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