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 "2019 AMC 12A Problems/Problem 14"

m (Fixed a LaTeX error)
m (Further improved formatting)
Line 12: Line 12:
 
<math>P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)</math>
 
<math>P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)</math>
  
−
by using the quadratic formula on the quadratic factors. Since the first four roots are all distinct, the term <math>(x^2 - cx + 4)</math> must be a product of any combination of 2 not necessarily distinct  factors from the set: <math>(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),</math> and <math>(x - [2 + 2i])</math>. We need the two factors to yield a constant term of 4 when multiplied together. The only combinations that work are <math>(x - [1 - i])</math> and <math>(x - [2 + 2i])</math>, or <math>(x - [1+i])</math> and <math>(x - [2-2i])</math>. When multiplied together, the polynomial is either <math>(x^2 + [-3 + i]x + 4)</math> or <math>(x^2+[-3-i]x+4)</math>. Therefore, <math>c = -3 \pm i</math> and <math>|c| = \boxed{\textbf{(E)}\sqrt{10}}</math>.
+
by using the quadratic formula on the quadratic factors. Since the first four roots are all distinct, the term <math>(x^2 - cx + 4)</math> must be a product of any combination of 2 not necessarily distinct  factors from the set: <math>(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),</math> and <math>(x - [2 + 2i])</math>. We need the two factors to yield a constant term of 4 when multiplied together. The only combinations that work are <math>(x - [1 - i])</math> and <math>(x - [2 + 2i])</math>, or <math>(x - [1+i])</math> and <math>(x - [2-2i])</math>. When multiplied together, the polynomial is either <math>(x^2 + [-3 + i]x + 4)</math> or <math>(x^2+[-3-i]x+4)</math>. Therefore, <math>c = -3 \pm i</math> and <math>|c| = \boxed{\textbf{(E) } \sqrt{10}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 19:19, 10 February 2019

Problem

For a certain complex number $c$, the polynomial \[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\]has exactly 4 distinct roots. What is $|c|$?

$\textbf{(A) } 2 \qquad \textbf{(B) } \sqrt{6} \qquad \textbf{(C) } 2\sqrt{2} \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \sqrt{10}$

Solution

The polynomial can be factored further broken down into

$P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$

by using the quadratic formula on the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of 2 not necessarily distinct factors from the set: $(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),$ and $(x - [2 + 2i])$. We need the two factors to yield a constant term of 4 when multiplied together. The only combinations that work are $(x - [1 - i])$ and $(x - [2 + 2i])$, or $(x - [1+i])$ and $(x - [2-2i])$. When multiplied together, the polynomial is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$. Therefore, $c = -3 \pm i$ and $|c| = \boxed{\textbf{(E) } \sqrt{10}}$.

See Also

2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12A_Problems/Problem_14&oldid=101770"