2007 AIME I Problems/Problem 8
Let this root be .
We then know that is a root of , so .
We then know that is a root of so we get: or , so is the highest.
We can trivially check into the original equations to find that produces a root in common, so the answer is .
Again, let the common root be ; let the other two roots be and . We can write that and that .
Therefore, we can write four equations (and we have four variables), , , , and .
The first two equations show that . The last two equations show that . Solving these show that and that . Substituting back into the equations, we eventually find that .
Since and are both factors of , which is cubic, we know the other factors associated with each of and must be linear. Let , where and . Then we have that . Equating coefficients, we get the following system of equations:
Using equations and to make substitutions into equation , we see that the 's drop out and we're left with . Substituting this expression for into equation and solving, we see that must be .
Notice that if the roots of and are all distinct, then would have four distinct roots, which is a contradiction since it's cubic. Thus, and must share a root. Let this common value be
Thus, we see that we have Adding the two equations gives us Now, we have two cases to consider. If then we have that On the other hand, if we see that This can easily be checked to see that it does indeed work, and we're done!
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