Difference between revisions of "2022 AMC 10B Problems/Problem 7"
MRENTHUSIASM (talk | contribs) (→Solution 4 (Similar to Solution 1): Sol 1 has the exact same idea. So, I consider removing Sol 4.) |
Lotusjayden (talk | contribs) m (→Solution 2) |
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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. | <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 <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 |
Revision as of 16:30, 26 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
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
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 ~
Solution 2
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 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 0 to satisfy all given conditions. Seeing something like
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
triangle, a
triangle, a
triangle, and a
triangle.
Multiply by two to account for negative k values (since k is being squared) and our answer is
.
Video Solution 1
~Education, the Study of Everything
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.