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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

(→‎Alternate Solution)
Line 6: Line 6:
 
<math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16</math>
 
<math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16</math>
  
−
==Solution==
+
==Solution 1==
 
Let <math>p</math> and <math>q</math> be the roots of <math>x^{2}+kx+36.</math> By [[Vieta's Formulas]], we have <math>p+q=-k</math> and <math>pq=36.</math>
 
Let <math>p</math> and <math>q</math> be the roots of <math>x^{2}+kx+36.</math> By [[Vieta's Formulas]], we have <math>p+q=-k</math> and <math>pq=36.</math>
  
Line 14: Line 14:
 
~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math>
 
~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math>
  
−
==Alternate Solution==
+
==Solution 2==
  
 
Note that <math>k</math> must be an integer. By the quadratic formula, <math>x=\frac{-k \pm \sqrt{k^2-144}}{2}.</math> Since <math>144</math> is a multiple of
 
Note that <math>k</math> must be an integer. By the quadratic formula, <math>x=\frac{-k \pm \sqrt{k^2-144}}{2}.</math> Since <math>144</math> is a multiple of
 
<math>4</math>, <math>k</math> and <math>k^2-144</math> have the same parity, so <math>x</math> is an integer if and only if <math>k^2-144</math> is a perfect square.
 
<math>4</math>, <math>k</math> and <math>k^2-144</math> have the same parity, so <math>x</math> is an integer if and only if <math>k^2-144</math> is a perfect square.
  
−
Let <math>k^2-144=n^2.</math> Then, <math>(k+n)(k-n)=144.</math> Since <math>k</math> is an integer and <math>144</math> is even, <math>k+n</math> and <math>k-n</math> must both be even. Assuming that <math>k</math> is positive, we get <math>5</math> possible values of <math>k+n</math>, namely <math>2, 4, 8, 6, 12</math>, which will give distinct positive values of <math>k</math>, but <math>k+n=12</math> gives <math>k+n=k-n</math> and <math>n=0</math>, giving <math>2</math> identical integer roots. Therefore, there are <math>4</math> distinct positive values of <math>k.</math> Multiplying that by <math>2</math> to take the negative values into account, we get <math>4*2=\boxed{8}</math> values of <math>k.</math>
+
Let <math>k^2-144=n^2.</math> Then, <math>(k+n)(k-n)=144.</math> Since <math>k</math> is an integer and <math>144</math> is even, <math>k+n</math> and <math>k-n</math> must both be even. Assuming that <math>k</math> is positive, we get <math>5</math> possible values of <math>k+n</math>, namely <math>2, 4, 8, 6, 12</math>, which will give distinct positive values of <math>k</math>, but <math>k+n=12</math> gives <math>k+n=k-n</math> and <math>n=0</math>, giving <math>2</math> identical integer roots. Therefore, there are <math>4</math> distinct positive values of <math>k.</math> Multiplying that by <math>2</math> to take the negative values into account, we get <math>4*2=\boxed{\textbf{(B) }8}</math> values of <math>k.</math>
  
 
pianoboy
 
pianoboy

Revision as of 08:25, 18 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 Formulas, 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 ~$\color{magenta} zoomanTV$

Solution 2

Note that $k$ must be an integer. By the quadratic formula, $x=\frac{-k \pm \sqrt{k^2-144}}{2}.$ Since $144$ is a multiple of $4$, $k$ and $k^2-144$ have the same parity, so $x$ is an integer if and only if $k^2-144$ is a perfect square.

Let $k^2-144=n^2.$ Then, $(k+n)(k-n)=144.$ Since $k$ is an integer and $144$ is even, $k+n$ and $k-n$ must both be even. Assuming that $k$ is positive, we get $5$ possible values of $k+n$, namely $2, 4, 8, 6, 12$, which will give distinct positive values of $k$, but $k+n=12$ gives $k+n=k-n$ and $n=0$, giving $2$ identical integer roots. Therefore, there are $4$ distinct positive values of $k.$ Multiplying that by $2$ to take the negative values into account, we get $4*2=\boxed{\textbf{(B) }8}$ values of $k.$

pianoboy

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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
2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_7&oldid=182241"