Difference between revisions of "2000 AIME II Problems/Problem 13"
(→Solution 2 (Complex Bash))
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== See also ==
== See also ==
Revision as of 21:42, 1 August 2020
The equation has exactly two real roots, one of which is , where , and are integers, and are relatively prime, and . Find .
We may factor the equation as:
Now for real . Thus the real roots must be the roots of the equation . By the quadratic formula the roots of this are:
Thus , and so the final answer is .
^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of with half of the polynomial's degree (in this case, divide through by ), and then to use one of the substitutions . In this case, the substitution gives and , which reduces the polynomial to just . Then one can backwards solve for .
Solution 2 (Complex Bash)
It would be really nice if the coefficients were symmetrical. What if we make the substitution, . The the polynomial becomes
It's symmetric! Dividing by and rearranging, we get
Now, if we let , we can get the equations
(These come from squaring and subtracting , then multiplying that result by and subtracting ) Plugging this into our polynomial, expanding, and rearranging, we get
Now, we see that the two terms must cancel in order to get this polynomial equal to , so what squared equals 3? Plugging in into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying , we see that it also works! Great, we use long division on the polynomial by
and we get
We know that the other two solutions for z wouldn't result in real solutions for since we have to solve a quadratic with a negative discriminant, then multiply by . We get that . Solving for (using ) we get that , and multiplying this by (because ) we get that for a final answer of
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