Difference between revisions of "2022 AMC 10B Problems/Problem 7"

Line 1: Line 1:
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 (ProblemsAnswer KeyResources)
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

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