# 2012 AMC 12B Problems/Problem 13

## Problem

Two parabolas have equations and , where and 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?

## Solution 1

Set the two equations equal to each other: . Now remove the x squared and get x's on one side: . Now factor : . If a cannot equal , then there is always a solution, but if , a in chance, leaving a out , always having at least one point in common. And if , then the only way for that to work, is if , a in chance, however, this can occur ways, so a in chance of this happening. So adding one sixth to , we get the simplified fraction of ; answer $D\box$ (Error compiling LaTeX. ! Missing number, treated as zero.).

## Solution 2

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.