# 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 .