Difference between revisions of "2010 AIME II Problems/Problem 7"
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| + | == Problem 7 == | ||
| + | Let <math>P(z)=x^3+ax^2+bx+c</math>, where a, b, and c are real. There exists a complex number <math>w</math> such that the three roots of <math>P(z)</math> are <math>w+3i</math>, <math>w+9i</math>, and <math>2w-4</math>, where <math>i^2=-1</math>. Find <math>|a+b+c|</math>. | ||
| + | == Solution == | ||
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>. | 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. | Since <math>a,b,c\in{R}</math>, the imaginary part of a,b,c must be 0. | ||
Revision as of 11:04, 3 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 
.
