2013 AMC 10B Problems/Problem 19

Problem

The real numbers form an arithmetic sequence with The quadratic has exactly one root. What is this root?

I didn't really understand this problem...

If there is one root, the discriminant is 0. Therefore, b^2-4ac=0. Let's define a as (b-x) and c as (b+x). If b^2-4ac=0, by substitution, b^2-4(b-x)(b+x)=0 and distributing, b^2-4b^2+4x^2=0. Thus, through simple arithmetic, 4x^2=3b^2 and 2x=\sqrt{3}\b. Then, the root, by quadratic formula, is -b +/- \sqrt{b^2-4ac}\/2a. Since the discriminant is assumed as zero, the root will be -b/2a, or -b/2(b-x), or -b/2(b-\sqrt{3}\b/2. This is equal to -b/2b-\sqrt{3}\b, or -1/2-\sqrt{3}\. Multiplying by the conjugate gives -2-\sqrt{3}\.