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?

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

Solution

t 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 $\boxed {D}$ .

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2013 AMC 10B Problems/Problem 19

Problem

Solution