Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution)
(→‎Solution)
 
(5 intermediate revisions by the same user not shown)
Line 17: Line 17:
 
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. However, it is not of the desired form in the problem, so we must factor the given expression to obtain the other 2 roots. From this point onwards, we assume that <math>z \neq -14i</math>.
 
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. However, it is not of the desired form in the problem, so we must factor the given expression to obtain the other 2 roots. From this point onwards, we assume that <math>z \neq -14i</math>.
  
−
Expanding <math>z(z+i)(z+3i)=2002i</math>, we have <math>z^3 + 4iz^2 - 3z - 2002i = 0</math>. We may factor it as <math>(z + 14i)(z^2 - 10iz - 143) = 0</math>.
+
Expanding <math>z(z+i)(z+3i)=2002i</math>, we have <math>z^3 + 4iz^2 - 3z - 2002i = 0</math>. We may factor it as <math>(z + 14i)(z^2 - 10iz - 143) = 0</math>. Since
 +
<math>z + 14i \neq 0</math>, we must have <math>z^2 - 10iz - 143 = 0</math>. Therefore, <math>z = \frac {10i \pm \sqrt{-100 + 4(143)}}{2} = 5i \pm \sqrt {118}</math>.
 +
 
 +
Since <math>a > 0</math>, we ignore the negative root. Therefore, <math>a = \boxed {\text{(A) } \sqrt {118}}</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}}

Latest revision as of 18:02, 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. However, it is not of the desired form in the problem, so we must factor the given expression to obtain the other 2 roots. From this point onwards, we assume that $z \neq -14i$.

Expanding $z(z+i)(z+3i)=2002i$, we have $z^3 + 4iz^2 - 3z - 2002i = 0$. We may factor it as $(z + 14i)(z^2 - 10iz - 143) = 0$. Since $z + 14i \neq 0$, we must have $z^2 - 10iz - 143 = 0$. Therefore, $z = \frac {10i \pm \sqrt{-100 + 4(143)}}{2} = 5i \pm \sqrt {118}$.

Since $a > 0$, we ignore the negative root. Therefore, $a = \boxed {\text{(A) } \sqrt {118}}$.

See also

2002 AMC 12P (Problems • Answer Key • Resources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12P_Problems/Problem_23&oldid=221455"