2013 Mock AIME I Problems/Problem 11
Contents
[hide]Problem
Let and
be the roots of the equation
, and let
and
be the two possible values of
Find
.
Solution 1
For some integer , let
be
.
Note that , so, to answer the problem, it suffices to know
and
.
Let . First, note the arbitrary decision in the expression
. Why must
be over
, but not
over
? From this observation, we can deduce that the aforementioned two possible values of this sum are
(WLOG let this be
) and
(WLOG let this be
). From these definitions and the knowledge from Vieta's Formulas that
, we can now combine fractions to get the following:
, so we look for a way to get a value for
. Fortunately, we can use Vieta again to see that
. Thus,
, and so
, or, by substitution and recalling that
,
. To get
, we think Newton Sums, which give us the following:
becomes
, so
.
Now that we have , we desire to find
. Note that Vieta gives us
, so, because
, by substitution
. By substituting our values for
and
into
, we see the following:
. To do this, we try to think of a function
whose roots are
and
.
will work. Using Newton Sums again and recalling that
, we see that:
and
, we have that
.
Thus, we can now find our desired answer:
Solution 2(Sketch)
This solution can also be solved by using Vieta's and getting ,
,
.
Then you can directly expand . And solve for the seperate parts.
A key step in the solution from there is realizing that , which can be used to simplify by a great deal.
~Bigbrain_2009