Difference between revisions of "2001 AMC 12 Problems/Problem 23"
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Use the quadratic formula to determine the form of the complex roots. | Use the quadratic formula to determine the form of the complex roots. | ||
− | <math> | + | <math>\frac{p+q-a\pm\sqrt{(a-p-q)^2-4\frac{4d}{pq}}}{2}</math> |
Now this is starting to look a lot like answers A and E. Noticing that the real part in each answer choice is <math>\frac{1}{2}</math>, <math>(k+n-a)=1</math> and <math>(a-k-n)^2=1</math>, and the imaginary part is positive. Furthermore, by Vieta's Formulas, we know that d must be a multiple of p and q, so <math>\frac{4d}{pq}</math> is a multiple of 4. Rearranging the expression, we get: | Now this is starting to look a lot like answers A and E. Noticing that the real part in each answer choice is <math>\frac{1}{2}</math>, <math>(k+n-a)=1</math> and <math>(a-k-n)^2=1</math>, and the imaginary part is positive. Furthermore, by Vieta's Formulas, we know that d must be a multiple of p and q, so <math>\frac{4d}{pq}</math> is a multiple of 4. Rearranging the expression, we get: | ||
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<math>/frac{1+i\sqrt{\frac{4d}{pq}-1}}{2}</math> | <math>/frac{1+i\sqrt{\frac{4d}{pq}-1}}{2}</math> | ||
− | The radicand therefore must be one less than a multiple of four, which is only the case in <math>{\frac {1 + i \sqrt {11}}{2}</math> or <math> | + | The radicand therefore must be one less than a multiple of four, which is only the case in <math>{\frac {1 + i \sqrt {11}}{2}</math> or <math>\boxed{A}</math>. |
== See Also == | == See Also == |
Revision as of 23:29, 12 October 2017
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.
Solution 3
After dividing the polynomial out by and
, where p and q are the real roots of the polynomial, we will obtain a quadratic with two complex roots. We can then use the quadratic formula to solve for these complex roots.
Let's start by using synthetic division to divide by
. Using this method, the quotient becomes
. However, we know that there should be no remainder because
is a factor of the polynomial, so
must equal 0, so
. When we divide the expression on the left by -p, we get
, so we can replace it in our original synthetic division equation with
.
We then want to synthetically divide by the next factor,
. Using the same method as before, we can simplify the quotient to
. Now for the easy part!
Use the quadratic formula to determine the form of the complex roots.
Now this is starting to look a lot like answers A and E. Noticing that the real part in each answer choice is ,
and
, and the imaginary part is positive. Furthermore, by Vieta's Formulas, we know that d must be a multiple of p and q, so
is a multiple of 4. Rearranging the expression, we get:
The radicand therefore must be one less than a multiple of four, which is only the case in ${\frac {1 + i \sqrt {11}}{2}$ (Error compiling LaTeX. Unknown error_msg) or .
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.