2010 AIME II Problems/Problem 7

Problem 7

Let $P(z)=z^3+az^2+bz+c$ , where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$ , $w+9i$ , and $2w-4$ , where $i^2=-1$ . Find $|a+b+c|$ .

Solution

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 imaginary part of c is 0 since it's the second easiest one to do: $x(x+6i)(2x-4-6i)$ . The imaginary part is $6x^2-24x$ , which is 0, and therefore $x=4$ , since $x=0$ doesn't work.

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

and therefore: $a=-12, b=84, c=-208$ . Finally, we have $|a+b+c|=|-12+84-208|=\boxed{136}$ .

Solution 1b

Same as solution 1 except that when you get to $x_1 = x$ , $x_2 = x+6i$ , $x_3 = 2x-4-6i$ , you don't need to find the imaginary part of $c$ . We know that $x_1$ is a real number, which means that $x_2$ and $x_3$ are complex conjugates. Therefore, $x=2x-4$ .

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2010 AIME II Problems/Problem 7

Contents

Problem 7

Solution

Solution 1b

See also