The discussion revolves around solving the inequality involving a fraction and a negative number, specifically the inequality ##\frac{a}{x^2} < -b##. Participants explore various approaches to isolate x and analyze the implications of different values for a and b.
Participants express disagreement on the simplification of cases regarding the signs of a and b, with some asserting that multiple cases must be considered while others argue for a reduction to two cases. The discussion remains unresolved regarding the best approach to take.
Participants highlight the importance of considering the signs of a and b, and the implications of multiplying by negative values, which may affect the direction of the inequality. There are also unresolved mathematical steps in the proposed transformations.
andrewkirk said:No. If it asks you to solve for x, it means you need to get x by itself on one side of the (in)equation. How would you go about doing that?
The four cases can be reduced to two: ab > 0 and ab < 0.davidmoore63@y said:You have to be more careful than that. Split in into 4 cases according to whether a and b are positive or negative. And a fifth case if b=0.
No, they can't. For example if a and b are both equal to -1, the inequality is [itex]-\frac{1}{x^2}< 1[/itex]. Multiplying both sides by the positive [itex]x^2[/itex], [itex]-1< x^2[/itex] which is true for all x. But if a and b are both equal to 1, we have [itex]\frac{1}{x^2}< -1[/itex] so that [itex]1< -x^2[/itex] and multiplying both sides by negative 1, [itex]-1> x^2[/itex] which is never true.Mark44 said:The four cases can be reduced to two: ab > 0 and ab < 0.