1962 AMC 12 Problem 21

Attempt Problem 21 of the 1962 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 1962 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

It is given that one root of 2x2+rx+s=0,2x^2+rx+s=0, with rr and ss real numbers, is 3+2i3+2i (i=−1).(i=\sqrt{-1}). The value of ss is:
undetermined
55
66
−13-13
2626
Answer: E
Concepts:complex numberVieta’s Formulas
Small hint:
A polynomial with real coefficients also has the conjugate root
Big hint:
Use the product of the roots and Vieta’s formula
Solution:
The other root is 3−2i.3-2i. Their product is (3+2i)(3−2i)=9+4=13. (3+2i)(3-2i)=9+4=13. Vieta’s formula gives s2=13,\frac{s}{2}=13, so s=26.s=26.
Therefore, the correct answer is E.
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