Difference between revisions of "2002 AMC 12P Problems/Problem 23"
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== Problem == | == Problem == | ||
− | + | The equation <math>z(z+i)(z+3i)=2002i</math> has a zero of the form <math>a+bi</math>, where <math>a</math> and <math>b</math> are positive real numbers. Find <math>a.</math> | |
− | <math> \ | + | <math> |
+ | \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} | ||
+ | </math> | ||
== 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. 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>. 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= | + | {{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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