2010 AIME II Problems/Problem 7: Difference between revisions
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set <math>w=x+yi</math>, so <math>x_1 = x+(y+3)i</math>, <math>x_2 = x+(y+9)i</math>, <math>x_3 = 2x+ | set <math>w=x+yi</math>, so <math>x_1 = x+(y+3)i</math>, <math>x_2 = x+(y+9)i</math>, <math>x_3 = 2x-4+2yi</math>. | ||
Since <math>a,b,c\in{R}</math>, the imaginary part of a,b,c must be 0. | |||
Start with a, since it's the easiest one to do: | |||
<math>y+3+y+9+2y=0, y=-3</math> | |||
and therefore: | |||
<math>x_1 = x+i</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math> | |||
Revision as of 10:51, 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:
,
,