Difference between revisions of "1962 AHSME Problems/Problem 21"

(Solution)
m (Solution)
 
(One intermediate revision by one other user not shown)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
{{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 <math>\overline{3+2i}=3-2i</math>.
 +
Now Vieta's formulas say that <math>s/2</math> is equal to the product of the two roots, so
 +
<math>s = 2(3+2i)(3-2i) = \boxed{26 \textbf{ (E)}}</math>.
 +
 
 +
==See Also==
 +
{{AHSME 40p box|year=1962|before=Problem 20|num-a=22}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:19, 3 October 2014

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 $s = 2(3+2i)(3-2i) = \boxed{26 \textbf{ (E)}}$.

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png