Difference between revisions of "2024 AMC 12B Problems/Problem 17"
(Created page with "==Solution 1== -10<= a, b <= 10 , a,b has 21 choices per Vieta, x1x2x3 = -6, x1 + x2+ x3 = -a , x1x2+ x2x3 + x3x1 = b Case: (1) (x1,x2,x3) = (-1,-1,6) , b = 13 not valid...") |
(→Solution 1) |
||
| Line 1: | Line 1: | ||
| + | ==Problem 17== | ||
| + | Integers <math>a</math> and <math>b</math> are randomly chosen without replacement from the set of integers with absolute value not exceeding <math>10</math>. What is the probability that the polynomial <math>x^3 + ax^2 + bx + 6</math> has <math>3</math> distinct integer roots? | ||
| + | |||
| + | <math>\textbf{(A)} \frac{1}{240} \qquad \textbf{(B)} \frac{1}{221} \qquad \textbf{(C)} \frac{1}{105} \qquad \textbf{(D)} \frac{1}{84} \qquad \textbf{(E)} \frac{1}{63}</math>. | ||
| + | |||
| + | [[2024 AMC 12B Problems/Problem 17|Solution]] | ||
| + | |||
==Solution 1== | ==Solution 1== | ||
Revision as of 00:45, 14 November 2024
Problem 17
Integers 
and 
are randomly chosen without replacement from the set of integers with absolute value not exceeding 
. What is the probability that the polynomial 
has 
distinct integer roots?
.
Solution 1
-10<= a, b <= 10 , a,b has 21 choices per Vieta, x1x2x3 = -6, x1 + x2+ x3 = -a , x1x2+ x2x3 + x3x1 = b
Case: (1) (x1,x2,x3) = (-1,-1,6) , b = 13 not valid
(2) (x1,x2,x3) = (-1,1,6) , b = -1, a=-6 valid
(3) (x1,x2,x3) = ( 1,2,-3) , b = -7, a=0 valid
(4) (x1,x2,x3) = (1,-2,3) , b = -7, a=2 valid
(5) (x1,x2,x3) = (-1,2,3) , b = 1, a=4 valid
(6) (x1,x2,x3) = (-1,-2,-3) , b = 11 invalid
probability = 
= 
