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2022 AMC 10B Problems/Problem 7: Difference between revisions

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Using Vieta's Formula, this states:
{{duplicate|[[2022 AMC 10B Problems/Problem 7|2022 AMC 10B #7]] and [[2022 AMC 12B Problems/Problem 4|2022 AMC 12B #4]]}}


<math>p+q=-k</math>
==Problem==
For how many values of the constant <math>k</math> will the polynomial <math>x^{2}+kx+36</math> have two distinct integer roots?


<math>p*q=36</math>
<math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16</math>


(Let <math>p</math> and <math>q</math> be the roots)
==Solution 1==
Let <math>p</math> and <math>q</math> be the roots of <math>x^{2}+kx+36.</math> By Vieta's Formula, we have <math>p+q=-k</math> and <math>pq=36.</math>


This shows that p and q must be the factors of <math>36</math>: <math>1, 36, 2, 18, 3, 12, 4, 9, 6</math> and its negative counterpart.
This shows that <math>p</math> and <math>q</math> must be distinct factors of <math>36.</math> The possibilities of <math>\{p,q\}</math> are <cmath>\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.</cmath>
Each unordered pair gives a unique value of <math>k.</math> Therefore, there are <math>\boxed{\textbf{(B) }8}</math> values of <math>k,</math> namely <math>\pm37,\pm20,\pm15,\pm13.</math>


We cancel out the <math>6</math> and <math>6</math> because the problem states that it wants distinct roots.
~Stevens0209


Thus, we have a total of <math>4</math> pairs and another <math>4</math> pairs (the negatives), which total us <math>4+4=\boxed{\textbf{(B) }8}</math>.
~MRENTHUSIASM
 
== See Also ==
{{AMC10 box|year=2022|ab=B|num-b=5|num-a=7}}
{{MAA Notice}}

Revision as of 17:10, 17 November 2022

The following problem is from both the 2022 AMC 10B #7 and 2022 AMC 12B #4, so both problems redirect to this page.

Problem

For how many values of the constant $k$ will the polynomial $x^{2}+kx+36$ have two distinct integer roots?

$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16$

Solution 1

Let $p$ and $q$ be the roots of $x^{2}+kx+36.$ By Vieta's Formula, we have $p+q=-k$ and $pq=36.$

This shows that $p$ and $q$ must be distinct factors of $36.$ The possibilities of $\{p,q\}$ are \[\pm\{1,36\},\pm\{2,18\},\pm\{3,12\},\pm\{4,9\}.\] Each unordered pair gives a unique value of $k.$ Therefore, there are $\boxed{\textbf{(B) }8}$ values of $k,$ namely $\pm37,\pm20,\pm15,\pm13.$

~Stevens0209

~MRENTHUSIASM

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America.

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