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2010 AIME II Problems/Problem 7: Difference between revisions

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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>.
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 imaginery part of c is 0, since it's the second easiest one to do:  
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 imaginery part is: <math>6x^2-24x</math>, which is 0, and therefore x=4, since x=0 don't work
<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>
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 $P(z)=x^3+ax^2+bx+c$, 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: $a=-12, b=84, c=-208$, and finally, we have $|a+b+c|=|-12+84-208|=\boxed{136}$.

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
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