Quandaries
and Queries 

Name: Jessica
Question: 

Hi Jessica, I want to consider [x^{2}9]/[x5] > 0. ([x^{2}9]/[x5] = 0 is easy. This can only way this can happen is if x^{2} = 9, that is x = ± 3.) For [x^{2}9]/[x5] > 0 to be true, either the numerator and denominator are both positive or they are both negative. This gives us two cases to examine. Case1: x^{2}  9 > 0 AND x  5 > 0 x^{2}  9 > 0 means x^{2} > 9, that is x > 3 or x < 3. At this point I would draw a diagram to help me "see" what is going on. The numbers x that satisfy x^{2}  9 > 0 are the numbers coloured green on the number line below x  5 > 0 means x > 5 and thus the numbers x that satisfy x  5 > 0 are the numbers coloured green in the number line below Thus the numbers that satisify x^{2}  9 > 0 AND x  5 > 0 are the numbers that are coloured green in both diagrams. These are the numbers x that satisfy x > 5. Case2: x^{2}  9 < 0 AND x  5 < 0 The numbers that satisfy x^{2}  9 < 0 are and the numbers that satisify x  5 < 0 are Thus the numbers that satisfy both inequalities are coloured green on both diagrams, and are hence Finally the numbers that satisify the inequality [x^{2}9]/[x5] > 0 fall into either case 1 OR case 2 and hence are the numbers that got coloured green by either the case 1 argument or the case 2 argument. Thus they are the numbers I hope this helps, 

