1992 AMC 12 Problem 28

Attempt Problem 28 of the 1992 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1992 AMC 12 solutions, or check the answer key.

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28.

Let i=−1.i=\sqrt{-1}. The product of the real parts of the roots of z2−z=5−5iz^2-z=5-5i is
−25-25
−6-6
−5-5
14\frac14
2525
Answer: B
Concepts:complex quadraticsquare root of a complex numberVieta
Small hint:
Apply the quadratic formula and write 21−20i=a+bi\sqrt{21-20i}=a+bi
Big hint:
Solve a2−b2=21a^2-b^2=21 and 2ab=−202ab=-20 to determine the real parts of the two roots
Solution:
The quadratic formula gives z=1±21−20i2. z=\frac{1\pm\sqrt{21-20i}}2. Since (5−2i)2=21−20i,(5-2i)^2=21-20i, the roots are 1+(5−2i)2=3−i,1−(5−2i)2=−2+i. \begin{aligned} \frac{1+(5-2i)}2&=3-i,\\ \frac{1-(5-2i)}2&=-2+i. \end{aligned} Their real parts have product 3(−2)=−6.3(-2)=-6.
Thus the correct answer is B.
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