2015 AIME II Problems/Problem 14
Let and be real numbers satisfying and . Evaluate .
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.
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
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 .
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 .
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