Difference between revisions of "2010 AIME II Problems/Problem 7"
Sinkokuyou (talk | contribs) |
Sinkokuyou (talk | contribs) |
||
Line 8: | Line 8: | ||
x(x+6i)(2x-4-6i), the imaginery part is: 6x^2-24x, which is 0, and therefore x=4, since x=0 don't work, | x(x+6i)(2x-4-6i), the imaginery part is: 6x^2-24x, which is 0, and therefore x=4, since x=0 don't work, | ||
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 |a+b+c|=|-12+84-208|=136. |
Revision as of 11:58, 3 April 2010
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:
x(x+6i)(2x-4-6i), the imaginery part is: 6x^2-24x, which is 0, and therefore x=4, since x=0 don't work,
so now,
and therefore:
, and finally, we have |a+b+c|=|-12+84-208|=136.