2013 AMC 10B Problem 19

Let the common difference be $d$, so $c=b-d$ and $a=b+d$. A quadratic with exactly one real root has discriminant $0$, so $b^2-4ac=0$. Substituting gives $b^2=4(b+d)(b-d)=4b^2-4d^2$, hence $4d^2=3b^2$. Since $b,d\ge0$, $2d=\sqrt3 b$. The double root is $\frac{-b}{2a}=\frac{-b}{2(b+d)}=\frac{-1}{2+\sqrt3}=-2+\sqrt3$. Thus, the correct answer is D .