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> | ||
− | + | 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> | |
− | 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> | + | 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. |
− | ~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math> | + | ~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta}zoomanTV</math> |
==Solution 2== | ==Solution 2== | ||
− | Note that <math>k</math> must be an integer. | + | 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. |
− | <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 | + | 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>. |
− | pianoboy | + | ~pianoboy |
==Solution 3 (Pythagorean Triples)== | ==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 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 | + | 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== | |
+ | 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>. | ||
+ | ~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 == | == 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.
Contents
Problem
For how many values of the constant will the polynomial have two distinct integer roots?
Solution 1
Let and be the roots of By Vieta's Formulas, we have and
It follows that and must be distinct factors of The possibilities of are Each unordered pair gives a unique value of Therefore, there are values of corresponding to respectively.
~stevens0209 ~MRENTHUSIASM ~
Solution 2
Note that must be an integer. Using the quadratic formula, Since divides evenly, and have the same parity, so is an integer if and only if is a perfect square.
Let Then, Since is an integer and is even, and must both be even. Assuming that is positive, we get possible values of , namely , which will give distinct positive values of , but gives and , giving identical integer roots. Therefore, there are distinct positive values of Multiplying that by to take the negative values into account, we get values of .
~pianoboy
Solution 3 (Pythagorean Triples)
Proceed similar to Solution 2 and deduce that the discriminant of must be a perfect square greater than to satisfy all given conditions. Seeing something like might remind us of a right triangle, where is the hypotenuse, and is a leg. There are four ways we could have this: a -- triangle, a -- triangle, a -- triangle, and a -- triangle.
Multiply by to account for negative values (since is being squared), and our answer is .
Solution 4
Since , that means there are possible factors of . Since violates the distinct root condition, subtract from to get . Each sum is counted twice, and we count of those twice for negatives. This cancels out, so we get .
~songmath20 Edited 5.1.2023
Video Solution (⚡️Lightning Fast⚡️)
~Education, the Study of Everything
Video Solution(1-16)
~~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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.