2013 AMC 10B Problems/Problem 19

Problem

The real numbers $c,b,a$ form an arithmetic sequence with $a\ge b\ge c\ge 0$ The quadratic $ax^2+bx+c$ has exactly one root. What is this root?

$\textbf{(A)}\ -7-4\sqrt{3}\qquad\textbf{(B)}\ -2-\sqrt{3}\qquad\textbf{(C)}\ -1\qquad\textbf{(D)}\ -2+\sqrt{3}\qquad\textbf{(E)}\ -7+4\sqrt{3}$

I didn't really understand this problem...

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

This would mean the answer would be \textbf{(B)}, but the answer key says \textbf{(D)}. Any criticism would be much welcomed.

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2013 AMC 10B Problems/Problem 19

Problem