1983 AIME Problems/Problem 3
Problem
What is the product of the real roots of the equation ?
Solution
Solution 1
If we were to expand by squaring, we would get a quartic polynomial, which isn't always the easiest thing to deal with.
Instead, we substitute for
, so that the equation becomes
.
Now we can square; solving for , we get
or
. The second root is extraneous since
is always non-negative (and moreover, plugging in
, we get
, which is obviously false). Hence we have
as the only solution for
. Substituting
back in for
,
![$x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.$](http://latex.artofproblemsolving.com/4/f/b/4fbfb8c3979ed14768e003b2256da6a1b2b0fd3e.png)
Both of the roots of this equation are real, since its discriminant is , which is positive. Thus by Vieta's formulas, the product of the real roots is simply
.
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
Reasoning as in Solution 1, the product of the roots is .
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |