1985 AHSME Problems/Problem 8
Problem
Let be real numbers with and nonzero. The solution to is less than the solution to if and only if
Solution
The solution to is , while that to is . The first solution is less than the second precisely if and multiplying this inequality by reversees the inequality sign, yielding .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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