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

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==Solution==
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I didn't really understand this problem...
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t is given that <math>ax^2+bx+c=0</math> has 1 real root, so the discriminant is zero, or <math>b^2=4ac</math>. Because a, b, c are in arithmetic progression, <math>b-a=c-b</math>, or <math>b=\frac {a+c} {2} </math>. We need to find the unique root, or <math>-\frac {b} {2a} </math> (discriminant is 0). From <math>b^2=4ac</math>, we have <math>-\frac {b} {2a} =-\frac {2c} {b}</math>. Ignoring the negatives, we have
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<math>\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} }</math>. Fortunately, finding <math>\frac {a} {c} </math> is not very hard. Plug in <math>b=\frac {a+c} {2}</math> to <math>b^2=4ac</math>, we have <math>a^2+2ac+c^2=16ac</math>, or <math>a^2-14ac+c^2=0</math>, and dividing by <math>c^2</math> gives <math>(\frac {a} {c} ) ^2-14( \frac {a} {c} ) +1 = 0</math>, so <math>\frac {a} {c} = \frac {14 \pm \sqrt {192} } {2} = 7 \pm 4 \sqrt {3} </math>. But <math>7-4\sqrt {3} <1</math>, violating the assumption that <math>a \ge c</math>. Therefore, <math>\frac {a} {c} = 7 +4\sqrt {3} </math>. Plugging this in, we have <math>\frac {1} {\frac {a} {4c} + \frac {1} {4} } = \frac {1} {2+ \sqrt {3} } = 2- \sqrt {3} </math>. But we need the negative of this, so the answer is <math>\boxed {D}</math>.
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If there is one root, the discriminant is 0. Therefore, <math>b^2-4ac=0</math>. Let's define <math>a=(b-x)</math> and <math>c=(b+x)</math>. If <math>b^2-4ac=0</math>, by substitution, <math>b^2-4(b-x)(b+x)=0</math> and distributing, <math>b^2-4b^2+4x^2=0</math>. Thus, through simple arithmetic, <math>4x^2=3b^2</math> and <math>2x=\sqrt{3}\/b</math>. Then, the root, by quadratic formula, is <math>\frac{-b\pm\sqrt{b^2-4ac}}{2a}</math>. Since the discriminant is assumed as zero, the root will be <math>\frac{-b}{2a}</math>, or <math>\frac{-b}{2(b-x)}</math> or <math>\frac{-b}{2(b-\sqrt{3}b/2)}</math>. This is equal to <math>\frac{-b}{2b-\sqrt{3}b}</math> or <math>\frac{-1}{2-\sqrt{3}}</math>. Multiplying by the conjugate gives <math>-2-\sqrt{3}</math>.
 
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This would mean the answer would be <math>\textbf{(B)}</math>, but the answer key says <math>\textbf{(D)}</math>. Any criticism would be much welcomed.
 
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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 23:07, 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?

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


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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