1983 AIME Problems/Problem 3
Contents
[hide]Problem
What is the product of the real roots of the equation ?
Solutions
Solution 1
If we expand by squaring, we get a quartic polynomial, which isn't always the easiest thing to deal with.
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 (plugging in as , we get , which is obviously not true)).So, we have as the only solution for . Substituting back in for ,
By Vieta's formulas, the product of the roots is .
Solution 2
We begin by noticing that the polynomial on the left is less than the polynomial under the radical sign. Thus: Letting , we have . Because the square root of a real number can't be negative, the only possible is .
Substituting that in, we have
And by Vieta's formulas, the product of the roots is .
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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