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

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I didn't really understand this problem...
 
I didn't really understand this problem...
  
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If there is one root, the discriminant is 0. Therefore, b^2-4ac=0. Let's define a as (b-x) and c as (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 -b +/- \sqrt{b^2-4ac}\/2a. Since the discriminant is assumed as zero, the root will be -b/2a, or -b/2(b-x), or -b/2(b-\sqrt{3}\b/2. This is equal to -b/2b-\sqrt{3}\b, or -1/2-\sqrt{3}\. Multiplying by the conjugate gives -2-\sqrt{3}\.
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If there is one root, the discriminant is 0. Therefore, <math>b^2-4ac=0</math>. Let's define a as (b-x) and c as (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 -b +/- \sqrt{b^2-4ac}\/2a. Since the discriminant is assumed as zero, the root will be -b/2a, or -b/2(b-x), or -b/2(b-\sqrt{3}\b/2. This is equal to -b/2b-\sqrt{3}\b, or -1/2-\sqrt{3}\. Multiplying by the conjugate gives -2-\sqrt{3}\.

Revision as of 21:36, 25 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 as (b-x) and c as (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 -b +/- \sqrt{b^2-4ac}\/2a. Since the discriminant is assumed as zero, the root will be -b/2a, or -b/2(b-x), or -b/2(b-\sqrt{3}\b/2. This is equal to -b/2b-\sqrt{3}\b, or -1/2-\sqrt{3}\. Multiplying by the conjugate gives -2-\sqrt{3}\.

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