2010 AIME II Problems/Problem 7

set $w=x+yi$ , so $x_1 = x+(y+3)i$ , $x_2 = x+(y+9)i$ , $x_3 = 2x-4+2yi$ . Since $a,b,c\in{R}$ , the imaginary part of a,b,c must be 0. Start with a, since it's the easiest one to do: $y+3+y+9+2y=0, y=-3$ and therefore: $x_1 = x$ , $x_2 = x+6i$ , $x_3 = 2x-4-6i$ 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, $x_1 = 4, x_2 = 4+6i, x_3 = 4-6i$ and therefore: $a=-12, b=84, c=-208$ , and finally, we have $|a+b+c|=|-12+84-208|=136$ .

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2010 AIME II Problems/Problem 7