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 , so we are looking for . Now, if we divide the second equation by the first one, we get ; then . Therefore, . Substituting for in equation 1, simplifying, and then taking the cube root gives us Finding by cubing on both sides and simplifying using our previous substitution, we get . Substituting this into the first equation and then dividing by , 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
Solution 7
Take and . Remark that The given equations imply that Substituting the first equation into the second, we have that , thus . Now and Thus
Solution 8
Let , then we can see , now, we see .
The rest is easy,
~bluesoul
Solution 9
Let's first put the left hand sides of the equations in factored forms. Doing this we obtain and . Now, we will subtract and add the equations to gather information on and . When we subtract the equations and clean it up via factoring, we yield , and when we add them, we yield . Now with some intuition, you should divide the equations to obtain . Now, we clean this up to obtain the following factoring of . This implies that . We plug that into the target expression to reduce it down to one variable, and get that target expression is . This means that if we can find a way to get , then the rest is trivial. We get by plugging in into . However, this time we only factor as because we particularly want a cubic degree on . Plugging in we get . Now lets plug this into our target expression to get .
~triggod, by MinecraftPlayer404
Solution 10 (Newton sums)
Let be roots of some polynomial . Then we have that . Now from the first equation we have that . For convenience, denote .
Define , where are roots of the above polynomial. Then we have that by Newton's formulas , where and . We desire . Now building up this recurrence;
Then we have that Plugging this value of into the second equation , yields . Now . To compute is now trivial, as returning this new value of into the second equation yields . Hence .
-th1nq3r
Video Solution by mop 2024
~r00tsOfUnity
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.