Difference between revisions of "2022 AMC 10B Problems/Problem 7"
MRENTHUSIASM (talk | contribs) (→Solution 4: Although Sol 4 finds the right answer, it assumes that the factors must be positive, and the order of the factors matters. These assumptions are not quite right ...) |
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~Education, the Study of Everything | ~Education, the Study of Everything | ||
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+ | ==Video Solution(1-16)== | ||
+ | https://youtu.be/SCwQ9jUfr0g | ||
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+ | ~~Hayabusa1 | ||
== See Also == | == See Also == |
Revision as of 13:32, 5 April 2023
- 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
namely
~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
.
Video Solution 1
~Education, the Study of Everything
Video Solution(1-16)
~~Hayabusa1
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.