Difference between revisions of "1974 AHSME Problems/Problem 2"
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==See Also== | ==See Also== | ||
{{AHSME box|year=1974|num-b=1|num-a=3}} | {{AHSME box|year=1974|num-b=1|num-a=3}} | ||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 11:42, 5 July 2013
Problem
Let 
and 
be such that 
and 
, 
. Then 
equals
Solution
Notice that and are the distinct solutions to the quadratic 
. By Vieta, the sum of the roots of this quadratic is the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, so in this case 
.
See Also
| 1974 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
