Problem:
What is the product of the real roots of the equation
Solution:
Substitute to simplify. Several choices work well; the following substitution, which eliminates the radical immediately, is perhaps best. Define to be the nonnegative number such that . (There is such a , for if were negative, the right-hand side of the original equation would be undefined.) So
Since , we have . That is, is a solution of the original equation iff (if and only if) , i.e., iff . Both solutions to this last equation are real (why?) and their product is the constant term, . (Note: by being careful about the sign of and by using iff-arguments we have avoided introducing any extraneous roots for .)
The problems on this page are the property of the MAA's American Mathematics Competitions