Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(→‎Problem)
Line 8: Line 8:
 
I didn't really understand this problem...
 
I didn't really understand this problem...
  
−
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>-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>.
+
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>{-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 20:46, 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 $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Since the discriminant is assumed as zero, the root will be $\frac{-b}{2a}$, or $\frac{-b}{2(b-x)}$ 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 $-2-\sqrt{3}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10B_Problems/Problem_19&oldid=51596"