2010 AIME II Problems/Problem 7: Difference between revisions
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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 10:29, 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 | ||
