2010 AIME II Problems/Problem 7: Difference between revisions
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<math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math> | <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math> | ||
now, do the part where the imaginery part of c is 0, since it's the second easiest one to do: | 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, | <math>x(x+6i)(2x-4-6i)</math>, the imaginery part is: <math>6x^2-24x</math>, 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: | and therefore: | ||
<math>a=-12, b=84, c=-208</math>, and finally, we have |a+b+c|=|-12+84-208|=136. | <math>a=-12, b=84, c=-208</math>, and finally, we have <math>|a+b+c|=|-12+84-208|=136</math>. | ||
Revision as of 10:59, 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:
, 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 
.
