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

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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?
  
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<math>\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10 \qquad</math>
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<math>\textbf{(A) }3 \qquad\textbf{(B) }4 \qquad\textbf{(C) }5 \qquad\textbf{(D) }6 \qquad\textbf{(E) }10</math>
  
 
==Solution==
 
==Solution==
  
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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>.
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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>.
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== Video Solution 1 ==
 +
https://youtu.be/aZdSO-4g40A
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~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
  
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{{AMC12 box|year=2018|ab=A|before=First Problem|num-a=2}}
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{{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

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