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Difference between revisions of "2013 AMC 10B Problems/Problem 19"

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The problem says "<math>a\ge b\ge c\ge 0</math>", but you assumed that <math>0\le a\le b\le c</math>. I know many people who made this mistake.
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The problem says <math>a\ge b\ge c</math>, but you assumed that <math>a\le b\le c</math>. I know many people who made this mistake.

Revision as of 18:58, 26 February 2013

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.


The problem says $a\ge b\ge c$, but you assumed that $a\le b\le c$. I know many people who made this mistake.

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