1983 AIME Problems/Problem 3
Problem
What is the product of the real roots of the equation ?
Solution
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
,
![$x^2+18x+30=10 \Longrightarrow x^2+18x+20=0.$](http://latex.artofproblemsolving.com/4/f/b/4fbfb8c3979ed14768e003b2256da6a1b2b0fd3e.png)
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |