Difference between revisions of "2013 AMC 10B Problems/Problem 19"
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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} | + | 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 20:44, 25 February 2013
Problem
The real numbers form an arithmetic sequence with The quadratic has exactly one root. What is this root?
I didn't really understand this problem...
If there is one root, the discriminant is 0. Therefore, . Let's define and . If , by substitution, and distributing, . Thus, through simple arithmetic, and . Then, the root, by quadratic formula, is . Since the discriminant is assumed as zero, the root will be , or or . This is equal to $-b/2b-\sqrt{3}\b$ (Error compiling LaTeX. Unknown error_msg) or . Multiplying by the conjugate gives .