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

Latest revision as of 00:09, 1 November 2023

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 $pq=36.$

It follows 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. Using the quadratic formula, $x=\frac{-k \pm \sqrt{k^2-144}}{2}.$ Since $4$ divides $144$ evenly, $k$ 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 $4\cdot2=\boxed{\textbf{(B) }8}$ values of $k$ .

~pianoboy

Solution 3 (Pythagorean Triples)

Proceed similar to Solution 2 and deduce that the discriminant of $x^{2}+kx+36$ must be a perfect square greater than $0$ to satisfy all given conditions. Seeing something like $k^2-144$ might remind us of a right triangle, where $k$ is the hypotenuse, and $12$ is a leg. There are four ways we could have this: a $9$ - $12$ - $15$ triangle, a $12$ - $16$ - $20$ triangle, a $5$ - $12$ - $13$ triangle, and a $12$ - $35$ - $37$ triangle.

Multiply by $2$ to account for negative $k$ values (since $k$ is being squared), and our answer is $\boxed{\textbf{(B) }8}$ .

Solution 4

Since $36 = 2^2\cdot3^2$ , that means there are $(2+1)(2+1) = 9$ possible factors of $36$ . Since $6 \cdot 6$ violates the distinct root condition, subtract $1$ from $9$ to get $8$ . Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get $\boxed{\textbf{(B) }8}$ .

~songmath20 Edited 5.1.2023

Video Solution (⚡️Lightning Fast⚡️)

https://youtu.be/WX871JJbdY4

~Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

Video Solution by Interstigation

https://youtu.be/_KNR0JV5rdI?t=679

Video Solution by Math4All999

https://youtube.com/watch?v=cnUq_Op3YzY&feature=shared

Video Solution by Gavin Does Math

https://youtu.be/1qO3eejxuPo

@@ Line 21: / Line 21: @@
 ~pianoboy
+==Solution 3 (Pythagorean Triples)==
+Proceed similar to Solution 2 and deduce that the discriminant of <math>x^{2}+kx+36</math> must be a perfect square greater than <math>0</math> to satisfy all given conditions. Seeing something like <math>k^2-144</math> might remind us of a right triangle, where <math>k</math> is the hypotenuse, and <math>12</math> is a leg. There are four ways we could have this: a <math>9</math>-<math>12</math>-<math>15</math> triangle, a <math>12</math>-<math>16</math>-<math>20</math> triangle, a <math>5</math>-<math>12</math>-<math>13</math> triangle, and a <math>12</math>-<math>35</math>-<math>37</math> triangle.
+Multiply by <math>2</math> to account for negative <math>k</math> values (since <math>k</math> is being squared), and our answer is <math>\boxed{\textbf{(B) }8}</math>.
 ==Solution 4==

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

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