2015 AIME II Problems/Problem 14
Contents
Problem
Let and
be real numbers satisfying
and
. Evaluate
.
Solution
The expression we want to find is .
Factor the given equations as and
, respectively. Dividing the latter by the former equation yields
. Adding 3 to both sides and simplifying yields
. Solving for
and substituting this expression into the first equation yields
. Solving for
, we find that
, so
. Substituting this into the second equation and solving for
yields
. So, the expression to evaluate is equal to
.
Note that since the value we want to find is , we can convert
into an expression in terms of
, since from the second equation which is
, we see that
and thus the value is
Since we've already found
we substitute and find the answer to be 89.
Solution 2
Factor the given equations as and
, respectively. By the first equation,
. Plugging this in to the second equation and simplifying yields
. Now substitute
. Solving the quadratic in
, we get
or
As both of the original equations were symmetric in
and
, WLOG, let
, so
. Now plugging this in to either one of the equations, we get the solutions
,
. Now plugging into what we want, we get
Solution 3
Add three times the first equation to the second equation and factor to get . Taking the cube root yields
. Noting that the first equation is
, we find that
. Plugging this into the second equation and dividing yields
. Thus the sum required, as noted in Solution 1, is
.
Solution 4
As with the other solutions, factor. But this time, let and
. Then
. Notice that
. Now, if we divide the second equation by the first one, we get
; then
. Therefore,
. Substituting
into
in equation 2 gives us
; we are looking for
. Finding
, we get
. Substituting into the first equation, we get
. Our final answer is
.
Solution 5
Factor the given equations as:
We note that these expressions (as well as the desired expression) can be written exclusively in terms of
and
. We make the substitution
and
(for sum and product, respectively).
We see that shows up in both equations, so we can eliminate it and find
, after which we can get
from the first equation. If you rewrite the desired expression using
and
, it becomes clear that you don't need to actually find the values of
and
, but I will do so for the sake of completion.
The desired expression can be written as:
Plugging in and
, we get:
- gting
Solution 6
Factor the first and second equations as and
, respectively. Dividing them (allowed, since neither are
), we have
or
Plugging into the quadratic formula and solving for
in terms of
we have
WLOG, let
Plugging into our first equation, we have
Plugging this result (and the one for
in terms of
) into our desired expression, we have
~ASAB
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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