1962 AHSME Problems/Problem 21

Problem

It is given that one root of $2x^2 + rx + s = 0$ , with $r$ and $s$ real numbers, is $3+2i (i = \sqrt{-1})$ . The value of $s$ is:

$\textbf{(A)}\ \text{undetermined}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ -13\qquad\textbf{(E)}\ 26$

Solution

If a quadratic with real coefficients has two non-real roots, the two roots must be complex conjugates of one another. This means the other root of the given quadratic is $\overline{3+2i}=3-2i$ . Now Vieta's formulas say that $s/2$ is equal to the product of the two roots, so $s = 2(3+2i)(3-2i) = \boxed{26 \textbf{ (E)}}$ .

Page

Toolbox

Search

1962 AHSME Problems/Problem 21

Problem

Solution