Difference between revisions of "1985 AHSME Problems/Problem 8"
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Revision as of 12:00, 5 July 2013
Problem
Let 
and 
be real numbers with 
and 
nonzero. The solution to 
is less than the solution to 
if and only if
Solution
The solution to is 
and the solution to is 
. The first solution is less than the second if 
. We can get rid of the negative signs if we reverse the inequality sign, so we have 
. All our steps are reversible, so our answer is 
.
See Also
| 1985 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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. 
