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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, <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+\sqrt{b^2-4ac}}\{2a}.</math> Since the discriminant is assumed as zero, the root will be <math>-b/2a</math>, or <math>-b/2(b-x)</math> or <math>-b/2(b-\sqrt{3}\b/2)</math>. This is equal to <math>-b/2b-\sqrt{3}\b</math> or <math>-1/2-\sqrt{3}</math>. Multiplying by the conjugate gives <math>-2-\sqrt{3}</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+\sqrt{b^2-4ac}\}\{2a}.</math> Since the discriminant is assumed as zero, the root will be <math>-b/2a</math>, or <math>-b/2(b-x)</math> or <math>-b/2(b-\sqrt{3}\b/2)</math>. This is equal to <math>-b/2b-\sqrt{3}\b</math> or <math>-1/2-\sqrt{3}</math>. Multiplying by the conjugate gives <math>-2-\sqrt{3}</math>.

Revision as of 20:43, 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 $a=(b-x)$ and $c=(b+x)$. 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 $\frac{-b+\sqrt{b^2-4ac}\}\{2a}.$ 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 $-b/2b-\sqrt{3}\b$ (Error compiling LaTeX. Unknown error_msg) or $-1/2-\sqrt{3}$. Multiplying by the conjugate gives $-2-\sqrt{3}$.

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