Difference between revisions of "2018 AIME II Problems/Problem 6"
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<math>\dfrac{36}{38} = \dfrac{18}{19}</math> | <math>\dfrac{36}{38} = \dfrac{18}{19}</math> | ||
| − | <math>18 + 19 = \boxed{ | + | <math>18 + 19 = \boxed{037}</math> |
==See Also:== | ==See Also:== | ||
Revision as of 19:28, 1 February 2020
Problem
A real number 
is chosen randomly and uniformly from the interval ![$[-20, 18]$](http://latex.artofproblemsolving.com/5/e/d/5ed3fd299f7f005431bef3d44f28cf5f7494b169.png)
. 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.
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. 
