Difference between revisions of "2001 AMC 12 Problems/Problem 23"
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So the two imaginary roots must multiply to give you an integer. | So the two imaginary roots must multiply to give you an integer. | ||
Taking the 5 answers into hand, we find that <math>\boxed{\frac {1 + i \sqrt {11}}{2}}</math> is our only integer giving solution. | Taking the 5 answers into hand, we find that <math>\boxed{\frac {1 + i \sqrt {11}}{2}}</math> is our only integer giving solution. | ||
+ | |||
+ | Note: The above solution is incorrect, as all the answers with the exception of <math>\textbf{(C)}</math> satisfy the above condition. | ||
== See Also == | == See Also == |
Revision as of 01:26, 29 December 2016
Contents
Problem
A polynomial of degree four with leading coefficient 1 and integer coefficients has two zeros, both of which are integers. Which of the following can also be a zero of the polynomial?
Solution
Let the polynomial be and let the two integer zeros be
and
. We can then write
for some integers
and
.
If a complex number with
is a root of
, it must be the root of
, and the other root of
must be
.
We can then write .
We can now examine each of the five given complex numbers, and find the one for which the values and
are integers. This is
, for which we have
and
.
(As an example, the polynomial has zeroes
,
, and
.)
Solution 2
By Vieta, we know that the product of all four zeros of the polynomial equals the constant at the end of the polynomial.
We also know that the two imaginary roots are a conjugate pair (I.E if one is a+bi, the other is a-bi).
So the two imaginary roots must multiply to give you an integer.
Taking the 5 answers into hand, we find that is our only integer giving solution.
Note: The above solution is incorrect, as all the answers with the exception of satisfy the above condition.
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.