax-b|+b$ into the linear functions $h(x)=ax-b+b$ and $k(x)=-ax+b+b$ . A linear function cannot be tangent to the parabola $f(x)$ at $2$ points since linear functions do not have turning points, so $h(x)$ and $k(x)$ are tangent to $f(x)$ at seperate points. Setting $f(x)=h(x)$ and the discriminant to $0$ yields $(a-3b)^2-4(a+b)=0$ and setting $f(x)=k(x)$ and the discriminant to $0$ yields $9(a-b)^2-4(a-b)=0$ . The latter equation is in quadratic form in terms of $(a-b)$ so solving it yields $(a-b)=$ $\frac{4}{9}$ . Substituting this into the former equation and using Vieta's formulas yields $162B=-64$ and $162A=144$ so $162(A+B)=144-64=80$ which is the final answer.