The original post by agripa had a pdf file of questions from Lithuania, but it has been deleted.
The problem printed there was: . Note that agripa has made a sign error in recopying the problem. The question appearing on the test paper is a concept question.
is only defined in the interval .
Looking at the graphs of both sides, one can easily see that the only solution is .
Do NOT cancel that x! (If x is negative then we would be dividing by a negative number. Not kosher here as that changes the < to a >.)
The factoring here is a bit rough but can be done using the quadratic formula:
The critical points here are , so we have four intervals on the real line to test in the original (that is to say, all the way back to the beginning!) inequality:
(Nope!)
(Check!)
(Nope!)
(Check!)
So the solution set for x is
(Note that there is a place where the domain of x is restricted. As this falls outside of the derived solution set I'm not going to worry over it.
(The graph is for the equivalent problem: . We are looking for where the expression is negative.)
-Dan