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 "2018 AMC 12A Problems/Problem 5"
 
(5 intermediate revisions by 5 users not shown)
Line 3: Line 3:
 
What is the sum of all possible values of <math>k</math> for which the polynomials <math>x^2 - 3x + 2</math> and <math>x^2 - 5x + k</math> have a root in common?
 
What is the sum of all possible values of <math>k</math> for which the polynomials <math>x^2 - 3x + 2</math> and <math>x^2 - 5x + k</math> have a root in common?
  
−
<math>\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10 \qquad</math>
+
<math>\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10</math>
  
 
==Solution==
 
==Solution==
 +
 +
We factor <math>x^2-3x+2</math> into <math>(x-1)(x-2)</math>. Thus, either <math>1</math> or <math>2</math> is a root of <math>x^2-5x+k</math>. If <math>1</math> is a root, then <math>1^2-5\cdot1+k=0</math>, so <math>k=4</math>. If <math>2</math> is a root, then <math>2^2-5\cdot2+k=0</math>, so <math>k=6</math>. The sum of all possible values of <math>k</math> is <math>\boxed{\textbf{(E) }10}</math>.
 +
 +
== Video Solution 1 ==
 +
https://youtu.be/aZdSO-4g40A
 +
 +
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
  
−
{{AMC12 box|year=2018|ab=A|before=First Problem|num-a=2}}
+
{{AMC12 box|year=2018|ab=A|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:56, 28 October 2022

Problem

What is the sum of all possible values of $k$ for which the polynomials $x^2 - 3x + 2$ and $x^2 - 5x + k$ have a root in common?

$\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10$

Solution

We factor $x^2-3x+2$ into $(x-1)(x-2)$. Thus, either $1$ or $2$ is a root of $x^2-5x+k$. If $1$ is a root, then $1^2-5\cdot1+k=0$, so $k=4$. If $2$ is a root, then $2^2-5\cdot2+k=0$, so $k=6$. The sum of all possible values of $k$ is $\boxed{\textbf{(E) }10}$.

Video Solution 1

https://youtu.be/aZdSO-4g40A

~Education, the Study of Everything

See Also

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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=2018_AMC_12A_Problems/Problem_5&oldid=179642"