Expanding the equation and combining like terms results in $2 x^{2}-(a+2 b+c) x+(a b+b c)=0$. By Vieta\textquotesingle s formula the sum of the roots is $\frac{-[-(a+2 b+c)]}{2}=\frac{a+2 b+c}{2}$. To maximize this expression we want $b$ to be the largest, and from there we can assign the next highest values to $a$ and $c$. So let $b=9, a=8$, and $c=7$. Then the answer is $\frac{8+18+7}{2}= 16.5$. 
Let $a, b$, and $c$ be three distinct one-digit numbers. What is the maximum value of the sum of the roots of the equation $(x-a)(x-b)+(x-b)(x-c)=0$? 
a: $$15$$ 
b: $$15.5$$ 
c: $$16$$ 
d: $$16.5$$ 
e: $$17$$ 