Difference between revisions of "2002 AMC 12P Problems/Problem 23"

m (Solution)
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== Solution ==
 
== Solution ==
Note that <math>2002 = 11 \cdot 13 \cdot 14</math>. With this observation, it becomes easy to note that <math>z = -14i</math> is a root of the given equation.
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Note that <math>2002 = 11 \cdot 13 \cdot 14</math>. With this observation, it becomes easy to note that <math>z = -14i</math> is a root of the given equation. Therefore, we may factor the given expression.
  
Expanding
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Expanding <math>z(z+i)(z+3i)=2002i</math>, we have <math>z^3 + 4iz^2 - 3z - 2002i = 0</math>.
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2002|ab=P|num-b=22|num-a=24}}
 
{{AMC12 box|year=2002|ab=P|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:46, 1 July 2024

Problem

The equation $z(z+i)(z+3i)=2002i$ has a zero of the form $a+bi$, where $a$ and $b$ are positive real numbers. Find $a.$

$\text{(A) }\sqrt{118} \qquad \text{(B) }\sqrt{210} \qquad \text{(C) }2 \sqrt{210} \qquad \text{(D) }\sqrt{2002} \qquad \text{(E) }100 \sqrt{2}$

Solution

Note that $2002 = 11 \cdot 13 \cdot 14$. With this observation, it becomes easy to note that $z = -14i$ is a root of the given equation. Therefore, we may factor the given expression.

Expanding $z(z+i)(z+3i)=2002i$, we have $z^3 + 4iz^2 - 3z - 2002i = 0$.

See also

2002 AMC 12P (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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