Difference between revisions of "2010 AIME II Problems/Problem 7"
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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>. | 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 == | == 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>. | |
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. | ||
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and therefore: <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math> | and therefore: <math>x_1 = x</math>, <math>x_2 = x+6i</math>, <math>x_3 = 2x-4-6i</math> | ||
− | + | Now, do the part where the imaginary part of c is 0, since it's the second easiest one to do: | |
− | <math>x(x+6i)(2x-4-6i)</math>, the | + | <math>x(x+6i)(2x-4-6i)</math>, the imaginary part is: <math>6x^2-24x</math>, which is 0, and therefore x=4, since x=0 doesn't work. |
− | + | So now, <math>x_1 = 4, x_2 = 4+6i, x_3 = 4-6i</math> | |
and therefore: <math>a=-12, b=84, c=-208</math>, and finally, we have <math>|a+b+c|=|-12+84-208|=\boxed{136}</math>. | and therefore: <math>a=-12, b=84, c=-208</math>, and finally, we have <math>|a+b+c|=|-12+84-208|=\boxed{136}</math>. |
Revision as of 21:56, 5 March 2012
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 imaginary part of c is 0, since it's the second easiest one to do: , the imaginary part is: , which is 0, and therefore x=4, since x=0 doesn't work.
So now,
and therefore: , and finally, we have .
See also
2010 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |