Difference between revisions of "2018 AMC 12A Problems/Problem 5"

m
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>.
+
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 ==
 
== Video Solution ==

Revision as of 23:29, 6 June 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

https://youtu.be/ba6w1OhXqOQ?t=2

~ pi_is_3.14

See Also

2018 AMC 12A (ProblemsAnswer KeyResources)
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