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

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Note that we can divide the polynomial by <math>a</math> 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  
 
Note that we can divide the polynomial by <math>a</math> 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  
 
<math>(x-r)^2 = x^2 - 2rx + r^2</math> where <math>1 \ge -2r \ge r^2 \ge 0</math>.
 
<math>(x-r)^2 = x^2 - 2rx + r^2</math> where <math>1 \ge -2r \ge r^2 \ge 0</math>.
−
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,
+
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 median. Thus,
 
<math>r^2 + 1 = -4r</math> and <math>r = -2 \pm \sqrt{3}</math>. Since <math>1 > r^2</math>, we easily see that <math>|r|</math> has to be between 1 and 0. Thus, we can eliminate <math>-2 - \sqrt{3}</math> and are left with <math>\boxed{\textbf{(D)} -2 + \sqrt{3}}</math> as the answer.
 
<math>r^2 + 1 = -4r</math> and <math>r = -2 \pm \sqrt{3}</math>. Since <math>1 > r^2</math>, we easily see that <math>|r|</math> has to be between 1 and 0. Thus, we can eliminate <math>-2 - \sqrt{3}</math> and are left with <math>\boxed{\textbf{(D)} -2 + \sqrt{3}}</math> as the answer.
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2013|ab=B|num-b=18|num-a=20}}
 
{{AMC10 box|year=2013|ab=B|num-b=18|num-a=20}}

Revision as of 14:04, 28 March 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?

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


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 median. 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.

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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