# Mock AIME II 2012 Problems/Problem 12

## Problem

Let . Assume the value of has three real solutions . If , where and are relatively prime positive integers, find .

## Solution

Let . Then and . From this, we have the system

Substituting the first equation into the second, we obtain

Plugging this into the third equation yields .

Thus, . Note that our three real roots multiply to . However, since , we need to multiply by , so our is

We need . Using vieta’s and making sure we count for each factor of we divided off, we have .

Our answer is , thus .

## Solution 2

Let and , where . Then, it is obvious that .

We first focus on the first equality: . This may be simplified using our logarithmic properties:

Now, let's focus on the last expression: note that,

We can equate all of these expressions:

Multiplying all expressions by gives us

Now, from our first equality we obtain

Since , we may safely divide by :

From the first and last expressions we have:

Equating our expressions for gives

Since , we may safely divide by :

By Vieta's formulas, we must have and . Dividing the former by the latter gives

and hence .

~FIREDRAGONMATH16