1983 AIME Problems/Problem 3
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Problem
What is the product of the real roots of the equation 
?
Solution
If we expand by squaring, we get a quartic polynomial, which obviously isn't very helpful.
Instead, we substitute 
for 
and our equation becomes 
.
Now we can square; solving for , we get 
or 
. The second solution is extraneous since 
is positive. So, we have as the only solution for . Substituting back in for ,
.
See also
| 1983 AIME (Problems • Resources) | ||
| Preceded by Problem 2 | Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
