Difference between revisions of "2010 AIME II Problems/Problem 7"
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so now, <math>x_1 = 4, x_2 = 4+6i, x_3 = 4-6i</math> | so now, <math>x_1 = 4, x_2 = 4+6i, x_3 = 4-6i</math> | ||
− | and therefore: <math>a=-12, b=84, c=-208</math>, and finally, we have <math>|a+b+c|=|-12+84-208|=136</math>. | + | and therefore: <math>a=-12, b=84, c=-208</math>, and finally, we have <math>|a+b+c|=|-12+84-208|=\boxed{136}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2010|num-b=6|num-a=8|n=II}} | {{AIME box|year=2010|num-b=6|num-a=8|n=II}} |
Revision as of 11:29, 4 April 2010
Problem 7
Let , where a, b, and c are real. There exists a complex number such that the three roots of are , , and , where . Find .
Solution
set , so , , .
Since , the imaginary part of a,b,c must be 0.
Start with a, since it's the easiest one to do:
and therefore: , ,
now, do the part where the imaginery part of c is 0, since it's the second easiest one to do: , the imaginery part is: , which is 0, and therefore x=4, since x=0 don't work
so now,
and therefore: , and finally, we have .
See also
2010 AIME II (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 | ||
All AIME Problems and Solutions |