2010 AIME II Problems/Problem 7
Contents
[hide]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 must be .
Start with a, since it's the easiest one to do: ,
and therefore: , , .
Now, do the part where the imaginary part of c is 0 since it's the second easiest one to do: . The imaginary part is , which is 0, and therefore , since doesn't work.
So now, ,
and therefore: . Finally, we have .
Solution 1b
Same as solution 1 except that when you get to , , , you don't need to find the imaginary part of . We know that is a real number, which means that and are complex conjugates. Therefore, .
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 | ||
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