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> |
+ | |||
+ | ~First | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=q2oc7n-n6aA | ||
+ | ~Shreyas S | ||
==See Also:== | ==See Also:== | ||
{{AIME box|year=2018|n=II|num-b=5|num-a=7}} | {{AIME box|year=2018|n=II|num-b=5|num-a=7}} | ||
+ | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:20, 27 March 2022
Contents
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.