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

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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>
  
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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>
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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>
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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>
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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> corresponding to <math>\mp37,\mp20,\mp15,\mp13,</math> respectively.
  
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~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math>
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~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta}zoomanTV</math>
  
 
==Solution 2==
 
==Solution 2==
  
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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
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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.
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<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.
 
  
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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>
+
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>.
  
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pianoboy
+
~pianoboy
  
 
==Solution 3 (Pythagorean Triples)==
 
==Solution 3 (Pythagorean Triples)==
  
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Proceed similar to Solution 2 and deduce that the discriminant of <math>x^{2}+kx+36</math> must be a perfect square greater than 0 to satisfy all given conditions. Seeing something like <math>k^2-144</math> 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 <math>9-12-15</math> triangle, a <math>12-16-20</math> triangle, a <math>5-12-13</math> triangle, and a <math>12-35-37</math> triangle.
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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.
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Multiply by two to account for negative k values (since k is being squared) and our answer is <math>\boxed{\textbf{(B) }8}</math>.
 
  
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==Solution 4 (Similar to Solution 1)==
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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>.
  
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Let <math>r_1</math> and <math>r_2</math> be the roots of the given polynomial. By Vieta's, <math>r_1+r_2=-k</math>, and <math>r_1r_2=36</math>.  
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==Solution 4==
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Since <math>36 = 2^2\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>.
  
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~songmath20  Edited 5.1.2023
  
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The <math>5</math> positive pairs of <math>(r_1, r_2)</math> that satisfy the second constraint are: <math>(1, 36), (2, 18), (3, 12), (4, 9),</math> and <math>(6, 6).</math>
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==Video Solution (⚡️Lightning Fast⚡️)==
 +
https://youtu.be/WX871JJbdY4
  
 +
~Education, the Study of Everything
  
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We multiply this number by <math>2</math> to account for the negative numbers that will multiply together to give us <math>36</math>, and subtract <math>2</math>, since <math>(6, 6)</math> is both a double root and is counted twice.  
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==Video Solution(1-16)==
 +
https://youtu.be/SCwQ9jUfr0g
  
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Therefore, the answer is <math>5\cdot2-2=\boxed{\textbf{(B) }8}</math> pairs.
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~~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
  
 
== See Also ==
 
== See Also ==

Latest revision as of 00:16, 9 November 2024

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

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

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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