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

Latest revision as of 13:28, 7 November 2025

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 $k,$ corresponding to $\mp37,\mp20,\mp15,\mp13,$ respectively.

~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

Solution 5 (Discriminant and Factor Rainbow)

Since to have two solutions, the discriminant must be greater than $0$ , $k^2 - 144 > 0$ so k must be greater than $12$ or less than $-12$ (since it's squared). Next, we know that when factoring, the two numbers must multiply to $36$ . So you can create a factor rainbow(remember from elementary school) which has the pairs of integers that multiply to $36( 1 2 3 4 6 6 9 12 18 36 )$ (you just look at the position and the one on the opposite side, for example $2$ is the second value, and $18$ is the second last value, and they multiply to 36). See that there are $4$ pairs of values which add up to greater than $12$ ( $6$ is equal to $12$ ), and we look at the negative values of them too(since k is squared) we get $\boxed{\textbf{(B) }8}$ total values.

~SIGMAMATHEMATICIAN

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 2: / Line 2: @@
 ==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?
+For how many values of the constant <imath>k</imath> will the polynomial <imath>x^{2}+kx+36</imath> have two distinct integer roots?
-<math>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16</math>
+<imath>\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 16</imath>
 ==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 <imath>p</imath> and <imath>q</imath> be the roots of <imath>x^{2}+kx+36.</imath> By [[Vieta's Formulas]], we have <imath>p+q=-k</imath> and <imath>pq=36.</imath>
-It follows 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>
+It follows that <imath>p</imath> and <imath>q</imath> must be distinct factors of <imath>36.</imath> The possibilities of <imath>\{p,q\}</imath> 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>
+Each unordered pair gives a unique value of <imath>k.</imath> Therefore, there are <imath>\boxed{\textbf{(B) }8}</imath> values of <imath>k,</imath> corresponding to <imath>\mp37,\mp20,\mp15,\mp13,</imath> respectively.
-~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math>
+~stevens0209 ~MRENTHUSIASM ~<imath>\color{magenta}zoomanTV</imath>
 ==Solution 2==
-Note that <math>k</math> must be an integer. Using the [[quadratic formula]], <math>x=\frac{-k \pm \sqrt{k^2-144}}{2}.</math> Since <math>4</math> divides <math>144</math> evenly, <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.
+Note that <imath>k</imath> must be an integer. Using the [[quadratic formula]], <imath>x=\frac{-k \pm \sqrt{k^2-144}}{2}.</imath> Since <imath>4</imath> divides <imath>144</imath> evenly, <imath>k</imath> and <imath>k^2-144</imath> have the same parity, so <imath>x</imath> is an integer if and only if <imath>k^2-144</imath> 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\cdot2=\boxed{\textbf{(B) }8}</math> values of <math>k</math>.
+Let <imath>k^2-144=n^2.</imath> Then, <imath>(k+n)(k-n)=144.</imath> Since <imath>k</imath> is an integer and <imath>144</imath> is even, <imath>k+n</imath> and <imath>k-n</imath> must both be even. Assuming that <imath>k</imath> is positive, we get <imath>5</imath> possible values of <imath>k+n</imath>, namely <imath>2, 4, 8, 6, 12</imath>, which will give distinct positive values of <imath>k</imath>, but <imath>k+n=12</imath> gives <imath>k+n=k-n</imath> and <imath>n=0</imath>, giving <imath>2</imath> identical integer roots. Therefore, there are <imath>4</imath> distinct positive values of <imath>k.</imath> Multiplying that by <imath>2</imath> to take the negative values into account, we get <imath>4\cdot2=\boxed{\textbf{(B) }8}</imath> values of <imath>k</imath>.
 ~pianoboy
@@ Line 24: / Line 24: @@
 ==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.
+Proceed similar to Solution 2 and deduce that the discriminant of <imath>x^{2}+kx+36</imath> must be a perfect square greater than <imath>0</imath> to satisfy all given conditions. Seeing something like <imath>k^2-144</imath> might remind us of a right triangle, where <imath>k</imath> is the hypotenuse, and <imath>12</imath> is a leg. There are four ways we could have this: a <imath>9</imath>-<imath>12</imath>-<imath>15</imath> triangle, a <imath>12</imath>-<imath>16</imath>-<imath>20</imath> triangle, a <imath>5</imath>-<imath>12</imath>-<imath>13</imath> triangle, and a <imath>12</imath>-<imath>35</imath>-<imath>37</imath> 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>.
+Multiply by <imath>2</imath> to account for negative <imath>k</imath> values (since <imath>k</imath> is being squared), and our answer is <imath>\boxed{\textbf{(B) }8}</imath>.
 ==Solution 4==
-Since <math>36 = (2\cdot3)^2</math>, that means there are <math>(2+1)(2+1) = 9</math> possible factors of <math>36</math>. Since <math>6 \cdot 6</math> violates the distinct root condition, subtract <math>1</math> from <math>9</math> to get <math>8</math>. Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get <math>\boxed{\textbf{(B) }8}</math>.
+Since <imath>36 = 2^2\cdot3^2</imath>, that means there are <imath>(2+1)(2+1) = 9</imath> possible factors of <imath>36</imath>. Since <imath>6 \cdot 6</imath> violates the distinct root condition, subtract <imath>1</imath> from <imath>9</imath> to get <imath>8</imath>. Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get <imath>\boxed{\textbf{(B) }8}</imath>.
 ~songmath20  Edited 5.1.2023
+==Solution 5 (Discriminant and Factor Rainbow)==
+Since to have two solutions, the discriminant must be greater than <imath>0</imath>, <imath>k^2 - 144 > 0</imath> so k must be greater than <imath>12</imath> or less than <imath>-12</imath>(since it's squared). Next, we know that when factoring, the two numbers must multiply to <imath>36</imath>. So you can create a factor rainbow(remember from elementary school) which has the pairs of integers that multiply to <imath>36( 1 2 3 4 6 6 9 12 18 36 )</imath> (you just look at the position and the one on the opposite side, for example <imath>2</imath> is the second value, and <imath>18</imath> is the second last value, and they multiply to 36). See that there are <imath>4</imath> pairs of values which add up to greater than <imath>12</imath>(<imath>6</imath> is equal to <imath>12</imath>), and we look at the negative values of them too(since k is squared) we get <imath>\boxed{\textbf{(B) }8}</imath> total values.
+~SIGMAMATHEMATICIAN
 ==Video Solution (⚡️Lightning Fast⚡️)==
@@ Line 47: / Line 52: @@
 ==Video Solution by Math4All999==
 https://youtube.com/watch?v=cnUq_Op3YzY&feature=shared
+==Video Solution by Gavin Does Math==
+https://youtu.be/1qO3eejxuPo
 == 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

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