1984 AHSME Problems/Problem 24
Problem
If and are positive real numbers and each of the equations and has real roots, then the smallest possible value of is
Solution
Since both of the equations have real roots, both of their discriminants are nonnegative. Therefore, we have
from the first equation, and
from the second.
We can square the second equation to get , and combining this with the first one gives , so . We can divide both sides by , since it is positive, and take the cubed root of that to get . Therefore, we have , and since is positive, we can take the square root of this to get . Therefore, , and the smallest possible value is .
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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