1966 AHSME Problems/Problem 5
Problem
The number of values of satisfying the equation is:
Solution
Since is in the denominator, . Simplifying,
Thus , which isn't in the domain of the equation. Thus there are no values of .
See also
1966 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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