Difference between revisions of "2010 AIME II Problems/Problem 7"

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<math>y+3+y+9+2y=0, y=-3</math>
 
<math>y+3+y+9+2y=0, y=-3</math>
 
and therefore:
 
and therefore:
<math>x_1 = x+i</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:
 +
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, <math>x_1 = 4, x_2 = 4+6i, x_3 = 4-6i</math>

Revision as of 11:56, 3 April 2010

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, $x_1 = 4, x_2 = 4+6i, x_3 = 4-6i$

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