Difference between revisions of "2022 AMC 10B Problems/Problem 7"
Stevens0209 (talk | contribs) m (→Solution) |
|||
Line 13: | Line 13: | ||
~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math> | ~stevens0209 ~MRENTHUSIASM ~<math>\color{magenta} zoomanTV</math> | ||
+ | |||
+ | ==Alternate Solution== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Let k^2-144=n^2. Then, <math>(k+n)(k-n)=144. 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.$ | ||
+ | |||
+ | pianoboy | ||
== See Also == | == See Also == |
Revision as of 23:20, 17 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.
Contents
[hide]Problem
For how many values of the constant will the polynomial have two distinct integer roots?
Solution
Let and be the roots of By Vieta's Formulas, we have and
This shows that and must be distinct factors of The possibilities of are Each unordered pair gives a unique value of Therefore, there are values of namely
~stevens0209 ~MRENTHUSIASM ~
Alternate Solution
Note that must be an integer. By the quadratic formula, Since is a multiple of , and have the same parity, so is an integer if and only if is a perfect square.
Let k^2-144=n^2. Then, k144k+nk-nk5k+n2, 4, 8, 6, 12kk+n=12k+n=k-nn=024k.24*2=\boxed{8}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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.