Difference between revisions of "2002 AMC 12P Problems/Problem 23"
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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. | + | 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 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 . Since , we must have . Therefore, .
Since , we ignore the negative root. Therefore, .
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 |
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