2018 AIME II Problems/Problem 6
Contents
[hide]Problem
A real number is chosen randomly and uniformly from the interval . The probability that the roots of the polynomial
are all real can be written in the form , where and are relatively prime positive integers. Find .
Solution
The polynomial we are given is rather complicated, so we could use Rational Root Theorem to turn the given polynomial into a degree-2 polynomial. With Rational Root Theorem, are all possible rational roots. Upon plugging these roots into the polynomial, and make the polynomial equal 0 and thus, they are roots that we can factor out.
The polynomial becomes:
Since we know and are real numbers, we only need to focus on the quadratic.
We should set the discriminant of the quadratic greater than or equal to 0.
.
This simplifies to:
or
This means that the interval is the "bad" interval. The length of the interval where can be chosen from is 38 units long, while the bad interval is 2 units long. Therefore, the "good" interval is 36 units long.
~First
Video Solution
https://www.youtube.com/watch?v=q2oc7n-n6aA ~Shreyas S
See Also:
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.