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2013 AMC 10B Problems/Problem 19

Revision as of 21:13, 19 March 2013 by Hikerboy (talk | contribs)

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?

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


Solution

Solution 1

It is given that $ax^2+bx+c=0$ has 1 real root, so the discriminant is zero, or $b^2=4ac$. Because a, b, c are in arithmetic progression, $b-a=c-b$, or $b=\frac {a+c} {2}$. We need to find the unique root, or $-\frac {b} {2a}$ (discriminant is 0). From $b^2=4ac$, we have $-\frac {b} {2a} =-\frac {2c} {b}$. Ignoring the negatives, we have $\frac {2c} {b} = \frac {2c} {\frac {a+c} {2}} = \frac {4c} {a+c} = \frac {1} {\frac {1} {\frac {4c} {a+c}}} = \frac {1} {\frac {a+c} {4c}} = \frac {1} {\frac {a} {4c} + \frac {1} {4} }$. Fortunately, finding $\frac {a} {c}$ is not very hard. Plug in $b=\frac {a+c} {2}$ to $b^2=4ac$, we have $a^2+2ac+c^2=16ac$, or $a^2-14ac+c^2=0$, and dividing by $c^2$ gives $(\frac {a} {c} ) ^2-14( \frac {a} {c} ) +1 = 0$, so $\frac {a} {c} = \frac {14 \pm \sqrt {192} } {2} = 7 \pm 4 \sqrt {3}$. But $7-4\sqrt {3} <1$, violating the assumption that $a \ge c$. Therefore, $\frac {a} {c} = 7 +4\sqrt {3}$. Plugging this in, we have $\frac {1} {\frac {a} {4c} + \frac {1} {4} } = \frac {1} {2+ \sqrt {3} } = 2- \sqrt {3}$. But we need the negative of this, so the answer is $\boxed {D}$.


Solution 2

Note that we can divide the polynomial by $a$ to make the leading coefficient 1 since dividing does not change the roots or the fact that the coefficients are in an arithmetic sequence. Also, we know that there is exactly one root so this equation is must be of the form $(x-r)^2 = x^2 - 2rx + r^2$ where $1 \ge -2r \ge r^2 \ge 0$. We now use the fact that the coefficients are in an arithmetic sequence. Note that in any arithmetic sequence, the average is equal to the mean. Thus, $r^2 + 1 = -4r$ and $r = -2 \pm \sqrt{3}$. Since $1 > r^2$, we easily see that $|r|$ has to be between 1 and 0. Thus, we can eliminate $-2 - \sqrt{3}$ and are left with $\boxed{\textbf{(D)} -2 + \sqrt{3}}$ as the answer.

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