2012 AMC 12B Problems/Problem 13
Two parabolas have equations y= x^2 + ax +b and y= x^2 + cx +d, where a, b, c, and d are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have a least one point in common?
Set the two equations equal to each other: + cx + d. Now remove the x squared and get x's on one side: ax-cx=d-b. Now factor x: x(a-c)=d-b. If a cannot equal c, then there is always a solution, but if a=c, a 1 in 6 chance, leaving a 1080 out 1296, always having at least one point in common. And if a=c, then the only way for that to work, is if d=b, a 1 in 36 chance, however, this can occur 6 ways, so a 1 in 6 chance of this happening. So adding one sixth to 1080/1296, we get the simplified fraction of 31/36; answer D.
Proceed as above to obtain . The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation has no solution if and only if and . The probability that is while the probability that is . Thus we have for the probability that the parabolas intersect.