2002 AMC 12P Problems/Problem 23: Difference between revisions
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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. However, it is not of the desired form in the problem, so we must factor the given expression to obtain the other 2 roots. | 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>. | ||
== 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 16:55, 1 July 2024
Problem
The equation 
has a zero of the form 
, where 
and 
are positive real numbers. Find 
Solution
Note that 
. With this observation, it becomes easy to note that 
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 
.
Expanding , we have 
. We may factor it as 
.
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 | |
These problems are copyrighted © by the Mathematical Association of America. 
