Difference between revisions of "2022 AMC 10B Problems/Problem 7"
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==Solution 4== | ==Solution 4== | ||
− | Since <math>36 = | + | 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 | ~songmath20 Edited 5.1.2023 |
Revision as of 23:46, 8 October 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
.
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
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.