1962 AHSME Problems/Problem 21: Difference between revisions
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==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 21:19, 3 October 2014
Problem
It is given that one root of 
, with 
and 
real numbers, is 
. The value of is:
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 
.
Now Vieta's formulas say that 
is equal to the product of the two roots, so
.
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
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